Statistcal Methods in Engineering and QA.pdf

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STATISTMETHODSin ENGINEERING -..u and A CE p~ W. • M. J()i.-.v 'M Statistical Methods in Engineering and Quality Assurance • Statistical Methods In Engineering and Quality Assurance PETER W. M. JOHN A Wiley-Intersciencc Publication JOHN WILEY & SONS, INC. New York • Chichester • Brisbane • Toronto • Singapore In recognition of the importance of preserving what has heen written, it is a policy of John Wiley & Sons, Inc. to have hooks of enduring value published in the United States pnnted on acid-free paper, and we exert our hest efforts to that end. Copyright © 1990 by John Wiley & Sons, Tnc. All rights reserved. Published simultaneously in Canada. Reproduction or translation of any part of this work beyond that permitted by Section 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Requests for permission or further information should be addressed to the Permissions Department, John Wiley & Sons, Inc. Library of Congress CalaJoging-in·Publication Dala John, Peter William Meredith. Statistical methods in enginec:ring and quality assurance I Peter W. M. John. p. cm ... (Wiley series in probability and mathematical statistics. Applied probability and statistics section) ). Includes bibliographical references (p. ISBN 0-471-82986-2 1. Engillc:ering- -Statistical methods. 2. Quality control-Statistical methods. 3. Quality assurance. T. Title. II. Series. T1\340.16" 1990 620'.0072- -dc20 90-3371 X CIP i09!l76543 Contents Preface I. Quality Assurance and Statistics xv 1 Ouality and American Business, 1 1.2 Competition, 2 1.3 The Reaction, 2 1.4 Total Quality, 3 Kaizen, 4 1.5 Why Statistics?, 4 1.6 1.7 The Future, 5 1.8 Gathering Information, 6 1.9 This Book, 7 1.10 Instructional Considerations, 7 1.11 Reprise, g 1.1 2. Descriptive Statistics 9 2.1 Introduction, 9 2.2 The Mean and Standard Deviation, 9 2.3 The Median and Quartiles, 11 Ishikawa's Seven Tools, 13 2.4 Histograms, 14 2.5 2.6 An Example of a Nonsymmetrical Distribution, Hi 2.7 Dotplots, 19 2.8 Stem-and-Leaf Diagrams, 20 2.9 Splitting Stems, 21 2.10 Box-and-Whisker Plots, 22 2.11 Pareto Diagrams, 24 v vi CONTENTS 2.12 3. Summary, 25 Exercises, 26 Discrete Variables, or Attributes 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19 3.20 3.21 Introduction, 29 Random Digits, 29 A Mathematical Model, 30 Some Basic Probability Theory, 31 The Coin Tossing Model, 33 The Binomial Distribution, 33 Unequal Probabilities, 35 The Binomial Distribution (General Case), 36 Mathematical Expectation, 37 The Variance, 38 Sampling without Replacement, 39 The Hypergeometric Distribution, 40 Conditional Probabilities, 41 Bayes' Theorem, 42 Bivariate Distributions, 44 The Geometric Distribution, 45 The Poisson Distribution. 46 Properties of the Poisson Distribution, 47 The Bortkiewicz Data, 48 Other Applications. 48 Further Properties, 48 Exercises, 49 4. Continuous Variables 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 29 52 Introduction, 52 Probability Density Functions, 52 The Cumulative Distribution Function, 54 The Uniform Distribution, 55 The Exponential Distribution, 56 The Quartiles of a Continuous Distribution, 58 The Memorylcss Property of the Exponential Distribution, 58 The Reliability Function and the Hazard Rate, 59 The Weibull Distribution, 60 The Gamma Distribution, 61 vii CONTENTS 4.11 4.12 4.13 4.14 4.15 4.16 The Expectation of a Continuous Random Variable, 62 The Variance, 64 Moment-Generating Functions, 64 Bivariate Distributions, 66 Covariance, 67 Independent Variables, 68 Exercises, 69 5. The Normal Distribution 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 Introduction, 72 The Normal Density Function, 72 The Tables of the Normal Distribution, 74 The Moments of the Normal Distribution, 75 The Moment-Generating Function, 76 Skewness and Kurtosis, 76 Sums of Independent Random Variables, 77 Random Samples, 79 Linear Combinations of Normal Variables, 80 The Central Limit Theorem, 80 The Chi-Square Distribution, 81 The Lognormal Distribution, 82 The Cumulative Normal Distribution Curve and Normal Scores, 83 Exercises, 86 6. Some Standard Statistical Procedures 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.12 72 Introduction, 88 The Three Experiments, 88 The First Experiment, 89 The Second Experiment, 89 The Third Experiment, 90 The Behavior of the Sample Average, 90 Confidence Intervals for the Normal Distribution, 90 Confidence Intervals for the Binomial Distribution, 94 Confidence Intervals for the Exponential Distribution, 96 Estimating the Variance, 97 Student's t Statistic, 100 Confidence Intervals for the Variance of a Normal Population, 101 88 5 8.2 7. 12 7.3 8. 8.3 7.16 6.13 6.6 7.4 7. 115 Null and Alternate Hypotheses. 134 The Duckworth Test.4 8.9 7.14 7. 120 The t-Test. 117 The Choice of Alpha.15 6. 129 Unknown Variances. 118 TIle z-Test. 131 Unequal Variances. 125 Multinomial Goodness of Fit. 127 Comparative Experiment. 110 Introduction. 114 The Decisions.13 7. 139 Comparing Binomial Populations. 113 An Example. 138 Comparing Exponential Distributions.8 7.9 8.1 7. 142 129 . 140 Chi-Square and 2 x 2 Tables. 141 Exercises.6 8. 136 Comparing Variances. 123 The Exponential Distribution.10 8.15 8. 108 Exercises.14 6..11 7.2 8.7 8. 119 The Type I I Error.1 8. 116 Acceptance and Rejection Regions. 104 Unbiased Estimators. 124 Chi-Square and Goodness of Fit.viii CONTENTS 6.Test.10 7. 124 Tests for the Percent Defective. 133 Wilcoxon's Two-Sample Test. 132 The Paired t.S 7. 105 Moment Estimators. 129 Comparing Two Normal Means with Known Variance. Chi-Square and Exponential Variables. 106 Maximum-Likelihood Estimates.12 113 Introduction. Hypothesis Testing 7.7 7. 116 One-Sided Tests.8 8. 102 Another Method of Estimating the Standard Deviation. 137 Confidence Intervals for the Variance Ratio. 122 Tests about the Variance.17 7. 126 Exercises.11 8. 179 Some Elementary Single Sampling Plans.4 11. 164 10.10 9. 164 10. 172 10.14 9. 180 Some Practical Plans.15 9. 177 Sampling by Attributes. 167 10. 171 10.8 9. 175 11. 158 Process Capability.1 9. 177 The Role of Acceptance Sampling.11 9. 158 I Charts. 156 Setting the Control Limits for an s Chart. 154 s Charts.9 9.6 9.4 Variable Sample Sizes.7 Introduction.2 Binomial Charts.5 9.1 1] .13 9.5 11. 183 177 .3 9.7 c-Charts. 148 An Example. 154 Average Run Lengths. 145 The x-Bar Chart. 146 Setting the Control Lines. Introduction. 165 10.12 9. 147 The R Chart. 144 Quality Control Charts.7 9. 151 Detecting Shifts in the Mean.6 The Arcsine Transformation. 149 Another Example. 178 Single Sampling Plans.1 144 164 Introduction. Acceptance Sampling I 11.2 11. 174 Exercises. 161 Control Cbarts for Attributes 10. 156 Alternative Control Limits. Quality Control Charts 9.16 10. 153 Alternative Stopping Rules.5 Interpreting Outlying Samples.8 Demerits.2 9.4 9.3 11.ix CONTENTS 9. 160 Exercises. 182 What Quality is Accepted? AQL and LTPD.3 Binomial Charts for a Fixed Sample Size. 169 10. 157 Nonnormality.6 11. 11 11.10 12. 197 MIL-STD-105D.6 12.7 12.11 12. 183 The Average Outgoing Quality. Acceptance Sampling II 12. Further Topics on Control Charts 13. 201 Exercises. 188 Exercises.16 12.9 12. 204 An Example.' Combined Rules. 204 . 192 An Example. 196 Military and Civilian Sampling Systems.2 12.1 13.12 11. 187 Continuous Sampling.8 12. 199 Tightened versus Normal Inspection. 187 Sampling by Variables.13 12. AOQ and AOQL.14 12. 195 An Example of a Two-Sided Test.18 The Cost of Inspection.2 13. 206 Moving Averages. 189 Double Sampling. 186 Rectifying Inspection Plans.5 13.15 12. 186 Chain Sampling Plans.17 12. 191 The Sequential Probability Ratio Test. 191 The Normal Case. 187 Skip Lot Sampling. 206 A Table of Values of ARL.CONTENTS X 11.. 206 203 . 200 Switching Rules. 202 13. 197 The Standard Procedure.12 12.6 189 Introduction. 193 Sequential Sampling by Attributes. 189 Sequential Sampling. 201 Acceptance Sampling by Variables.10 11. 194 Two-Sided Tests. 194 An Example of Sequential Sampling by Attributes. 188 12.8 11.14 ILlS Choosing the Sample Size.5 12.13 11.3 12.9 11.4 12. 184 Other Sampling Schemes.1 12.3 13. 199 Reduced Inspection. 203 Decision Rules Based on Runs.4 13. 198 Severity of Inspection. 7 14.20 15. 248 More about the Example.5 14.Tests.7 Arithmetic-Average Charts. 231 Confidence Intervals for a Predicted Value. 212 13. 243 The Method of Adjusting Variables. 215 14. 213 13. 227 Estimating the Variance. Introduction.16 14. 237 Exercises. 228 Sums of Squares. 214 Exercises.11 EWMA and Prediction.6 15.15 14.10 14. 231 Influential Points. 238 Multiple Regression 15.9 14. 207 13. 221 The Method of Least Squares.18 14.13 14. 228 t. 225 A Simple Numerical Example.3 14. 226 The Estimate of the Slope. 229 Standardized Residuals. 223 An Algebraic Identity.12 14. 249 243 . 247 Properties of the Estimates.3 15.1 14.7 217 Introduction. 217 The Correlation Coefficient.8 14.8 Cumulative Sum Control Charts. 218 Correlation and Causality.2 14.19 14.4 15.5 15. 232 Residual Plots.9 A Horizontal V-Mask.10 Geometric Moving Averages. 232 Lines through the Origin. 209 13. 225 The Error Terms. 235 Transformations.11 14.17 14.2 15.xi CONTENTS 13.4 14. 245 More Than Two Predictors. 233 Other Models.1 15.6 14.14 14. 248 A Geological Example. Bivariate Data: Fitting Straight Lines 14. 219 Fitting a Straight Line. 220 An Example. 244 The Method of Least Squares. 265 Which Treatments Differ'?.7 16. 279 Several Factors. 264 The Analysis of Variance.6 16.14 15.13 15.3 17. 249 The Second Run.12 15.5 17.13 16.4 17.8 17.8 15. 295 284 .1 17. 276 Interaction. 294 Normal Plots. 273 Orthogonality.9 15. 252 Fitting Polynomials. 280 Exercises. 270 Quantitative Factors. 255 Testing for Goodness of Fit.2 16.12 16. 259 Exercises.xii CONTENTS 15.5 16.9 17.11 16.1 16.10 16.14 17. The Printout. 253 A Warning about Extrapolation.3 16. 264 The One-Way Layout. 276 Interaction with Several Observations in Each Cell. 284 Experimenting on the Vertices of a Cube. 293 Estimating the Variance. 285 An Example of a 22 Experiment. 267 The Analysis of Covariance. The Analysis of Variance 16.6 17.2 17. 288 Thc Regression Model.11 15.8 16. 256 Singular Matrices. 267 The Analysis of Covariance and Regression.10 264 Introduction.7 17. 280 Design of Experiments: Factorial Experiments at Two Levels 17.9 16. 287 Three Factors. 278 Three Factors. 257 Octane Blending.4 16. 290 Yates' Algorithm. 292 An Example with Four Factors. 260 Introduction.15 16. 286 The Design Matrix. 258 The Quadratic Blending Model. 271 The Two-Way Layout. 274 Randomized Complete Blocks.10 15. 296 Fractions with Eight Runs.4 19.3 19.2 18.14 18.6 19. 304 Exercises.18 17. 318 The L(1S) Lattice. 299 The L(J6) Lattice.16 17. 325 Factors at Five Levels. Taguchi and Orthogonal Arrays 19. 333 The Three Stages of Process Design.8 IS. 316 Thirteen Factors in 27 Runs.5 IS. 328 19. 333 A Strategy for Parameter Design.13 17.4 18. 323 Graeco-Latin Squares. 311 Fractions: Four Factors in Nine Points. 331 Terminology. 321 Latin Squares. 335 Inner and Outer Arrays. Fractions.14 17.13 18. 300 Screening Experiments. 335 Signal-to-Noise Ratios. 312 Three Factors in 27 Runs. 313 Four Factors in 27 Runs. 298 The L(8) Lattice.15 Factors with Three Levels.7 18. 324 Five Factors at Four Levels in 16 Runs. 333 Parameter Design Experiments.12 17.2 19.11 1S. 304 Fractions with 12 Points.5 19.1 \8.10 18.6 18.9 18. 328 Exercises.3 18. 324 Four-Level Factors and Two-Level Factors. 301 Fo1dover Designs.20 18.1 19.15 17.7 311 Introduction.xiii CONTENTS 17. 337 331 .11 17. 297 Seven Factors in Eight Runs.17 17.19 17. 305 Design of Experiments: Factorial Experiments at Several Levels 18. 319 A Nonorthogonal Fraction. 319 The L(36) Lattice.12 18. 303 Resolution V Fractions. 311 Two Factors. 323 Hyper-Graeco-Latin Squares. 296 Three Factors in Four Runs. 356 367 . 354 F Distribution.9 19. 343 References 347 Tables 351 I. 351 I-Distribution. 340 Tolerance Design. 353 Chi-Square Distribution. 341 Exercises. 341 Finale. Ill. IV. 338 The Confirmation Experiment.8 19.10 19.xiv CONTENTS 19.11 An Example of a Parameter Design Experiment. Index The The The The Normal Distribution. II. It reflects more than 30 years of teaching statistics to engineers and other scientists. Why is it so exciting to have these capabilities available for the PC? Before the 1950s. from 1957 to 1961. Almost equally satisfying is the burst of technical capabilities for the implementation of quality assurance methods by engineers at plant level-a direct result of the wide accessibility and growing power of the personal computer.Preface This book is designed to give engineers an introduction to statistics in its engincering applications. we could not perform such tasks as multiple regression. with particular emphasis on modern quality assurancc. As a practical matter. Now. People felt inhibited about using the company's computers unless they had a relatively big problem and an expert to help them. Under some accounting procedures. statistical calculations were a forbidding chore. both in formal courses at universities and in specialized courses at their plants. Given that hackground. a key element of modern quality assurance. using them could be intimidating and often complicated. Then came the mainframe. Since returning to academia in 1961. it was also costly. My own interest in quality assurance evolved from my long-standing specialization in the design of experiments. That aspect of the design of experiments has become the basis of off-Line quality control. which began with my experience as the first research statistician in the laboratories at Chevron Research Corporation. after three decades of ever-accelerating progress in both hardware and software. which meant that the engineer had to work through the computing section with its programmers and keypunch operators. Although the mainframes facilitated wonderful progress. augmented by broad consulting experience. The examples in the text cover a wide range of engineering applications. the personal computers that most engineers have on their xv . you can imagine how satisfying I find American industry's current surge of interest in quality assurance. I have devoted much of my research to fractional factorial experiments. including both chemical engineering and semiconductors. v W. in addition. and should. I intend this book to help us both to do that. Hence. In just a few seconds. Moreover. I also use Minitab in much of my research. of course. Texa. and most of them use it on our instructional VAX. which I find more convenient. Not only has the PC freed engineers from the mainframes for everyday computations. Parts of the last chapters on the design of experiments may look deceptively easy. most engineers have little or no occasion to deal with the mainframe that may be miles away from their plants. using Minitab for calculations and Statgraphics for figures.xvi PREFACE desks can do more easily and quickly all that the mainframes did in the late 1950s. When I have more complicated problems. However. This book is only an introduction to a broad field. they. M. it has lately become available in a PC version. Even though engineers now can and should perform many of the applications independently. are available in versions for ~he PC with a hard disk. PETER AIL~ti/l. Now engineers can. use statistical methods and statistical thinking on an everyday basis. both to collect their data and to analyze them. a few words with a statistician before you get started may save you hours of pain later. too. to a word processor for composition. JOHN . they can apply standard statistical procedures easily and quickly. We are at a stage in our technological development when engineering and statistics can walk together hand in hand. All of the computations that I have made in writing this book have been carried out on such a PC. today's engineers can make plots that normally would have taken their fathers all day. but better still. Then. the engineers themselves can keep their data filt(s on their own disks. As to the more complicated problems. It has the great advantage of being very easy for the students to use. I hope that it will persuade engineers to use statistical thinking as a standard operating procedure. I use Minitab on the mainframe for my teaching at the University of Texas. This is particularly advantageous with graphical procedures. 1 have been able only to touch upon some of them. I can turn to SAS or to GUM. it is even better to have convenient access to a statistician with whom to discuss problems and seek advice. as a matter of routine. Statistical Methods in Engineering and Quality Assurance . We will be beaten not by price.1. we will be unable to compete in areas that once we dominated. where the government stilI insisted on its use. industrial engineering. It is not too harsh to say that American industry became interested in quality control during World War 11 because the government insisted upon it. One hoped that they were able to spot most of the defective parts. worked on the development of acceptance sampling. the inspectors inspected the lots. they probably would not have known the expression and would have said something about being sure that we kept on doing things as we had always done them. QUALITY AND AMERICAN BUSINESS If. If you had asked them what quality control was. of whom there were only a few. academic statisticians. but by inferior product quality. Inc CHAPTERONE Quality Assurance and Statistics 1. and Enoch Farrell. Industry lost interest as soon as the war was over. mainly at Princeton and Columbia. Without dedication to total quality in our manufacturing companies. 30 years ago. During World War 11. Two of the main techniques in quality assurance-control charts and acceptance sampling-<iate back to the i930s at Bell Labs. But they no longer worked on quality control. you had asked engineers or executives what quality assurance was. as we are engaged in a life and death struggle with foreign nations that hardly existed as competitors 30 years ago. bad. the answer would have been something about Shewhart control charts and government sampling plans. Some managers 1 . F Dodge. That is what quality control and quality assurance were. and they knew the conventional wisdom. quality has beome a vital part of American industry. M. In some industries. or indifferent. which were then either scrapped or reworked. accepted some. except for those companies that were engaged in defense work. good. H. John Copyright © 1990 by John Wiley & Sons. After the war. that group spread out over the country to form the core of the growth of statistics in American universities. They were the work of such pioneers as Walter Shewhart. and rejected others. Today. The topic was relegated to a few lectures in departments of.Statistical Methods in Engineering and Quality Assurance Peter W. the production people churned out the parts. called back from the wilderness. Once the production department shoveled the product out of the plant door. So-called new models might appear annually with cosmetic changes.2. Once Americans listened to them. but worked better and lasted longer. Juran. American manufacturers then preferred to focus on cost control rather than quality control. Then people began to realize that the explanation was that the Japanese had done some things right. For a time. Juran had also been preaching quality control in this country and in Japan for many years. but often little real improvement over the previous year's product. Some did not even bother to control costs. The term total qu. 1. Deming emphasizes that quality requires the complete dedication of management. inflation psychology was rampant and extra costs could be passed through to the customer with no great difficulty-especially if that customer was Uncle Sam and the manufacturer had a cost-plus contract. It meant that service after the sale was often subject to cavalier neglect. THE REACTION The reaction came a few years ago. the division of functions was simple. 1. Warranty and service were pushed down to the level of the local retailer or the local distributor. televisions. began producing cars.2 QUALHY ASSURANCE AND STATISTICS resented the expense and the hassle of this. an American statistician who had started the Japanese on the road to quality in 1950. This meant two things. who had earned an international reputation for shoddy workmanship before 1940. The Japanese. chips. Marketing sold the product. etc. The number of Japanese cars on our highways and television sets in our homes increased by leaps and bounds. that were not only cheaper than ours.. Engineers began to talk of quality assurance rather than quality control. It also meant that there was really very little feedback from the final buyer. One of those things was to listen to William Edwards Deming. Deming. the user. became a prophct in his own country. COMPETITION Then came the Japanese shock. and would havc been happier to ship the product as is and let the buyer beware. In some companies. M.3. to the manufacturing department concerning what was wrong with the items and how they could be improved. it became marketing's problem. He summarizes this requirement in "Fourteen Points for . but was being completely ignored in his homeland. Some Americans tried to attribute the increase in imports from Japan entirely to the alleged fact that Japan's wages and standard of living were lower than ours.ality entered the jargon of manufacturing. a radical change began to take place in much of this nation's industry. and to the American engineer J. It emphasizes that quality is a matter of dedication of the whole company and all its personnel. TOTAL QUALITY The first of Deming's points sets the tone for total quality." The final point is "Put everybody in the company to work in teams to accomplish the transformation. You also have your own customers within the plant-the people at the next stage of the production cycle. One of the leading converts to total quality. much maligned idea that the objective is customer satisfaction: to supply the customer with good product. No step in the process stands alone. from the CEO down to the employee who sweeps the corridors. Houghton.4." of which the first two are "Create constancy of purpose toward improvement of product and service. the CEO of Corning Glass Company. James R. Do it right the first time. Total quality means a dedication to quality by the entire company. All workers are part of a team working together to achieve quality. No worker is an island. The word "customer" does not just mean the end recipient who pays for the product. and then doing it right-the first time." The first of these two objectives has far-reaching social implications. we are in a new economic age. and not just a function of a few inspectors stationed one step before the customer shipping area.TOTAL QUALITY 3 Management. ." Then he boiled it down even further to two objectives: "Quality is everyone's job. having the tools to do it right. " and "Adopt the new philosophy. summed it up more concisely in 1986: "Quality is knowing what needs to be done. It demands a realization that quality is the concern of everybody. All steps are combined in the process. It emphasizes the old. " 1. 4 QUALITY ASSURANCE AND STATISTICS This whole concept of the workers in a plant or a company as a team is revolutionary to many people. The Taylor model is sometimes called management by control.5. Total quality involves workers in a cooperative effort.6. Why the pervasive use of statistical methods? Where does statistics come into this picture other than just in teaching people how to draw Shewhart charts? The distinguished British physicist Lord Kelvin put it this way: . insertions and modifications from middle management. who writes on excellence in manufacturing. The new thrust in management requires a fundamental change of style. American manufacturers have tended either to go for the big leap or to stop caring about improving the process when they think it is good enough to do the job. sometimes with unwise." The Japanese call this kaizen. The division head sets goals or quotas for each of his department heads: increase sales by 10%. 1. 1. Simply ordering the workers to do this or that is now seen to be counterproductive. and perhaps unauthorized. It is certainly not a part of the Frederick Taylor model of scientific management that has dominated managerial styles in American manufacturing since it was introduced at the beginning of this century. Its essence is the breakdown of the manufacturing process into pieces that are then bolted together in a rigid hierarchical structure. These dicta end up as quotas or work standards at the shop level. Tom Peters. KAIZEN The fifth of Deming's points is "Improve constantly and forever every activity. meaning continuous searching for incremental improvement. decrease warranty costs by 15%. There is an expression "overengineering" that is used disparagingly for wasting time fussing at improving a product once it has reached a marketable state. with initiative encouraged and responsibility shared from top to bottom. The CEO gives his division heads a profit objective and down the line it goes. and so on. The key factor is the pervasive use of statistical methods. WHY STATISTICS? How is kaizen achieved'! It is achieved by integrating the research and development effort with the actual production facility and devoting a lot of time and money to "getting the process right" and then continually making it better. points out that some authorities consider this Japanese trait the single most important difference between Japanese and American business. If quality and productivity are to improve from current levels. with his disciples. changes must be made in the way things are presently being done. 1. There have been well-publicized success stories in other industries besides semiconductors. . Will American manufacturing. then lose its appetite for quality assurance and go back to its old noncompetitive ways until another crisis hits'? There are grounds for increased optimism. Hunter of the University of Wisconsin.THE FUTURE 5 "When you can measure what you are speaking about and express it in numbers. companies are achieving outstanding results from their emphasis on quality. Key examples may be found in the steel industry. once back on track. whether or not this is perhaps a passing fad. your knowledge is of the meager and unsatisfactory kind." But perhaps the most cogent summation is that of W. Taguchi. It has been won by several of the companies that have become household words in this country. and. We should like to have good data to serve as a rational basis on which to make these changes. you know something about it. Ford. named after the late Secretary of Commerce. Quality assurance now has official recognition. G. less famous. We mentioned Corning Glass earlier. 2.7. Motorola. The twin question must then be addressed: what data should be collected. Every year the Japanese present their prestigious Deming Award to the company that makes the greatest impact in quality. The United States has now established the Malcolm Baldrige award for quality." Quality assurance pioneer Deming was trained as a statistician. It was awarded for the first time in the fall of 1988. in our gloomier moments. who. vigorously advocates his program of statistically designed experiments. but when you cannot measure it. We could add Westinghouse. how should they be analyzed? 4. Among the winners was a major semiconductor producer. Now Japan's principal leader in quality control is an engineer. G. where old dinosaurs are being replaced by small speciality firms that have found niches in the market for high-quality items. of which more will be said in the last three chapters. Statistics is the science that addresses this twin question. once collected. THE FUTURE Some of us who have been involved in the quality revolution have asked each other. when you cannot express it in numbers. Many other. 3. and a host of other major companies. who said: "1. They necessarily learned to conduct experiments in its presence. in the presence of variability. The second. Lately. down to the grass roots of American manufacturing. There are data out there. the data should be obtained in an organized way so that their yield of information shall be as rich as possible. Manufacturers demand quality from their suppliers and train them to produce the quality parts that they need. more telling statistic came from James Houghton. It should be a long while before they forget it. Brazil. he has tended to increase that earlier figure of 85%. We are working. we must learn more about them. When an engineer nudges the temperature at one stage of a plant by a degree or so. developing the mathematical theory. she is doing an experiment. We cannot just sit in our offices. The statistician can help to extra<-1: all the information that is to be found in the data. Fewer than 15% are worker-related. For them it was the reality of variability in weather. the Wall Street Journal carried an advertisement from Ford giving a list of their suppliers in the United States and other parts of the world-Mexico. When another engineer uses a catalyst bought from another manufacturer. They may not realize it. Engineers conduct experiments all the time. in weights of pigs from the same litter. Quality pays. and. Shewhart's great contribution was to educate engineers about the existence of variability. Liechtenstein. in soil fertility. and then turn to the computers to solve the differential equations. American management has been learning that statistic the hard way. Two statistil:s can be cited in support of this principle. and more-who had won Preferred Quality Awards for the quality of the parts that they supplied to the automotive manufacturer. On the day that this paragraph was drafted. and management is learning that.6 QUALITY ASSURANCE AND STATISTICS Concern about quality is becoming pervasive. Quality is working its way down to the small subsubeontraetor.8. Juran has said for some time that at least 85% of the failures in any organization are the fault of systems controlled by management. Customers demand and expect quality from their manufacturers. whether we like it or not. he is doing an experiment and getiug some data. And that is before taking into account sales lost through customer dissatisfaction. They have to be gathered and analyzed. Agricultural scientists already kncw about natural variability. 1. if . GATHERING INFORMATION If we are to improve our manufacturing processes. Better still. In some companies the cost of poor quality-the cost of the twin demons SCRAP and REWORK -runs somewhere between 20 to 30% of sales. That procedure has become the basis of modern experimental design. operating conditions." The design people have made one batch of the prototype device under lahoratory conditions.e. The last chapter. The last six chapters point the reader toward active quality assurance. Although based on the work of Fisher and Frank Yates. and others. TillS BOOK There are 19 chapters in this book. The instructor who has only one semester will necessarily make some choices. The sampling procedures tell whether the incoming or outgoing batches of product meet the specifications. Cuthbert Daniel. How can the process be improved'! Chapters fourteen and fifteen arc devoted to regression. 1. The semiconductor industry is full of horror stories about devices that have been designed miles away from the future fabricators and "tossed over the transom.. Chapters seventeen and eighteen are about factorial experiments. That means carrying out experiments at the manufacturing level. will help to choose a designed experiment to yield good data. INSTRUCTIONAL CONSIDERATIONS There is enough material in this hook for a two-semester course. Those topics deal with passive quality assurance. Chapters nine through thirteen deal with the traditional topics of quality controlcharts and acceptance sampling. This is the procedure for designed experiments developed by G. The charts tell whether the process is moving along under statistical control. The Taguchi method combines the ideas developed by Fisher. a procedure that Sir Ronald Fisher introduced 60 years ago to design and analyze agricultural experiments. . Chapters two through eight can be taken as a course in probability and statistics designed for engineers. these chapters focus on the developments of the past 40 years in industrial applications by George Box.10. Box. preferably optimum (whatever that really means). It is now used increasingly in the west as well. and others with some new insights and emphases by Taguchi to form a systematic approach to conducting on-line experiments. Now let the manufacturing engineer make their design manufacturable! But how'? The manufacturing engineer has to learn more about the process and to find good.9. Chapters sixteen through nineteen are concerned with off-line quality control. is mainly devoted to the so-called Taguchi method. where it is widely used. nineteen.INSTRUCfIONAL CONSlDERA TIONS 7 given the opportunity. i. Chapter sixteen is an introduction to the analysis of variance. Taguchi in Japan. They consider how to conduct experiments to determine the best conditions for the operation of the process. fitting models to the data by the method of least squares. 1. and communicating the results to the user so that valid decisions can be made. In the last part of the book. . some instructors may choose to leave out all of the material on acceptance sampling in chapters eleven and twelve. one could postpone the last half of chapter eighteen and some of the specialized topics in the two chapters on regression. judicious cuts could be made in the assignment of introductory segments. You cannot make good decisions about quality without good information. 1. The emphasis in the later sections would depend on the focus of the course.11. any more than a general can make good decisions on the battlefield without good intelligence. analyzing that data correctly. he could omit some of the more advanced and specialized topics in chapters three through eight. REPRISE Why do engineers need to learn statistics? We have mentioned some reasons earlier. or at least to omit the sections on sequential sampling in the latter chapter. If the students already know some statistics.8 QUALITY ASSURANCE AND STATISTICS For example. Others may elect to omit the advanced topics on control charts in chapter thirteen. preferably by properly designed experiments. We have to gather good data and interpret it properly. However. Those techniques for designed experiments are fundamental to the improvement of manufacturing processes. any modern course in statistics for engineers should include the material in chapters seventeen and nineteen. That is what statistics is all about: gathering data in the most efficient way. For example. Less thorough coverage could be accorded the chapter on analysis of variance. Statistical Methods in Engineering and Quality Assurance Peter W. which at first glance convey little. these are only 80 numbers. we discuss how one might summarize and present a large sample of data.- 9 . in some sense. we make this set of 80 numbers more intelligible in two ways. The name mean is short for the arithmetic mean..1 shows the daily yields (percent conversion) in 80 runs in a chemical plant. which is written as i.2. we show how the points can be presented graphically hy a histogram and also in several other ways. As they stand. There arc two ways of doing this. This used to be a chore. The other is by using graphical methods. We use this data set to iIIustrate the ideas throughout this chapter. 2. Inc CHAPTERTWO Descriptive Statistics One picture is worth more than ten thousand words.1. Table 2. Then. In the next sections. THE MEAN AND STANDARD DEVIATION The word statistic is used to describe a number that is calculated from the data. First. It is the simple average of the data. John Copyright © 1990 by John Wiley & Sons. 2. INTRODUCTION In this chapter. but nowadays computers have made it easier.1. We can calculate from the sample certain important statistics like the mean and the standard deviation that. Two of the most commonly used statistics for describing a sample arc the mean and the standard deviation. x"' the mean. or x-bar. we derive some numbers (statistics) that summarize the data points. An obvious candidate is the average of the 80 observations. calling for hours of tedious tabulation. M. is defined as _ "" XI fl' X=L. characterize the data. If there are n observations denoted by x I ' x 2 ' ••• . One is by words and figures.. For any observation x" the quantity dev x.3 69.5 68. When JJ. x-bar = 71.7 68. to denote values that the variables actually take.0 69.5 69. t 76.438.0 70.1.0 69. or (1"2. x. n. or center of gravity.3 73. (We say more about the criteria for "good" estimates in later chapters.0 77.3 68.5 77. We try to use uppercase \etters. Its average is the population mean..2 68.0 73.4 66. .9 66.9 71. observations.5 65.0 68.8 76. we do not know JJ-.5 70.7 76. The set of all possible daily yields of the plant forms a statistical population. .7 73. The population mean is often unknown and has to be estimated from the data.is known.5 77.3 72.1 72. .4 75.6 69.8 69.4 70.6 63.9 74.9 76.2 68. to denote the names of variables and lowercase. If the sample has not been taken under unusual or biased circumstances.0 80.5 75.9 70.1. Then the variance is estimated by the sample variance A The divisor is n .5 72.5 64.6 74.1. denoted by the greek letter JJ-.. we could also have written x = ~.1 73.JJ. Yields of 80 Runs in a Chemical Plant 71.3 70.5 70.5 73. the variance is estimated from the sample by n (The hat denotes that the expression is an estimate from the data.1 rather than n because x-bar has been used instead of JJ-.1 73.) Usually. represents the center of a distribution of mass along the x-axis. -. = x.5 69. Our set of data is a sample of 80 observations from that unknown population.9 67.4 61.4 74.8 73.8 77.7 68.0 71. . the sample mean provides a good estimate of JJ-.2 70.8 70. The variance and standard deviation of x are measures of the spread of the data. It is denoted by V(X). 2.3 69.5 78.9 71.7 68.x..7 69.1 73.10 DESCRIPTIVE STATISTICS Table 2.4 68.) The mean represents the center of the data in the same way that the centroid.4 68.8 76.4 66.7 60. as x.1.1 74.is the deviation of the ith observation from the population mean.5 72. the deviation from the sample mean.7 72.5 75.9 70.1 72. Le.6 74.1 72.6 75. The average of the squares of the deviations is the variance of X.8 71. but it is very hard to be absolutely consistent in this notation. X. so we redefine dev x.7 69.4 68.4 73.2 where the summation is over the values i = 1. For the set of data in table 2. the sample median is clearly defined. the median is defined as X=X(t+l)' If n = 2/. 1. A less ambiguous definition follows.. The first quartile. with only five ohservations.35. J. Just as the median divides the data into two halves. and take the middle observation-the third one.. u. 4 = 60. The sample median is a number with the property that there are as many points above it as below it. Let u = (N + 1)/4. I. For example. THE MEDIAN AND THE QUARTILES The median is another measure of the center of the data. If n is odd. too low. Approximately one-quarter of the points are below it and approximately three-quarters are above it.4. the points are arranged in ascending order. is the square root of the variance.1. The median is the second quartile.925. It is usually estimated by s. Q I ' lies to the left of the median.. 1.(_I) = 68.X(20» 3x Similarly.25(x(21) -. _ x= X(I) + X(I+l) 2 In the example of table 2. 9. If we only have four observations. when N = 80. 8. 4. and 9.2. . 3u (20) + x. we use linear interpolation.75 and so Q3 is defined by . any number 4. Suppose that the data are written down in order and that X(I) is the ith largest observation. u = 20.8. For example. = (X(40) + X(41» 12 = 71. 4. 8. 2.II THE MEDIAN AND HIE QUARTILES We see later that dividing by n would give a biased estimate-an estimate that is. on the average. and 9. and so we define QI = x(2() + O. the quartiles divide it into quarters. For this set of data. ~ 8 will satisfy the definition. If n = 2t + 1. and 2. s = 3. we arrange them in ascending order.3.25. then and If u is not an integer. If u is an integer. We define Q I and Q 3 in the following way. The standard deviation.. although we use a different method when we come to set up Shewhart charts.79. As before. Others interchange Q1 and Q3' so that Q I > Q3' High school counselors may say that you have to be "in the first quartile" of your high school class if you are to go on to college. the trimmed mean (TRMEAN) and the standard error of the mean (SEMEAN). The machine prints out the mean. median. quartiles. as is shown in table 2.. One sets the data in the first column and types the command DESCRIBE Cl. or MINI CI.175 SEMEAN 0. they define and Q . the standard deviation. They mean that you must be in the top 25% of the class.438 MIN 60.12 DESCRIPTIVE STATISTICS QJ = X(/iO) + O.925 STDEV 3. but they are not universal. and the minimum are called order statistics. Statistics such as the median. Some other packages interpolate differently when u is not an integer. maximum. The points that divide the data into ten Table 2. Some authors refer to all the observations in the bottom quarter as constituting the first quartile. We have defined a quartile as a particular value that separates one-quarter of the data from the remainder. the data and the results of calculations are contained on an imaginary work sheet.175. Those are the definitions that the Minitab package uses. Description of the Yield Data N 80 MEAN 71. and minumum values and two other statistics that arc introduced later.3. the maximum. it takes only one or two typewritten commands to read a data set into the work sheet. These statistics are easily obtained on personal computers. If you are in the habit of storing your data in computer files. = x (hO) +X 2 (hI) = 74.1.700 MEDIAN 71. There is another possibility for confusion about quartiles.75(x(61) - X(60) + 3X(61) 4 X(60» = 74. In the Minitab package.05. Individual statistics can be obtained by typing MAXI Cl.788 03 74.546 01 68.000 TRMEAN 71.1. They place the quartile midway between the observations immediately above and below it.423 .350 MAX 80. etc.3. With N = 80. the quartiles. or bar graph. D. stratification.. to present the breakdown of the grades as a histogram. The tally chart has long been used for counting how many objects in a group fall into each of several classes. . They involve no deep mathematics and arc rarely mentioned in mathematical statistics courses. The Pareto diagram is another type of bar chart.11. and graphs. it is discussed in section 2. 3. Scatter plots demonstrate the relationships between two variables. after tallying. making the sct of five look like a gate.4. 5. For a batch of incoming students at a university. whose use is enhanced by modern computers. After you have gone through all the papers. one can plot the entrance examination scores in English and mathematics with English along the vertical axis and mathematics along the horizontal. B. If the first student got B. 2. histograms. or for graduates and undergraduates. or F? The idea is simple. . If the second student failed. 7. cause-and-effect diagrams. 2.ISHIKAWA'S SEVEN TOOLS 13 groups are called deciles. B. graphs. you make a vertical mark in the B row. it is exceeded by 5% of the data. 4. but their visual impact is simple and directly useful. and there are the totals for the various grades. The first four marks are vertical. and so on. It would be quite appropriate. The counting is usually made easier by counting in groups of five. How many of the components that were rejected failed for which of several defects? How many votes were cast for each of several candidates? How many students in a large class got A. Pareto diagrams. 6. scatter plots. you make a mark in the F row. The tally chart is only one step removed from the histogram.. His seven tools are the following: 1. Each student is a dot on . These are simple graphical tools. you make rows for A. and then you take the exams one by one. tally sheets (or check sheets). using linear interpolation if needed. On a piece of paper. ISHIKAWA'S SEVEN TOOLS Ishikawa (1976) proposed a kit called "The Seven Tools" for getting information from data. you count the number of marks in each row. the fifth is a diagonal mark through those four. Control charts fall under the last category. C. The ninety-fifth percentile may be defined as the observation in place 95(N + 1)/100. and we can also use percentiles. You might also choose to stratify the data by making separate charts for men and women. which is the topic of the next section. Fishhonc diagram.1) to 61.8. There is no agreement on .0 occurs in both intervals..0 (=60.5. men. the graph. showing that scores in the two subjects are highly correlated. The resulting diagram looks like the backbone of a fish.0 to 63.4. It was obtained by typing the command HISTOGRAM C1. using one symbol for engineers and another for arts students.llnll1g MACIIINI~~ Mixer -~---r---------. figure 2.0 (=60.4. The program has arbitrarily elected to use intervals of length 2. that we are concerned about the number of defective items being produced. It contains one observation..0 centered on the even integers. As we have phrased it. and then lines for subsubareas. Minitab prints the histogram sideways. or do they fall into the shotgun pattern that suggests independence? Again... A rectangle is then erected on each interval. materials.. Then we draw diagonal lines running to it to denote major areas that lead to that defect. and so on.0 and also contains one observation. 2.. We draw a horizontal line on a piece of paper to denote the defect.1. Would that suggest a pattern? A cause-and-effect diagram looks like the diagram that students in English grammar classes used to draw to analyze sentences.:ure 2. and maintenance..g. The range of values taken by X is divided into several intervals. PAINT Fi.0 . for example. machines. The second interval runs from 61. Is there a pattern? Do the points tend to lie on a line.E OJo~.0 + 1).:l. The area of the rectangle is proportional to the number of observations in the interval.14 DESCRIPTIVE STATISTIC'S INSPEcnON Tn. one could use stratification. 61. HISTOGRAMS One way in which a sample can be represented graphically is by using a histogram. 60. the value 61.7. e..S.1.1 shows the histogram for the data of table 2.1. Suppose. Into each of these diagonal lines we draw lines for subarea'. usually of equal width. An example is shown in figure 2. and so these cause-and-effect diagrams arc commonly called fish bone diagrams. The first interval runs from 59.1.----"""'7-~ [)EFECIS Rcpcunncn Trauung PWPI. 0 that are treated differently in the two packages. Statgraphics suggests 16 intervals for this data set. 80. This difficulty is illustrated by figure 2. Some authors suggest counting them as being half a point in each interval. The histograms of samples of normal data like that in table 2. 66.0. this too happens with samples of data from symmetric distributions.1.1 are usually almost symmetric. Statgraphics has 13 in the former interval and 18 in the latter. Several other things should be noted about these histograms.1. distribution. The number of intervals is also arbitrary. so that the intervals are 59.0.. 74.44 and 71. It is not a question of one writer being correct and the other wrong.I. It has a slightly different shape. handling this.0 < x ~ 63. The difference between the two is a matter of choice. but it gives the user freedom to choose 10. 62. There is an observation at the boundary.0 . Statgraphics puts it in the lower interval. which shows another histogram for the same data set obtained with Statgraphics.. Minitab puts it in the upper interval. 68.0 < x"'. The interval with the most observations is called the modal interval. Minitab puts points on the boundary in the higher interval. In this case.. there are two observations at x = 73.2. respectively) are close to one another.15 HISTOGRAMS MIDDLE OF INTERVAL 60. 64. 70. or normal. . 72..S. We have already noted that the sample mean and the sample median (71. For example. 69.0. J. Minitab has 12 observations in the interval centered on 68 and 19 in the interval centered on 70. 7R. it is a symmetric distribution.5. Similarly. Other programs put them in the lower interval. the modal interval is centered on 70. 61. The distribution of the yields appears to follow the bell-shaped curve that is associated with the Gaussian. Most software packages can give you intervals of your own choosing by a simple modification in the histogram command. NUMBER OF OBSERVA nONS 1 * I 2 4 12 19 12 14 9 5 1 * ** **** ************ **** **** *********** ************ ************** ********* ***** * Figure 2. 61. Histogram for table 2. The normal distribution is discussed in later chapters. 76.35. figure 2. 2. _ . The prices are listed in descending order. In this case.-- . shows those two houses appearing as outliers. It is obvious that the two highest priced houses are markedly more expensive than the others. Frequency histogram.16 DESCRIPTIVE STATISTICS I I ____ J ___________ 15 I I ~-- .2 76. we might have doubted that ..8 72.- 10 5 ---------I--~ I I I I I I _1.6..1.4 67. I I I i <:r J: .2.l. Suppose that we can regard this list as typical of a realtor serving clients in the middle and upper income brackets in those days.6 81 Yields Figure 2. 59 63.6._ _ _ _ _ _ _ _ _ . there is a clear explanation of such outliers--there are always a few expensive houses on the market for the wealthy buyers._I .5. If these had been measurements of the yield of a chemical plant.. _ .6. The histogram. AN EXAMPLE OF A NONSYMMETRICAL DISTRIBUTION Some years ago a realtor listed in the Austin American-Statesman newspaper the asking prices of all the houses that he had on his list. They are given in thousands of dollars in table 2. 900 156. The histogram also shows that the distribution is not symmetric. This is because the median is much less affected by outlying observations. It is skewed with a long tail to the bottom.000 104. 80.950 195.950 94. . 260.950 they were valid and gone back to the records to check whether mistakes had been made in recording the observations.000 61.900 52.500 51.900 60.000 74.950 53.450 57.500 62.500 122. for an ordinary person coming to town.000 69.500 77.500 66.91X) 39.17 AN EXAMPLE OF A NONSYMMETRICAL DISTRIBUTION Table 2.950 51. Each provides an estimate of a central value of the distribution.950 92. 160.200 76.1. the coach has less effect on the median. according to whether or not the head football coach was included in the sample. On the other hand.950 74. If one takes a small sample of faculty incomes at an American university.000 144. 180. 140.000 139.500 96.900 87. 60.000 49. 120.000 79. the median is a better estimate of how much one should expect to pay for a house than the mean.950 71.500 95. For a nonsymmetric distribution.500 129. We saw earlier.500 88.950 56.500 59. 240.500 55. 100.000 79.6. Histogram of house prices.000 54.900 71.950 275.500 134. one can argue about what is meant by a central value.000 65.6.950 69.500 114. MIDDLE OF INTERVAL 40.950 178.900 129.500 94.500 125.950 70.950 124. the mean will vary considerably.950 119. 200. House Prices (in Thousands of DoUars) 208.000 89. in the case of a sample from a symmetric distribution.950 67.000 149.000 110. 280.500 62.900 73. NUMBER OF OBSERVA TIONS 3 *** 18 ****************** 15 *************** 7 ******* 8 ******** 4 **** 2 ** I * 1 1 0 0 2 * * ** FIgure 2.1.500 74.900 225. that the sample mean and median are close to one another.000 154.000 93.500 85.950 48. A case can be made that. 220. In this example. . her highest and lowest MIDDLE OF INTERVAL 3.6 3.lJ25. Each competitor is evaluated by a panel of several judges. 71.8 5.25%.998. among other statistics.430. the four highest and the four lowest observations were dropped. even when estimating the population mean for a normal distribution. where the observations are close together.4 4.3 gave. a small change in the middle of the data.0 4. The amount of trimming is arbitrary. This is obtained by dropping the largest 5% of the observations and the lowest 5% and finding the mean of the remaining 90%. a decrease of 6%. some packages trim as much as the upper and lower 20% of the data.f::. When the two most expensive houses are removed.0 5. such as Olympic gymnastics. the mean drops to $91. The DESCRIBE subroutine that was llsed in section 2. Although the sample mean can be shown to be a more precise estimate mathematically when all the observations are valid.2 4.6. the 5% trimmed mean of the yield data. which is symmetric. some of which may be erroneous observations.546 as opposed to the mean of the whole set.4 NUMBER OF OBSERVATIONS 1 1 10 12 12 7 X 6 2 1 * * ** * *** * ** * ************ ************ ** **** * * * ** * * ** ****** ** * Figure 2. it is more sensitive to outliers.2. The median is said to be a more robust statistic.6 4. The median decreases to S78. the trimmed mean is 71. respectively. It can be argued that the median itself is essentially a 50% trimmed mean because all the observations are trimmed away save one! Sports fans wiII note that in some athletic contests. Histogram of natural logarithms of house pril:es. The median moves only from (xpl) + x(32»/2 to (x(3() + x(3I)/2.982 i and == $79.18 DESCRIPTIVE STATISTICS The mean and the median of the house prices are . a loss of only 1.8 4. Indeed. The reason is that the calculation of the median never involved the outlying values directly. some statisticians prefer the median over the sample mean. some scientists go further and use trimmed means.925 . or to provide a more rohust estimate of J-L. This is why.: $97.2 5. the rating of each competitor is a trimmed mean. Figure 2. I selected the third row on purpose to illustrate this point: there may actually have been an improvement in the variance during that period. Figure 2.7.00 •• ---. but the third row has less variation than the first row. Figure 2. It is.19 OOTPLOTS scores arc discarded (trimmed). we can try a transformation.6./ 76.. 72..1. This second histogram is more like a bell curve than the first. It has been suggested that this is a precaution against outliers caused by unduly partisan judges.7.00 72.00 • _+_ .00 • • ••• • • • -+ -..1.j 70.0() I<lgure Z. If it is important to have a variable that has a bell-shaped distribution.1 shows a dot plot of the last 10 yields in table 2. • • •• •• ·1 --.00 .1 on the same horizontal scale.2 shows dotplots of the first and third rows of table 2..00 Figure Z. 2..- I Row:1 78. Even when we omit the two highest houses. Dotplot of row 8.i • ••••• -f--. of course...00 •• I 74. We have no reason to expect that the housing prices should follow the normal bell-shaped curve.-.2.7. we see immediately that the averages of the two TOWS arc about the same.00 • .. her rating is the average of the remaining scores.---.00 • •• .. Dotplots arc also useful when comparing two samples. It is very easy to make and is useful when there are fewer than 30 points..00 80. DOlplot (If rows 1 and 3 -+ _. It is a simple diagram that can be made on any piece of paper without any fuss..-. DOTPLOTS The dotplot is a simple graphical device that is rather like a quick histogram. but it seems to have been temporary . The engineer draws a horizontal scale and marks the observations above it with dots.1.i- ~·i 64. the curve is still skewed. • . i 68.7.00 68.l. important that the engineer decide on the level of trimming before looking at the data and not after seeing the observations and forming an opinion. RowX I 76.2 shows the histogram of In(price) with the two highest houses omittcd.7. One possibility is to replace the price by its natural logarithm. Row . -~-~ 60. Stcm-and-Ieaf display of house prices. W. 48. 48 and 49 become stem 4 with leaves 8 and 9.8. 1 3 12 21 (11) 30 26 20 19 l6 11 9 7 5 5 4 4 Figure 2. In the example. In ascending order. and 280 at the end as high values. 280 .1. with the command STEM C1.1. The histogram only has diamonds. 54. The last digit of each trimmed observation is called a leaf. 39 is leaf 9 on stem 3.0000 1 2 REPRESENTS 12. Tukey (1977).000. or rectangles. We begin by rounding the price of each house down to the number of thousands by trimming off the last three digits.1. respectively. separates 225. 53. 195 is leaf 5 on stem 19.. 51. stars. On the other hand.8. This means STEM-AND-LEAF DISPLAY LEAF DIGIT UNIT = 1. 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 9 89 112345679 012256799 01134446799 5789 234456 4 049 24599 49 49 46 8 5 HI 225. 52. 49. STEM-AND-LEAF DIAGRAMS A variation of the histogram was introduced as a tool in exploratory data analysis by J. 275. This can be considered a point in favor of the stem and leaf because it shows more information without adding more clutter to the picture.20 DESCRIPTIVE STATISTICS 2. 51. the digits are written as leaves. In the stem-and-leaf diagram. there has to be a stem every $10.6.275. It is called the stem-and-leaf diagram. The diagram is shown in figure 2. . The other digits form the stems. the stems have to come away from the trunk at unit intervals.. the data are now 39.. There are two main differences between the histogram and the stem-andleaf diagram. The Minitab program.8. We now illustrate the procedure for the house price data in table 2. or may not.1 illustrates this for the chemical data in table 2.000. we start to count from the high end of the data. (11) indicates that the stem has eleven leaves.1.000. In this example.0000 t 2 REPRESENTS 12. 7* contains observations 70 and 71. which makes it hard to determine the general shape of the distribution of the observations. This makes it easier to compute the median. Tukcy has a method for splitting stems into five smaller stems. . the stem-and-Ieaf diagram may have too many intervals. with a maximum at about $71. 8* 45 6667 888888888889999999999 ()(}()()()()OOO 11111 2222222333333333 444445555 666667777 8 o Figure 2. which may. the stem and leaf draws attention to the fact that the curve may have two maxima. there are 11 observations to go. we have counted 21 observations from the low end of the data.1. At the end of stem 12 (the beginning of stem 13). the histogram suggests a smooth curve. LO 60 2 6* 3 5 6T 6F 6S 6.950 is less alarming to a potential customer than $101. If there are not many data points.000 range. The seventh stem contains the median. SPLITTING STEMS If there are too many leaves on a stem. 7F contains 74 and 75.9. Once we have passed the median. Note that stems 16 and 18 have no leaves. the diagram may not have room for them. Figure 2.1. The seventh stem has been split into five shorter stems. There is only one house in the $100.9. The first column of the figure gives running totals of the number of observations from the ends.9. This is a matter of sales tcchnique-$99. By the time we have reached the end of the sixth stem. Each yield is rounded down to an integer by trimming the last digit. be a good idea.000 to $110. 7T contains 72 and 73. 7* 7T 7F 7S 1 3 9 30 (14) 36 20 11 2 1 7.21 SPLl1TING STEMS that there tend to be more intervals in the stem-and-Ieaf diagram than in the histogram. 2.1. Stcm-and-lcaf display for table 2.1. 7S contains 76 and STEM-AND-LEAF DISPLAY LEAF DIGIT UNIT = 1. 35.7. + * 72 68 M 80 76 Yield in percent Figure 2. 61.1. I.925 = 5.2.10.1. T with 2 and 3.QI) and QI . with 8 and 9.111ds of dollars Figure 2. The H-spread is 74. Recall that for this data set.7.05 and 74. contains 78 and 79.875 = 82. finally. the right-hand whisker goes as far as the largest observation. I.175. respectively.10.10. lies beyond the fence. the quartiles. Box-and-whisker plot of chemical yields in table 2. In the box plot of the houses data. it is classified as a low value and marked with an asterisk. The median is denoted by +. The two vertical ends of the box are the quartiles. and so the fences arc 68.175 + 7. In this example. The whiskers protrude in either direction until they reach the next observed value before the fence. which Tukey calls hinges. S with 6 and 7. the three outliers are all on the high side. This plot emphasizes the spread of the data. F with 4 and 5. --I * + ** i 50 100 150 200 250 Prices inlhoui>.5( Q 3 . which is shown in figure 2. QJ . The left-hand whisker goes as far as the second lowest observation. 60. It shows the median.0.925 .05. BOX-AND-WHISKER PLOTS Another of Tukey's graphical representations is the box-and-whisker plot. 2.10.68. The box is the rectangle in the middle of the diagram. which is considered to be unusually low by a criterion described in the next paragraph.925. or H-spread. One observation.1. and.8.25.Q I is called the interquartilc range. . the median is 71. and QJ = 74. The asterisk at the far left denotes the lowest value.2. The horizontal lines coming out from either side of the box are called whiskers.6.175 -.1. 7. which is 80.10.5( Q 3 .875 = 61. or hinges. . QJ + 1. The box-and-whisker plot of the chemical data is shown in figure 2.QI) are called the upper and lower fences. Box-and-whisker plot for houses data in table 2. QI := 68.7.22 DESCRIPTIVE STATISTICS 77. Notice Tukey's choice of symbols for the last digit of the integer: * with 0 and 1. 60. and the high and low values. 3. The right whisker stops at the fourth highest observation.840 . MulLiple box-and-whisker plots for three samples. 76 72 64 - I -. the left whisker.0 "'igure 2.0 75.240.lgure 2.900. and the fences are at $62.10. The hinges are Q\ = $62. .. Multiple box-and-whisker plots.000 + $90.4.000.840. $195. therefore. We recall that the median is $79.-- Row . leaving the three highest values marked by asterisks.0 69.-------.400 and $123..10.$90..925. stops at $39.240 =-$27. The lower fence is negative and we have no observations below it.000.0 72. Q3 = $123.23 BOX-AND-WHISKER PLOTS i- + ** 8 I + [- + ------~I--------~I------~I--------+I--------~------Cl 63.0 66.240 = $213. " and introduced Pareto analysis. An obvious step is to reduce the number of items produced that do not conform to specifications. Which should we work on first? J. argued that only a few types of defect account for most of the nonconforming items. In most processes. He called this phenomenon "the rule of the vital few and the trivial many. there are several types of defect that could cause an item to be rejected as nonconforming. Table 2. one of the pioneers of quality assurance.11.1. is much less than the variance for the other two samples. We are not going to be able to eliminate every type of defect at once. Juran. We inspect some of the items produced and set aside those that are nonconforming or defective. Note that there Table 2. and 8 of the data in table 2. One should be careful not to read too much into the appearance that the variance for sample 3.24 DESCRIPTIVE STATISTICS It is sometimes convenient to make box and whisker plots of several samples on the same scale.l gives a set of data from a group of 97 nonconforming items from a production run. as manifested by distance between the hinges. Analysis of Defects Type of Defect Insulating varnish Loose leads Solder joint A Solder joint B Resistor 1 Resistor 2 Capacitor Number Reported Percent Cumulative Percent 54 40 39 20 9 7 29 15 7 5 40 68 83 90 95 5 4 99 2 1 100 .3 shows such a multiple plot for rows 1. With small samples.1 t.1. 2.11. multiple box-and-whisker plots may raise more questions than they can answer.10. The essence of the analysis is to record the number of each type of defect found in the nonconforming items that were rejected and then to make a chart rather like a histogram that will emphasize clearly the important types. Remember that there are only ten observations in each of these samples. Figure 2. in a figure.4 goes a step further and shows.10. which he named in honor of the Italian economist and sociologist Alfredo Pareto. M. PARETO DIAGRAMS One of the main purposes of a quality assurance program is to reduce the cost of poor quality. Figure 2. considered as three separate samples.3. box-and-whisker plots for each of the eight rows considered as separate samples.1. . The sets consisted of observations on a single variable. _ J ____ ..89. In later chapters. stem-and-Ieaf plots. and box-andwhisker plots.6 .- .2S SUMMARY 100 136 I ~. are more than 97 defects recorded because some items had more than one. Pareto chart.. I I I 81.. SUMMARY In this chapter. _____ ___ .8 . followed by x-y plots for bivariate data and residual plots in the two chapters on regression. Control charts appear in chapters nine.. and the solder on joint A between them account for over 80% of the defects.1. _ I I t 80 60 I I ~ (j ~ .11.7 I 108. With a few easy commands on a personal computer. ~_l." Not many years ago.4 CL1 Cl2 Cl3 Cl4 CL5 CL6 d:: Cl7 Figure 2. . I .11.1. each of these plots gives the viewer a simple visual summary of the data that provides a picture of "how the data looks. such as percent conversion or housing prices. The methods used in the main part of the chapter were histograms. loose leads.L I - .12. ten. The corresponding chart is figure 2. allowing for false starts and interruptions along the way. it took an hour or more to calculate the mean and variance and to . The table shows that insulating varnish.9 . we have discussed some methods of presenting data sets graphically..I I . we mention other types of plots. and thirteen.J~ ____ . J. ~ c ~ ~ 40 54. 2.. 0 . - - -~. dotplots. 8 16. 15.7 15. for convenience.4 14.6 17.1 15.5 ]2.0 14.8 14.3.6 15. are 1831 1839 1828 1800 1831 1813 1850 1817 1816 1832 1781 lR32 1874 1578 1876 1821 1847 1823 1848 . Always have a good look at your data set before you try to do anything complicated with it! EXERCISES 2. Make a histogram of the data and calculate the mean. and standard deviation.6 13. 2. During World War II.6 15.0 12.1 14.4 14.2 16.4 14.2 tS. Verify that in exercise 2.3 14.3 15.4 13.8 17.5 13.5 14. there would be little change in the values of the mean.1 13.2.5 14. The reported thicknesses.6 17.2 17.8 13. The following data set is made up of measurements of the thickness of 100 strips of mica in onethousandths of an inch.9 12. An engineer takes a sample of 19 wafers and measures the oxide thickness.1 15.9 15.1 15.8 16.8 16.4 14.0 15.2 13.0 12. and even longer to make a histogram. Verify that you would get the same results in exercise 2.3 13.5 11.5 15.2 L4.2 15.3 18.7 18. median. you subtracted 10.7 16. there is no excuse not to make them a routine first step in any analysis.1 12. thin slices of mica were used as dielectric material in condensers for radios.0 13.9 17.5 11.2 14.8 12.4 19. and standard deviation if you rounded off each observation to the nearest integer.6 16.6 16. Perhaps that was an excuse for a busy engineer not taking the time to do it.5 15.1 16.2 14.7 17. 2.3 18.4.2 16.2 2.0 14.0 15.1 if.4 16.0 t3.6 13.9 15.1 15.2 10.26 DESCRIPTIVE STATISTICS locate the maximum and minimum and potential outliers in a large set of data.9 15.0 14.7 14.4 14.6 13.5 13.8 15.8 16.6 15.5 16.8 13.5 14.4 13.7 17.5 15.1.7 15.6 14.3 16.S 16.7 16.4 15. in angstroms.6 13.8 13.1.0 14.4 15.0 from each observation before making your calculations (and added it back to obtain the mean and median).6 15.3 13.0 18. Now that these simple graphics take only a few minutes at the terminal that is conveniently situated on one's desk.3 16.9 15. median. 7.27 EXERCISES Obtain the mean. in exercise 2.10.5. Explain why the maximum observation is flagged as a possible outlier. The observation 1781 is the only observation below 1800. Does this set of data approximate a bell-shaped curve? The observations arc the reciprocals of the obser4 vations in exercise 2.4.7 multiplied by 10 and rounded.6. Make a box-and-whisker plot of the data in exercise 2. Make a histogram of the following set of 50 observations and use it to find the median and the quartiles. median. 2. On the other hand. 2. Make a dotplot and a box-and-whisker plot.9.S. that 1578 is an invalid observation and that he is justified in throwing it out. the observation 1578 is obviously an outlier. the median is unchanged and the trimmed mean and quartilcs increase slightly. . Suppose that after investigation. standard deviation. Does the data set seem to have the traditional bell curve (the normal distribution)? 180 215 179 193 247 222 189 262 202 211 183 219 193 188 187 182 184 17] 201 185 160 207 208 249 231 209 241 185 243 ]89 285 258 264 223 175 218 214 183 211 257 228 231 187 219 204 155 199 202 189 205 2.4 (after 1578 has been excluded) and use it to find the median and the quartiles. standard deviation. Make a box-and-whisker plot of the data in exercise 2. hut the minimum observation is not. 2. and the quartiles for the remaining ]8 wafers. trimmed mean. the trimmed mean. median. the engineer decides. 2.7. there are appreciable changes in both the mean and the standard deviation. Make a stem-and-Ieaf diagram for the 18 observations in exercise 2. Do that and recalculate the mean.9. make a new dotplot and a new box-and-whisker plot. In both plots. Find the hinges and fences. Make a histogram of the following set of observations and use it to find the median and the mean. 56 47 56 52 40 45 53 38 50 47 55 46 52 53 53 55 54 58 50 54 63 48 48 40 43 48 41 54 41 53 35 39 38 45 57 46 47 55 47 39 44 65 43 53 46 49 50 50 53 49 2. and the quartiles of the data. why is it not starred as an outlier in the box-and-whisker plot? Note that when we drop the outlier. In a week.11.12. and 25% of the components lasted more than twice as long as specified. This sounds encouraging. with an average of 20 defects per machine A2 F 67 B 5 G 15 C 54 o H13 I 10 12 E 8 K 14 Make a Pareto diagram that provides a graphical illustration of the importance of focusing on machines C and F. A plant has ten machines working a single shift. Sixty of the components are put on test and their lifetimes are given below.28 DESCRIPTIVE STATlSnCS 2. . Find the mean and median lifetimes of the components tested. together with the standard deviation and the quartiles. An electric component is specified to have an average life of 1000 hours. The average exceeds the specification by 27%. the following numbers of defects were noted for the ten machines. Make a histogram of the data. 15 213 434 722 1108 1576 2217 3159 72 241 452 723 1171 1601 2252 3215 74 270 464 877 1316 1631 2316 3832 99 285 504 959 1328 1657 2684 100 302 526 1024 1410 1893 2858 127 361 652 1051 1451 1975 2862 195 366 666 1075 1463 2059 2873 195 430 704 1087 1529 2076 3088 4295 2. but what fraction of the data fails to meet the specification? This suggests that it is not enough just to look at the sample average. acceptable or not acceptable. A simpler alternative is sampling by attributes.. X is said to be a discrete random variable. .1..1 shows 320 digits taken from the Austin phone book. but unusable if its length is outside that interval. the inspector may classify each rod as either good or bad.2. Continuous variables are discussed in the next chapter. Inc CHAPTER THREE Discrete Variables. Y. Y will not take the same value every time. perhaps. i. The inspection can now proceed in two ways. take values anywhere in a certain range. In the jargon of quality control. M. Hence. 1. An inspector takes a sample of rods from the production line and measures the lengths. The inspector can record the actual length. either too long or too short.e. RANDOM DIGITS Table 3. The reason that I chose the third digit rather than the last was that I was not sure whether the phone company held back some numbers ending in zero for test purposes.Statistical Methods in Engineering and Quality Assurance Peter W. It takes only integer values 0.. Thus. and then I took the third digit after the prefix in each number. INTRODUCTION Suppose that we have a process for manufacturing metal rods and that we are concerned about the uniformity of the rod lengths. X. . the inspector is sampling by variables. in the sample. A rod may be usable if its length lies in the interval 109 mm :s Y:s 111 mm. of each rod. The customary terms are defective or nondefective. 29 . or Attributes 3. 2.2. Because the rods will not have exactly the same length. in which case it is said to be a continuous random variable. as opposed to values over a continuous range. the last digit but one. I took the right-hand column from several pages (for convenience). John Copyright © 1990 by John Wiley & Sons. The inspector may record the number of defective rods. It may. Y is a random variable. 3. We now develop some theoretical background for discrete variables. another tenth must be 2. We do not claim that.1. and. We may also say that X has a particular probability distribution.1. OR ATIRIBUTES Table 3. for the probability that X takes the value zero. A MATHEMATICAL MODEL Suppose that we regard this data as a sample of the output from a process for producing digits.2. X takes the values 0.. Table 3. 1. 3. more generally..3. . The probability distribution of the random variable X in this example is called the multinomial distribution. or P(i). = P(9) = 0.. . A plot of them appears as Table 3. 9 has probability 0. for the probability that X takes the value i. exactly one-tenth of the digits in our sample must be O.. or law.1.'requencies of OccurreO(:es of Digits o 40 1 2 36 29 3 33 4 5 6 7 1:\ 9 29 34 32 30 31 26 . it is. .30 DISCRETE VARIABLES. 320 Digit.. This implies that each value 0. 1. It is a generalization of the binomial distribution that is discussed later. therefore. P(X = i). one-tenth must be 1. Some go further and call it a uniform discrete distribution because all the probabilities are equal.9 according to a probability model. in the sense that each of the ten values is equally likely to occur.. In this example.1 shows the actual frequencies of occurrence of the digits.3. the probability law is simple.1. Taken from the Phone Book Number of Zeros 1620 3076 9104 6773 4114 2269 7510 4305 1516 8796 6761 4255 7063 94H5 1573 1589 4879 2140 8518 6H07 6089 0325 8966 6136 2476 6699 5270 3992 6298 6370 5308 4241 4741 0722 2024 3073 7673 3733 6653 8068 8350 1449 5518 7781 5272 1663 3902 3472 0546 9131 5659 0158 0863 2118 t085 0480 7957 0229 2440 1042 2953 8593 9706 5054 1057 8404 0037 8534 8008 2191 1938 8227 7 4 6 1301 5198 4590 6081 5253 5343 8764 1013 6 5 3 4 5 It is reasonable to think that the digits ought to be evenly distributed. as a consequence of this probability law. and so on. The probability model for the process is given by P(O) = P(l) = P(2) = . We write P(X = 0). 2. called a random variable. Let X denote the value of a digit. . That is clearly not so..3.. or P(O). 31 SOME BASIC PROBABILITY THEORY o 2 3 "lgure 3. Out of the lotal of 320 digits in the sample. tossing a die can be regarded as an experiment with six outcomes. the number 0 should occur four times. It has ten possible results. In only one of the rows are there fewer than four zeros.1.3. The deviations of the frequencies from 32 that we see in this particular set of data are not unreasonable. figure 3.3. the set of possible results of the experiment is . In section 7. Similarly. We should also expect that out of the 40 digits in each row. or outcomes. On the other hand. each digit should have occurrred about 32 times. It occurs more often in the first row than in any of the others. The experiment consists of picking a digit. and yet 0 occurred 40 times and 9 appeared only 26 times. we expect that each digit should appear about one-tenth of the time. 4 FI'~qu~ncics 5 6 7 8 9 of digits. SOME BASIC PROBABILITY THEORY We can also look at this from the point of view of an experiment. we present a statistical test for the hypothesis that this set of data did indeed come from a process with equal probabilities. Is this unreasonable? The answer to that question can be shown to be no.4.1. In the general case. ]5. Is that unreasonable? 3. not (13 + 4)/52. It follows that P(E C ) ==}.. In this case. X = 1 or 3 or 5 or 7 or 9.4. or both. we can consider an ordinary deck of playing cards. They are nonnegative and their sum is unity.2. which consists of the points 0.1) Example 3. EI n E2 is the set (1. ].20. The union of E. not necessarily in equal portions. and E2 be drawing an ace.4. i.7. or. What is the probability that a part will have no defects? Let El be the event that a part has the input defect. in the digit example. it occurs only if both EI and E2 occur. One can picture a I-pound jar of probability jam being ladled out.16. El U E2 is the set (0.32 DISCRETE VARIABLES. 30% have an output defect.e. and 10% have both defects. or product. For a discrete variable. We are given that P(E J ) = 0. The intersection. is a set of points in S. Some might get no jam at all. either an input defect or an output defect. it occurs if either E. peEl) === 13/52 and P(E2 ) ==4/52. 7. An example in which S has an infinite number of points appears in section 3. Of the parts offered by a supplier. a probability measure. Suppose that E2 is the event that X is odd. An event is said to occur if the result. A simple part in an electronic system can have two kinds of defect. or outcome. is assigned to each point of S. It contains six points. 20% have an input defect. E e is the event X>3. Its probability is 0. 1. The following formula expresses this: C (3. The probability that the event occurs is the sum of the probabilities of the points in the set.2. 9) with probability 0. it is the set of points of S that are not included in E and corresponds to the event that E does not occur.6 = I . The ace of spades is the only card in EI n E 2 . Let EI be the event of drawing a spade. The latter answer is wrong because the ace of spades has been counted twice. It consists of 52 cards.4. U E2 • It is the set of all points that are in either El or E 2 . We can now define unions and intersections of events. OR ATIRffiUTES called the sample space and is denoted by S. P(E 1 ) = P(X s 3) = P(O) + P(l) + P(2) + P(3) = 0.1. A probability. of the experiment is one of the points in the set. 5. which occur in both sets. and let E2 be the event that a part has the output defect. .P(E). An event. As another example. peel n E 2 ) = 1/52. of which 4 are aces and 13 arc spades. 2. the event El might be the event X S 3. 3) with probability 0. but peEl U E 2 ) = 161 52. The event E is the complement of E. Thus. In computing the probability of EI U E 2 . For example. 3. In this example. These probabilities satisfy two requirements. of EI and E2 is written as El n E 2 • It is the set of points that are in both EI and E 2 . among the various points. The number of points in S may be either finite or infinite. E.P(E). and E2 is written E. or E2 occurs. S is a set of discrete points. and 3. those points have zero probability. more precisely. we must be careful not to count twice points I and 3. 000. The random variahle recorded in Table 3. X. and P(E J n E 2 ) = 0. X = I if the coin shows a head.30 .20 + 0. the same as the probability that the digit will be odd. 5. The probability that a digit from the set 0.1. 8).5. that is observed in table 3.40 .5. Alternatively.2. 6. The first entry in the table is 3.1 is the number of even digits in a group of 4.1 is 1620. and let X denote the number of heads that appear. and X = 0 if the coin shows a tail. since the first set of four digits in Table 3. The probability that a part will have at least one defect is.30. Let X take the value 1 if the digit is even.4. 3.5. 8. 9) or even (0. 2. The mathematical model for drawing a digit at random from that set of possible values is the same as the model for tossing a fair coin. 1.1.0.5. 9 will be even is one-half. THE COIN TOSSING MODEL Table 3. 320 Digits Taken in Groups of 4 (Number of Even Digits) Numher of Twos 3 2 2 3 2 2 2 1 2 1 0 I 3 0 2 I 4 I 1 4 3 3 1 2 3 1 2 2 2 2 3 2 I 2 4 1 I 2 4 1 2 1 2 2 2 2 0 2 3 3 2 2 2 3 2 0 4 1 I J 2 3 2 0 2 2 1 3 4 2 4 2 3 1 3 1 1 3 1 2 6 8 7 11 . and the value 0 if the digit is odd. 3. THE BINOMIAL DISTRIBUTION The random variable.2.1.7.5. toss a fair coin once. 0 3.1 is based on the data in Table 3. 6.60. This time the digits have been classified as odd (1.5. 4.ins three even numbers.33 THE BINOMIAL DISTRIBUTION P(E 2 ) = 0. P(E J U E 2 ) = 0. which conta. This is the simple coin tossing model.40 = 0. It follows that the probability that a part will be free of defects is 1.10. In both examples.1 is the number of even digits in each set of four. 4. the behavior of X follows the probability law P(O) == P(I) == 0. from equation 3. 7.10 = (l. 2. 3. This is equivalent to observing the number of Table 3.6. 5. HHTT. Frequencies of Occurrence of Value. HTHH.6. THHT. THTH.1. We multiply the probabilities of tails on the individual tosses.. I. P(O) == ( ~)(! )( !)( !) = -h = P( 4) . TITT 40 __ JI _______ . HTTT THHH.1.. HHHT. P(O). 2. Frcqucncies of values of X.. OR ATfRlBUTES Table 3._. 3. but not with equal probabilities. . each has probability 1116.34 DISCRETE VARIABLES. The points arc HHHH. The numbers of times that these five values occurred arc shown in table 3. HHTH. 4.1 and in figure 3. TTHH. TI'HT. nTH. THTT. HTTH.6. The probability. X takes the five values 0.. .. They arc equally likely.6..i 10 o Figure 3. S. __ I __ 30 ~ I 20 .6. The sample space. has 16 points. of X X= 0 1 Observed frequency We expect 5 5 21 20 3 2 32 15 30 20 4 7 5 heads that occur when a coin is tossed four times. I. that X takes the value 0 is the same as the chance of tossing four tails in succession. . HTHT. 7. while X = 3 at THHH. 4/16. Example 3. replace all the nonzero digits by ones. 3. A fair coin is tossed five times. X = 3 requires three heads and two tails. the statistician thinks of a process. three places out of five for the heads can be found in 5!/(3!2!) = 10 ways. represented by zeros. THTI. HTHH. THTH. P(3) = 10/32. X = 0 occurs only at the point lTITT.l. TTHH. If the data were given in the form of a sequence of zeros and ones. What are the probabilities that X = 0. respectively. An alternative version of the sample space would be the five points. 1/16. 11TH. exactly two heads occur at six points: HHTT. P(l) == P(3) = 4/ 16. The random variable X is said to have a binomial distribution. which takes the . The probability of a head at each toss is 0. we should expect X = 2 to occur (6/16)(80) == 30 times. HHTH. corresponding to the values taken by X with probabilities 1/16. and 90% that are nondefcctive. 0 3. TfI-IT. 1.6. Hence. We can think of it as representing the output from a production process that produces 10% defective parts. 4/16.2. called success or failure. 4. 2. X = 3. HTHT." in which each trial has exactly two outcomes. 6/16. This topic is developed further in the next two sections. or a "sequence of trials. Thus.5. and so P(2) == 6/16. Exactly one head in four tosses occurs at each of the following four points: HTIT. THHT. In a set of 80 observations. HHHT. P(O) = 1/32 = P(5). That is a special case of a more general model. Again. Associated with each trial is a random variable X. Finally. and X == 5? 5 There are 2 == 32 points in the sample space. HTTH.35 UNEQUAL PROBABILITIES x = 0 for the point TlTf.1. Now we have a sample of data from a process that produces 10% zeros and 90% ones. One might ask the following question. 0. how would you decide whether the process is operating under control at a 10% defective rate? In the general case. UNEQUAL PROBABILITIES The binomial distribution was introduced in the previous section from the coin tossing model. we use the data in table 3. This time. p = q = 0. As a practical matter. It would not be reasonable to argue that.36 DISCRETE VARIABLES.. The idea that the trials are independent with constant probability of success is important to the argument. and let X denote the total number of successes in all n trials. If we toss a fair coin nine times and observe tails each time. In the coin tossing model with "heads" as the successful outcome. If we were to classify a defective as a failure. which is often denoted by q.e. the probability of a head on the tenth throw is still 0.10. and so P(l) = 4pq3 . OR ATIRIBUTES value one if the trial is a success and the value zero if the trial is a failure. This would lead one to conclude that p is considerably less than 0. It is not reasonable to bet in favor of heads on the tenth throw.10 in the example and q = 0. We have described what is known to statisticians as a sequence of Bernoulli trials.8.90. We saw in table 3. We argued.5. 3. a case could be made for the argument that a sequence of nine tails in succession is strong evidence that the original assumption of p = 0.50. we multiplied p2q2 by 6. the number of ways of choosing . X is said to have a binomial distribution with parameters nand p. and q = 0. The probability of success is the same for each trial and is denoted by p. By generalizing this for p ¥. which is the number of sequences with H twice and T twice. THE BINOMIAL DISTRffiUTION (GENERAL CASE) Suppose that we have a series of n independent Bernoulli trials with probability p of success at each trial. The probability of failure at each trial is 1 .5 is false. the probability of a sequence with one head and three tails is pq3. that X == 1 occurs for four sequences. which is the topic of chapter seven.5. i.1 an example of a binomial variable with n = 4 and p = 0.5. That argument takes us into the area of hypothesis testing and decision making. Indeed.5. To compute P(2).0.90. the example ill the first paragraph of this section would have p = 0. Each sequence has probability 1/16. named after the Swiss mathematician James Bernoulli (1654-1705). each of which has the letter H once and the letter T three times. since the probability of one head and nine tails in ten throws is greater than the probability of ten tails. and with the probability of any part being defective denoted by p. quality control engineers are more interested in focusing upon the number of defectives produced by a process than on the number of nondefectivcs. they would have p = 0. The system is postulated to have no memory.5 and to bet on tails for the tenth toss. They apply this model with defectives classified as successes. and so P(l) = 4/16. for example. Thus. the chance of a head on the tenth is greater than a half.p. In a series of Bernoulli trials. We do state that if we were to throw a fair die for hours on end. (3. the values of 11 may be rather large. It is denoted by E(X).9. denote the number of successes on the ith trial: P(X.:: h+ ~ + ~ + ~ + ~ + r. The mathematical expectation. let X. x) =:: )' x. Let X be the number that shows on the top face.1) where the summation is made over all values x that X takes. weighted according to the probability law. =:: 1) =:: p E(X. Then £(X) ::. ::= 0)::= q .-x (x ::.: x!(n .) =:: and (l)(p) P(X. to use the Poisson distribution as an approximation to the binomial.5.9. we should not be surprised to see a few more or a few less. (3. we should expect to have 32 zeros. If we wish to have x successes in n trials.17-3.5 . In the language of permutations and combinations. and it is hoped that the values of p are small. 4. =:: 3. 6. Mathematically.19.I( n _ x. 2. each with probability 1/6. In practice.9. the formula for E(X) is £(X) ::. then the average number of points per throw would be approximately 3. and the probability that X = x is . We do not believe that we shall ever see the die showing three and one-half points. or expected value.37 MATHEMATICAL EXPECfATlON two places out of four for the successes.2) . For a discrete variable. Consider throwing a single die. Under those circumstances.p q P) . 5. 3. and adequate. it is defined by the center of gravity of the probability distribution. x n. of a random variable X is the long-term average of X. MATHEMATICAL EXPECTATION We said earlier that out of 320 random digits. X takes the values I.x)! In quality control work. The Poisson distribution is discussed in sections 3. this is the number of combinations of two objects out of four.: 2: xP(x) . it is easier. the number of sequences available is n! C(n.. + (O)(q) =:: p. 3. Definition. and E(X) = p. whence V(X) + [E(X)]2 = p .p2 = p(l. then Y == 3X.10. X is the sum of n independent variables XI' •• .1) E(X)f} . (3. If a is some constant. OR AITRIBUTES For a binomial distribution. For a single toss of a coin. respectively. and if Y = a 1 X J + a 2 X 2 + . p . are random variables with expectations m 1 . (3. E( X2) = 12 ..10.4) This is merely a matter of change of scale. and so X 2 •• E(X) = E(X + £(X 2 ) + .· .10. E(aX) = a£(X) .2) [E(X)l2 . THE VARIANCE We are now able to give a more formal definition of the variance of a random variable. . If X has been measured in yards and if Y is the corresponding measurement in feet. " then ••• . we note that EUX .9. This definition is compatible with the definition that was given in chapter two.p) = pq ..9. m 2 . The variance of a random variable X is V(X) = (J"2 = £ {[X - (3.9.38 DISCRETE VARIABLES.. == np J ) .. (3. In practice. and so E(Y) = 3£(X). If XI' X z .3) The following properties of expectations are easily verified.5) 3.E(X)f} = E(X2) = E(X2) - 2£(X)' £(X) (3. P + 0 2 • q :::. we can develop properties of the variance that parallel equations (3. This was the basis for deriving the binomial distribution. of which only two are defective. the probability of drawing a defective item remains constant because the composition of the batch from which the inspector is choosing is unchanged. However. for example. of defectives in the sample. p. = npq . The alternative is sampling without replacement. the sample can be drawn in two ways.3) . stretching a piece of yarn until it breaks in order to determine its strength. for the second item. If a and b are constants. we have sampling with replacemelll. then V(X) = V(X 1 ) + V(Xz) + ..39 SAMPLING WITHOUT REl)LACEMENT For independent variables. Clearly.4) and (3. or batch. it is quite possible for the same item to be counted more than once before the inspector has reached the quota.1O.. V(aX + b} = aZV(X) (3. where N is the size of the . and the next item is drawn from the remainder of the batch.5).11.10. which is independent of the parts that have been sampled so far.. If the inspector returns each item to the batch after examination and then draws the next item from all N members of the batch as before. With this scheme.4) Applying equation (3. of N items and observes the number. and if Xl' X 2 ' ••• are independent random variables with expectations m I ' •. " then m2. If the first item were nondefective. of which three arc defective.5) o 3. {J = 3/6. we should have p = 2/5. there would now be five items left in the batch. SAMPLING WITHOUT REPLACEMENT In acceptance sampling by attributes. we should have. if the testing of the items is destructive. the denominator is N-.• . We have argued until now that the probability of drawing a defective part at any stage of the sampling is a constant. If the first item chosen were defective.4). When the first item in the sample is chosen. and if Y = a1X 1 + a 2 X 2 + .10.1. we see that if X is a binomial variable with n trials and probability p. However.10.9. Suppose that a batch consists of only six items. X. p = 3/5. an inspector draws a sample of n items from a lot. as. In computing p for thc second item. The inspector sets each item aside after it has been tested. (3.9. sampling with replacement is impossible. respectively. (3. for the second item. 3. n) ways.1.40 DISCRETE VARIABLES. of which three arc defective. We wish to find the probability law. THE HYPERGEOMETRIC DISTRIBUTION Example 3.d. On the other hand. we can choose a sample of three out of six in 20 ways. The sample of n items contains x out of d defectives and n . or distribution. . and 0 so the desired probability is 9/20. Returning to the batch of six items.x of N . What is the probability that it will contain two defectives and one nondcfective? We can choose two defectives out of three in three ways. or the hypergeometric distribution.9991 44. suppose that a large university has 45. the numerator depends on the result of the first drawing.d. In the general case. and the number of defectives are not very small. hence. It can be shown that the sum of these probabilities for the various values of X is indeed unity and that E(X) = ndl N. Let x be the number of defective items in the sample. and a sample of n items out of N in C(N.12.000 are men and 15. When a second student is chosen.x) C(N. both of which are close enough to 113 for practical purposes. n .1) The algebraic expression in the numerator resembles an expression that appears in classical mathematics in a term in the hypergeometric series. but a model based on the change in batch size and content. which is the number of combinations of two items out of three. The correct model would not be the binomial model. the name hypergeometric for this distribution. N. of whom 30. Altogether. so that P(x) = C(d. for P(x). the probability of selecting a woman will he either 14. That model is called the hypergeometric law.333363.x nondefectives out of N . even if we sample without replacement.d nondefectives.x) ways. Similarly.001144. When the batch size. OR AITRIBUTES original sample. n) (3.999. 3 x 3 = 9 give us the sample that we seek. n .din C(N .dare nondefective. Of these 20 ways. That is a simple example of the hypergeometric distribution. the binomial model is still a good enough approximation to the real world to justify its use. we can choose one nondefective out of three in three ways. suppose that we have N items of which d are defective and N . x)C(N .3333185. x) ways. or 15. n . The probability that a student chosen at random will be female is 1/3.000 arc women. and that we draw a sample of n items.000 students. We can choose x defectives out of d in C(d.12.999 = 0. suppose that a random sample of three items is drawn.12.: 0. the number of combinations of three out of six. 1) Substituting the values given above..1. We sce from the data that P(A) == 0. An inspector examines a component and finds that the first resistor is defective. In 40% of the components.41 CONDITIONAL PROBABILITIES 3. What is the probability that the second resistor will be defective also'? Let A be the event that the first resistor is defective and B the event that the second is defective.67 . (3. and B is the event that the second item is defective. given that event A has occurred. p(AIB') = 3/5. the second resistor is defective. In 12% of the components. and p(AIB) = 2/5.40.12. Event B can occur in two distinct ways. we define P( B IA) as the ratio P( A B) / P( A). C Similarly.048. If P(BIA) = PCB). the first resistor is defective.c. A denotes the event that the first item drawn is defcctive. PCB) = P(A)P(BIA) + P(A')P(BIA') .13. either it is preceded by A or else it is preceded by A c. P(BIA) = P(AB) peA) 0. both resistors are defective. PCB) = 0. 0 . i. (3. A is the event that the first item is nondefective. Let AB denote the joint event that both items are defective (we drop the symbol n). in C this case. This is an example of the basic law of conditional probability: P(AB) == P(A)P(BIA) = P(B)P(AIB) . we calculate P(B) == n)0 ) + n)( 0 = } . The components each contain two resistors.2) Mathematically. They would have been independent only if we had P(AB) = (0. CONDITIONAL PROBABILITIES Suppose that in the previous problem.13. Hence. We writc this as P(BIA) == 2/5. the event that A did not occur.12)(0. and P(AB) = 0. The notation P(BIA) means the conditional probability of event B. If the first item is defective. In 8% of the components.13. The two events are not independent. Let A' denote the complement of A. then p(AIB) = peA) and P(AB) == P(A)P(B).08 = o. the ratio has no meaning if peA) = O. we say that the events A and B are (statistically) independent. A supplier is manufacturing electronic components.40) = 0.08. Example 3. the probability that the second item is also defective is 2/5. P(BIA ) == 3/5. In that case. L2 == 0.13. We see that P(AB) == 1/5 == P(A)P(BIA). It is the probability of the occurrence of A before we know whether or not B has occurred.13. OR ATfRlBUTES 3.14. and applying equation (3. after his death. P(A)P(BIA) p(AIB) = P(A)P(BIA) + P(AC)P(BIA") More generally. It enables us to make use of prior knowledge in our probabilistic calculations. but you cannot read the serial number.- peA IB) - P(AB) _. It is our revised value of the probability of the occurrence of A in the light of the additional information that B has occurred. he can only supply 80% of your need. occurred would provide no information about the occurrence of A. if the first experiment has a set of outcomes A the theorem becomes P(A.14. and Bayes' formula. that the component came from Dallas. What is the probability that A has occurred also? We wish to calculate the conditional probability p(AIB).1). Suppose that there are two experiments.13.13. You pick a component at random.•.1). It is defective. . Bayes' rule. .1 leads us in the direction of Bayes' theorem. and so you have to buy the remaining 20% from another supplier in Houston who makes 10% defectives. You purchase most of them from a supplier in Dallas who is on record as producing only 5% defectives. If you did not .2). Your company uses a large numher of a certain electronic component.IB) = P(AJP(BIAJ 2: P(A (3. Let B be the event that the component is defective. Thomas Bayes.2) h )P(BIA h ) In equation (3.14.1. If A and B are independent events. The theorem seems at first to be just a routine exercise in manipulating formulas. P(AIB) is the posterior probability of A. BAYES' THEOREM Example 3. Example 3.14.42 DISCRETE VARIABLES. The second has outcomes Band B". On the other hand. We notice that B has occurred. It is called by various authors Bayes' theorem. The first has outcomes A and A". What is the probability that it came from the supplier in Dallas'? Let A denote the event. From equation (3. the knowledge that B has. or has not. P(A)P(BIA) PCB) - PCB) . (3. Its importance has been realized in the last half century. peA) is the prior probability of A.1) I' A 2' . and that probability would remain unchanged. or outcome.14. and published in 1763. Unfortunately. It was proved by the Rev. in some cases. By equation (3.13.43 BAYES' THEOREM know whether the component was defective.2.001)( 1. Suppose that one young male in a thousand has a certain disease that has no obvious unique symptoms. By equation (3.999)(0.20)(0.1).l66~L Only one-sixth of the men who test positive actually have the disease! 0 P( A /lJ) is strongly influenced by the rate of false positives and the rareness of the disease. that a patient has a disease when that is not so. P(AIB) = (0.005) + (0. This can lead to unfortunate consel\uences. Example 3. the knowledge that the component is defective reduces the odds that it came 0 from Dallas and increases the chance that it came from Houston. this means that the test may indicate.005995 . Some diagnostic tests are not perfect. If P(A) were 0. (0.80. Applying Bayes' theorem. in this example. It is important to notice the assumption that the young man was chosen at . applying Bayes' theorem.10) = 0.000) P( A / B ) == ---0.0001 (one in ten thousand) and the rate of false positives were the same. A young man. If we assume that there are no false negatives.000.005.8~?ci~·0~' == 0. What is the probability that he actually has the disease'? Let A be the event that the patient has the disease.67 .13. PCB) == (0. Because of the comparatively good quality record of the Dallas supplier.1).05) + (0. the prior probability is peA) == 0. Another example occurs in medical statistics. In medicine. This is the overall rate of defectives for the supply of components entering your plant.14.0(0) == 0.001)( 1. P(B/A) = 1.005995 .80)(0. you would argue that peA) = 0. They may give a small percentage of false positives. and that the routine test has probability 0. takes the test and is rated positive. Let B be the event that he tests positive.001. = O.06. then. P(B/A") = 0. P(A/B) would fall to about 2%. PCB) = (0.005% (one-half of one percent) of false positives. chosen at random from the population. 44 DISCRETE VARIABI. Denote the two variables by XI and X 2 • In our example. we use data from a similar example to illustrate some of the basic ideas of bivariate distributions. This might be an appropriate model for routine medical exams for all the potential inductees to the armed forces under conscription or of a large company that routinely gives medical examinations to all potential employees as a condition of employment. BIVARIATE DISTRIBUTIONS An object can sometimes be classified in two ways. In that case. it contains "" observations. The probability that XI takes the value i and X 2 Table 3.15." The corresponding probabilities are called the marginal probabilities. Now there are two variables: location and quality. The reader with experience of test procedures can think of similar examples in industrial work. In this section. In section 3. 1: zero if the test result is OK. The table has four cells. There are N observations altogether. we saw an example in which a component could be classified on the one hand as having come from Dallas or Houston. OR ATTRIBUTES random from the whole population.1 Strength Length OK Defective Total OK Defective Totul 90 10 100 30 120 20 30 50 150 . X 2 = j.1.ES. The data can be arranged in a 2 x 2 table: two rows and two columns. 3. The totals have been written at the rightmost column of the table and along the bottom row-"in the margins. Xl pertains to length and X 2 pertains to strength. The ijth cell corresponds to XI :::: i. whether or not he had the disease. Each variable can take two "values. and on the other as being defective or nondefectivc. Eaeh test classifies the rod as either OK or defective. These data show the output of a plant that makes rods and is not very good at it. It would not be so appropriate if he was chosen because he was already in a doctor's office or a clinic as a consequence of feeling ill. and one if the rod is defective. They take the values 0.15.14." We usually prefer to say that each has two levels. This is an example of a bivariate distribution.15. Each rod is subjected to two tests: length and strength. A sample of 150 rods gives the results that are shown in table 3. Every man in the population had the same chance of being chosen. The data have been modified to make the arithmetic simpler. P(A) might be larger. : 101150. X has a binomial distribution with parameters P and N. The mathematical argument is that we inspect N items and find X defectives. POt:::.16. its . for every i and every j . (3. we have Poo ::= 0.=P. The marginal totals are denoted by n i . is the total for the jth column.12. We discuss this type of problem in more detail in chapters six and seven.1) In the example. The events are independent if P(AB) = P(A)P(B). In this example.3) In the example. This suggests that there might be a tendency for parts that fail one test to fail the other also. We return to this point when we discuss the chi-square test for 2 x 2 tables later.15. inspect them.33 is the probability that a rod is defective in length. P.4S THE GEOMETRIC DISTRIBUTION takes the value j is p" = n. and B be the event that it passes the strength test. and only if. THE GEOMETRIC DISTRIBUTION Suppose that we wish to obtain an estimate of the percentage of nonconforming.15. The conditional probabilities are defined in terms of the cell probabilities and the marginal probabilities: P(X2 = jlX t = i) = p" . Poo = 90/150. With independence. Are the two types of defects independent or do they tend to go hand in hand? Let A be the event that a bar passes the length test. or defective.·p·. (3. and n.2) It is sometimes appropriate to ask whether the two variables are independent. The corresponding marginal probabilities are n. 11... is the total for the ith row of the table. and Pt. n.. and then argue that the percentage of defectives in the batch is a good estimate of the actual process percentage./ N. and =N n. Pt. An obvious procedure is to take a batch of items. p. items produced by a manufacturing process.60 ~ ( ~ )( ~ ) . we should have expected to have had (2/3)(4/ 5)( 150) = 80 parts passing both tests. = 0. = 201150..=N"' (3.. We rephrase that and say that the variables are independent if. p.15. in section 8. P.. 3. Pw = 30/150. q =E =1 p . the cost of obtaining the estimate. He was investigating the death by accident of soldiers in the Prussian cavalry. This means that we have to have a sequence of X-I failures followed by the lone success. Its sum is -p- 1. and so we estimate p by X I N. The Poisson approximation is commonly used by engineers in acceptance sampling. THE POISSON DISTRIBUTION The last discrete distribution that we shall discuss is the Poisson distribution. 2. In that case.16.. r _l)p'qx-r (3. and is infinite. Mathematically. The sample space. We can show that £(X) = 1/1' and V(X) = q/p2. 3. It is featured in the c charts in chapter ten. is a geometric series with first term I' and common ratio q.. .17. Bortkiewicz found a use for Poisson's distribution in his research. This procedure fixes the number of items inspected and.2) with E(X) = rip and V(X) = rqlp2.1) The infinite series P(l) + P(2) + P(3) + . In 1898. Poisson's mathematical contribution was to obtain an approximation to the binomial distribution when p is small.. This aspect of the Poisson distribution is one of the topics of the chapters on acceptance sampling (chapters 11 and 12). Poisson (1781-1840). We may then estimate I' by 11 X. we derive the geometric distribution from a sequence of Bernoulli trials with the probability of success being p. L. OR ATTRIBUTES expected value is E(X) = Np. The probability distribution of X is called the geometric distribution. We continue until we have had r successes. and let X be the number of trials until the first success. He observed ten corps of cavalry over a period of 20 years and . 3. S.. named after the famous French mathematician D. Let X be the number of items inspected (including the defective).16. consists of all the positive integers 1.46 DISCRETE VARIABLES. The negative binomial distribution is an extension of the geometric distribution. The negative binomial variable is merely the sum of r independent geometric variables. and so (3. Another approach to the problem would be to inspect items one at a time until we found a defective. P(x) = C(x -1. thus. . It is known to some mathematicians as the Poisson limit law. 17. we have 23 = e-IJ-II. we can replace q by unity and write VeX) = lip = t:(X). . but we can write np = I-L and usc Poisson's approximation to the binomial. 1. This says that X has a discrete distribution with X = 0. p. 2. Summing equation (3. and VeX) = npq... r ( 0+ I +" + r ~ + ~ + . We do not know either n or p. (3.1) We recall that O! = 1. E[X(X . of soldiers. X. .1) from x = 0 to infinity. each of whom had a small probability.47 PROPERTIES OF THE POISSON DISTRIBUTION recorded for each of the 200 corps-years the number of men who died from being kicked by a horse.17. we can estimate I-L by taking several samples and letting 3. What prohability law did his data follow? One may make this kind of loose argument. ) 2! 3! = I-L . The number of deaths. Each corps contained about the same number. (3. has roughly a binomial distribution.3) . It.18.e I-L' IJ- x! .. with E(X) = Itp. Since E(X) = J-L.. of dying from such an accident.1) Similarly. (3.2) It follows that and so (3. This is the Poisson distribution. PROPERTIES OF THE POISSON DISTRIBUTION We now confirm that equation (3.1)\ = J-L2 .1) defines a probability distribut}on.18.1S. and P(X = x) = p(x) '-'. J-L. 3.17. If p is small. It has a single parameter.18. . and so p(O) = e IJ-. and so on. Eventually. it began to be a popular and useful model for situations in which we count the number of occurrences of some event that we regard as happening at random with a certain average frequency. and had no place in industry.1.7 to be zero. Similarly. We estimate fJ. Out of 200 observations. There were 200 observations with a total of 122 deaths. Poisson Distribution of Deaths x 0 1 2 3 4 Frequency of Observations fx 109 0 65 22 3 I 65 200 122 -- 44 9 4 Fitted Value 108.3. 3.1 0. Examples are the number of cancerous cells per unit area of a section of tissue in a biopsy.1. FURTHER PROPERTIES It can be shown that if XI' X 2 arc independent Poisson variables with . To make the fit to the data. The p(O) = e' O 61 = 0.21.54335) = 108.19.19. the number of breaks in a spool of thread. Industrial examples include the number of meteorites that collide with a satellite during an orbit. The last column of fitted values is explained in the following. we should expect 200(0.48 DISCRETE VARIABLES. 3. we substitute 0. Other examples in the literature have inspectors taking sheets of pulp and counting the number of dirt specks greater than 0. The Poisson model gives a very good fit to this set of data. OR AITRIBUTES Table 3. or the number of chromosome interchanges in cells. and the number of failures of insulation in a length of electric cable. this model seemed to be used only for rare events.20. such as the number of deaths of centenarians in any month in a big city.61e.54335. the expected frequency of x = 1 is (200)(O. or the number of ships arriving per day in a seaport.1 mm 2 in area or counting knots in sheets of plywood.by x = 0.7 66.u.19.61 ) = 66.61.6 3.2 4. THE BORTKIEWICZ DATA The data arc given in table 3. OTHER APPLICATIONS For quite a while.3 20. or the number of misprints per page in a book.61 for fJ-. 4. there are 36 possible results (1.2). A device consists of six lamps in parallel.EXERCISES 49 parameters fLl and 1-'2.. (e) Y> 8. . X k are independent Poisson variables. (b)Y=l1. The probability that any given lamp works is 0. (a) exactly two arc defective? (b) At least two are defective? 3. Suppose that the thicknesses of slices of mica made by a machine are . each with probability 1136. their sum. (c) Y is odd. The following results are reported by the registrar of a college. (d) the two numbers are the same.95. What is the probability that at least one of the lamps works? 3.2. 40% of the freshmen failed mathematics and 30% failed English. I). (g) both (c) and (d) occur simultaneollsly. (f) both (d) and (e) occur simultaneollsly.(6. Two-thirds of the devices that we make are defective..3. There is light as long as at least one of the lamps works.). to taste. Last year. What are the probabilities of the following events? (a) Y. in a sample of six devices. Ten judges are each given two brands of coffee.1. A and B. This is the additive property of the Poisson distribution. What percentage of the students passed English but failed mathematics? 3. EXERCISES 3.6). . It follows that if X" X 2 . A certain electronic device is rather difficult to make. 3. ••• . What is the probability that. What is inconsistent in that report? The amended report from the registrar said that 50% of the freshmen failed at least one of the two subjects. has a Poisson distribution with parameter fLl + 1-'2. XI + X 2 . (1. 80% failed at least one of the two subjects.= 7. then Y has a Poisson distribution with parameter k E (fL.6. Let Y denote the sum of the numbers on the two die. If there is no real difference between the two brands.S. what is the probability that exactly five judges will prefer A and exactly five will prefer B? 3. each with parameter fL" and Y is their sum. When two die are thrown. At the end of the day.1 to present the data. 3. Both components have failed in 4% of the devices. because the first inspector only replaces the device that he tests if it is good. A batch of eight devices contains three defectives. Construct a table likc table 3. Eight of them are thicker than specified. what is the probability that it came from machine A? 3.9. OUf process produces 4% defective items. Suppose that an item is declared to be defective after testing. he puts it back. 3. it also erroneously declares 1% of the good parts to be defective.15. An inspector tests one device.0015 in.10. if it is OK. Items are testcd one at a time until the defective is found.so DISCRETE VARIABLES. tests one device. A batch of ten items is known to (. and that the manufacturer specifies a median thickness of 0. An inspector chooses two of them at random without replacement and tests them.7. What is the probability that he chooses a defective device? Note that this is not the same as sampling with replacement. What is the probability that it is defective? If it is defective. he throws it out. A quality control engineer measures ten slices. Let X be the . What is the probability that it actually is defective? Repeat the exercise assuming that the tester catches only 95% of the defectives and misclassifies R% of the good items. Another batch of eight devices also contains three defectives. A piece of test equipment dctects 99% of thc defective parts that it sees.12. 3. B makes 300 items a day with 5% defective. A device has two critical components. OR ATrRIBUTES symmetrically distributed about the median thickness. An item is taken from the bin and tested.lt. If it is defective. B has failed also? Suppose that the results came from examining 1000 devices. Failures in the device are accompanied by failures in one or both components. A and B.~ontain exactly one defective. What is the probability that as many as seven slices out of ten will be thicker than the amount specified if the slicing machine is working properly? 3. Machine A makes 200 items a day with 4% defective. Are the events (a) A has failed and (b) R has failed independent? What is the conditional probability that if A has failed. and C makes 400 a day with 2% defective. A plant has three machines. Component A is found to have failed in 6% of the devices. component 8 in 8%. Unfortunately. too. A second inspector comes along later and he.8. What is the probability that both are defective? 3. the outputs of the three machines are collected in one bin. 5. 3. Generalize your result to the case where there is one defective in a batch of n items. Let X denote the number that is shown when a single die is thrown.50 that the value will be above the specified value. What is the probability that he will unnecessarily nag the machine for overhaul if the process is on specific(ltion? 3. 4. each with probability 1/6.1.14.13. Calculate E(X).EXERCISES 51 number of items tested. 2.50% of a value below the line. He decides that he will flag the machine for overhaul over the weekend if there arc either seven (or more) out of ten points above the line that week or seven (or more) out of ten below the line. X takes the values ]. and that the probability will be 0.17. Calculate E(X) and VeX). A plan for examining lots of incoming items calls for an inspector to take a sample of six items from the lot. Calculate £(X) and V(X) for the geometric distribution. A quality control engineer takes two samples each day from a production line and computes the average of some characteristic. Calculate E(y) and V(Y).16. If the vendor is producing items with 10% defectives. what is the probability that a lot will pass inspection'! You may assume that the lot size is so large that it is appropriate to use the binomial distribution rather than the hypergeometric distribution. 3. . If the process is on specification. 3. there is a probability p = 0. 3. which is the center line on his chart. 3.15. He will reject the lot if there is more than one defective item in the sample. and 6. Let Y denote the sum of the numbers on two die as in exercise 3. is defined in terms of intervals. for example. Now we picture the probability being sprcad over the interval in a layer whose thickness may vary. we considered discrete variables whose sample spaccs are sets of discrete points. The density at the value X = x is denoted by f(x). In theory.2. might. X. In this chapter. That is the topic of the next chapter. we could have variables for which S consists of a number of disjoint intervals. We turn now to continuous variables whose sample spaces are intervals. PROBABILITY DENSITY FUNCTIONS The thickness of the layer of probability jam in our analogy is formally called the probClbility density limetion. 4. One can also have hybrid variables that take values over both intervals and discrete points. The probability that X takes a value in thc interval a < x < b is dellned by the integral Pea < x < b) 52 = f f(X) dx .1. distribution.Statistical Methods in Engineering and Quality Assurance Peter W. The sample space. A continuous random variable. (4. sometimes continuous and somctimes discrete. INTRODUCTION In the previous chapter. To consider them would take us beyond the scope of this book. The picture of the contents of a l-lb jar of probability jam being allotted in lumps changes. Events are subintervals or collections of subintervals of S. The reader who is pressed for time may prefer to postpone some of the sections at the end of the chapter and then refer back to them as needed when reading the later chapters. M. John Copyright © 1990 by John Wiley & Sons. S. We shall usually consider random variables for which S is either the whole line from -00 to +00 or the half line from 0 to +00. is the normal. take any value in the interval 4 < x < 8. or distributions. as one might spread jam unevenly over a piece of bread.1) . we discuss the exponential and gamma distributions. Inc CHAPTER FOUR Continuous Variables 4.2. The most important of the continuous probability laws. or Gaussian. The probability density functions usually contain constants.1). The portions where f(x) == 0 make zero contribution to the value of the integral. The integral that evaluates the probability is the same in both cases.:: J1 3 2 dx = 4 .53 PROBABILITY DENSITY FUNCTIONS Since the probability of an event has to be nonnegative and the probability that X takes a value somewhere in the whole space S is unity. the integral shrinks to zero. The probability density function is proportional to x.2. and p. This implies that the probability that X takes any single value. where k = 1/2.:. where k is a constant to be calculated. such as (J in the uniform and exponential distributions that follow in later sections of this chapter.2) for all x in S. their values are unknown. We have f(x) = kx.2. It is convenient to adopt the convention that f(x) is zero for values of X that are not in S. A random variable takes values in the interval 0:5 x:5 +2.3) can be written as the integral from -00 to +00. If. Find f(x) and calculate the probability that x> 1.0). the following conditions are imposed on the density function: f(x):2: 0 (i) (4. One advantage is that formulas such as that in the condition of equation (4. In practice. in equation (4.2.2. It is customary to denote them by Greek letters. (ii) (4.0) . They are called the parameters of the distribution.3) where the integral is taken over the whole sample space. and one of the main problems in statistics is to obtain estimates of the parameters from a set of data.0. Note that it would have made no difference had we asked for P(X:2: 1. and u in the normal distribution. as 0 opposed to the set of values in an interval.2. we take the limit as b approaches a. .2. Example 4.3) fkXdx=l. From equation (4.1. (2 X P(x> 1. is zero. 1 ) f(x) dx . For the .S4 CONTINUOUS VARIABLES 4. the prohability density function defines a graph y = f{x) that lies on or above the x-axis. i. (4.5).1) defines probabilities in terms of integrals. For the left-hand portion of the triangle.1} may be modified to read Pea < X < b) = F(b) . (4..4) (4. (ii) F(x) is a nondecreasing function. Alternatively. (iii) F( -00) = 0.5) Example 4. F(b) 2:: F(a). I'(x)::: x L 0 X dx == x2 "2 .3.3. F( +~) = 1. .3. A random variable has density function =x F(x) O:sx!Sl.S:s x :s 1.3. ( 4. We may take the lower limit as zero because f(x) right-hand portion. we may define F(x) by F(x) = P(X:s x) . =0 Find F(x) and evaluate P(O. =2-x elsewhere. It is the area under the curve between the lowest value in Sand x.F(a) .2.) The defintion of eyuation (4. The graph of y = .3.1.3. F(x).2. if b > a. =0 for x:5 O. The cumulative distribution function of X (we usually omit the word cumulative) is defined by F(x) = f".e.0.f(x) is an isosceles triangle with a vertex at the point (I. (There may be cases when the graph of y is a collection of pieces of curves or a step function. I). satisfies the following conditions: (i) F(x) 2:: 0 for all x. The distribution function.3) (4.2) and equation (4. or areas under the curve. THE CUMULATIVE DISTRmUTION I-UNCTION In general.3. There arc two common generalizations of the uniform distribution on (D. ( 4.5) = F( 1. has the distribution that we used before in example 4.F(0.2. each of which has a uniform distribution on (0.0) has f(x) 1 =0 O<x<O. elsewhere. their sum. 1). It is immaterial whether we use the open interval or the closed interval in the definition.1) This density function obviously satisfies the conditions of section 4. o Some writers prefer to develop the theory by starting with the distribution function and then introducing the density function as its derivative.55 THE UNIFORM DISTRIBUTION (X I F(x) J = {) Xdx + JI (2 (2 - x) dx = 0. (4. has 1 f(x) = ---2 .01 elsewhere. 4.1.4. f(x) I = =0 O<x<l.2) We will see latcr that if XI and X 2 are two independent random variables. 1).5) . X = X. + X 2 . the uniform distribution on the intcrval HI < x < 02 with 0.5 _ (2 ~x )2 X)2 =1-~-2~ P(O. 1. The uniform distribution on the interval (0. The distribution function is F(x) = () x<O.3 ) ° .5 s X:::.5) = ~ --- ~ = ~ . More gencrally. =0 elsewhere.4. =x O<x<l. For this distribution. THE UNIFORM DISTRIBUTION The simplest continuous distribution is the uniform distribution on the interval (0. =0 °> 2 (4.1).5 + 0.4.4. =1 x 2= I .3. 5._ _ oj..1) The probability density function for an exponential distribution with (J = 100 is shown in figure 4....1...2..2) is clearly satisfied._____ . +1 ______ ....- I - - - - - _0- I IL ______ _ I I I I I I I I I I I I ---t~--------~-- 2 I I I 100 200 300 400 500 x Figure 4. ______ ~----------I----------~---------~---------~------- I . because._ I I I I 4 + _. Indeed.5..___ I . when x 2: 0.1.. I I I I I I 1 I . we have f(x) > O. 81 = -8 and 8z = +8.+ . 600 .e. -. It is a good model for the lifetimes of electronic components that do not receive the unusual stresses of sudden surges in current or voltage and have a constant failure rate... (X 10..___ . 4. (4. The condition of equation (4.. Probability density function for an exponential distribution.I I I I t.5... i..5. it is sometimes called the constant failure rate (CFR) model.56 CONTINUOUS VARIABLES A special case occurs when the interval is symmetric about zero. The density function is defined as X 2:0... _ I - _______ I I ~ I ~ LI _______ ...__ I 1 f-.... THE EXPONENTIAL DISTRIBUTION The exponential distribution is commonly used in reliability work....3) 10 ~~~-r~~r-~~-r-r'-~~~~-r~~r-~~-r-r'-~ I I I I 8 I I . ... Even if one .. i.5.3% of the components will fail before 500 hours.F( 15(0) = e --\.5.3) is satisfied by showing that F( +00) Integrating f(x) gives us F(x) = 1- e.1).5 = 0.5 = 0. 22. Whether one uses equation (4.04978. there are no planes left. What percentage of them will fail before 500 hours? What percentage will last for more than 1500 hours? We have F(x) = 1- e-x/JOOO • O F(500) = I .1. Only 5% of the wings will last for the required length of time! 0 There is a difficulty with this example. The lifetimes of electrical components made to a certain set of specifications have an exponential distribution with an average lifetime (0) of 1000 hours.57 THE EXPONENTIAL DISTRIBUTION We confirm that equation (4.1) or equation (4. Example 4.5.1) as follows: 1:( ) = ae -ax J' X x>o. 39. what should (J be? It was reported in the press that the average lifetime appeared to be only 2000 hours. The new parameter a is the death rate. 0 Example 4.5. Specifications call for 95% of the center sections of the wings of a new type of fighter plane to last for 6000 hours. It is the population average lifetime of the components.5.3) is that we have written a instead of 1/8. as x tends to infinity. the rate. I . That is not a very practical solution because by the time that the last wing has failed.e. We will stay with equation (4.e.3) The only difference between equation (4. measured in components per unit time. what percentage of the wings will last for the desired 6000 hours? We have 8:. It will be shown later that the parameter 8 has a physical meaning. (4.5.0. at which the components are failing.393. so that F(6000) = 1 .1.2. 95% of the wings will have failed.2) The limit.3% will last longer than 1500 hours. (4.5. How do you estimate the average lifetime of the wings? One possibility is to wait until all the wings have failed and then calculate the average.5.1) and equation (4. By 6000 hours. If that report is correct.5. of F(x) is indeed unity as required.: 2000.3) to define the distribution is a matter of personal preference.223. If the lifetimes have an exponential distribution.5. One can rewrite equation (4.x'O • = 1. The quartiles. When 8 is doubled.69326. half the wings had failed? This statement implies that the median is 2000. Find the median and the quartiles of the exponential distribution. THE MEMORYI~ESS PROPERTY OF THE EXPONENTIAL DISTRIBUTION An important property of this probability model is that the system has no memory. Using this value for 0.5.25 and 0.28778 and Q3 = 1. What if the reporter had said that after 2000 hours. Example 4.1.5.5. If a lamp has lasted until now.6932 = 2885 hours. We see here that the parameter lJ is a scale factor. QJ and QJ' arc the values for which F(x) = 0. The median is obtained by solving the equation l-e"X.7. the last one might take quite some time. 4. Perhaps the reporter was confusing the median with the mean. so that we can estimate 0 as 2000/0. We return to this example in the next section.5) = -In(2).2 «(:ont.o=0. we can calculate that about 12% of the wings would last for 6000 ~~. The median of X is the value for which F(x) = 0.5. Example 4.6. -ilo = In(0. THE QUARTILES OF A CONTINUOUS DISTRIBUTION The quartiles are simply defined in terms of the distribution function. the 0 median and the quartiles are doubled too. the probability that it will last for another 100 hours is the same as the probability of lasting another 100 hours . respectively. 0 4. Similar calculations show that QI = 0.38630. Taking natural logarithms of both sides.58 CONTINUOUS VARJABLES were to take a sample of the planes and wait until they all failed. or e- xl8 = 0.).6. whence i = 6 In(2) = 0.75. 8.8% of the items will survive past time 0. and so only 36. and B be the event that it lives at least until time a + b. We have to show that p(BI A) = P(X> b). We end up with the formula [(x) [(x) h(x) = 1 . R(x)=e. Let X denote the lifetime of a component. or the probability that it would have lasted for 100 hours from the time that it was first switched on.2). B occurs.368. and B: X> a + b. = P(B) = ea +" .12. Let A be the event that the component lives at least until time a. Thus. and only if.F(a) = eO . given that it has lasted until time x. F(x) . R(x).X1'J .2) R(O) = 0. The reliability function. by equation (3. 4.59 THE RELIABILITY FUNCTION AND THE HAZARD RATE was last week. so that AB is the same as B. P(A) P(AB) = P(X > a) = 1.8. ( 4. Going through all the stages of the calculation and taking the limit is similar to computing a derivative using the step process. P(AB) P(BIA) = P(A) I> =e = 1- F(b). we have A: X> a. It is. It is defined in the following way: h(x) dx is the incremental probability that the component will fail in the time period x to x + dx. Events A and B can occur if. defined by R(x) = 1 _. Then.• (4. The hazard rate hex) is the instantaneous failure rate.F(x) = R(x) . for all values of a I and a 2 • This can be established by using some of the prohahility concepts that were introduced in the previous chapter. (4.8. The computation of hex) uses the conditional probabilities. Formally. therefore.3) . THE RELIABILITY FUNCTION AND THE HAZARD RATE Two other probability functions are commonly used by reliability engineers. is the probability that an item will survive to time x.8.1) For the exponential distribution (constant failure rate). 2) and If we set 1 = a. h(l) = {31(3-1 .9. . Readers who are interested in reliability will also want to read about the lognormal distribution in section 5...9.x == exp(-x)..12.exp(-/(3). THE WEIBULL DISTRIBUTION We have just seen that the exponential distribution has a constant failure rate and a constant hazard ratc_ We now modify thc hazard rate to make it time-dependent. This is the basic Weibull distribution. we obtain a more general density function: f(t) ~ ( t )Il-I exp [( =:. = e. If we take the simplest form of the exponential distribution. Suppose that the lifetime of a piece of equipment has a Weibull distribution with a = 15 years and {3 = 2.9. R(t) = exp(-t(3).9..XIO h(x) = e --xIO = () ..60 CONTINUOUS VARIABLES The exponential distribution has a constant hazard ratc: (1/0)e. 4.1 )Il] (4. we have R(a) = e.1) = 1 . Example 4. Like the parameter () in the exponential distribution.I ::= 0.368. (4. f(x) and replace x by t iJ . it is the lifetime by which 63.2% of the devices have failed. If we write Iia instead of just I.. These items are warranted . the element of probability becomes dP = f(x) dx = exp( - t f3 ){3t(3 1 dr • and the probability density function of t is f(l) and F(t) = {31(3-1 exp(-t(3).1. a is called the characteristic life. 61 THE GAMMA DISTRIBUTION to last for 3 years. What percentage is still alive at the end of the warranty period? R(t) =exp[ R(3) -(£rJ. The density function is (4. Only 4% of the items do not survive the warranty period. we keep applying the reduction formula until we arrive at f(m) = (m - 1)!r(I) = (m .1) . When m is not an integer.4) with a > 0 and p > O.10. o 4. The chi-square distribution.I)! . (4. we cannot evaluate the gamma function by the methods of elementary calculus. (4. When a = 1. (3)(2)(1) .I)! as f(m).10. the distribution is skewed with a maximum . The gamma distribution is defined for a continuous random variable that takes nonnegative values.10. it is conventional to define (m . When a > 1.2)(m .3) where (m . the gamma distribution reduces to the exponential distribution with {} = p. The sum of the lifetimes and also the average lifetime have gamma distributions.I)! denotes the factorial (m .x dx m>O.3)·· .9608. is a special case of a gamma distribution. = exp[ -( :5 rJ = 0. (4.10. we have the reduction formula f(m) = (m - l)f(m . which is very important in later chapters. THE GAMMA DISTRIBUTION Suppose that we put on test n items that have exponential lifetimes.1 ) [ntegrating by parts. That is the usual definition of the factorial. The name gamma comes from the gamma function and is defined by f(m) = foOL x m .2) When m is an integer. When m is not an integer.10.l)(m .1e. 0.9.62 CONTINUOUS VARIABLES 0.0 and f3 = 5.015 I -.10. 4.Q1 0. (mode) at x = f3(ex -1) and a long tail on the right as x goes to infinity.1) . we defined the expectation of a discrete random variable by the formula of equation (3. THE EXPECTATION OF A CONTINUOUS RANDOM VARIABLE In section 3.::.1).1 I I I I I I 1 I 0.11.----- -I I I I I I I I I I I I I I .>~ xf(x) dx .0 is shown in figure 4. For a continuous variable.03 .10._____ J. 0.• --------t-. The probability density function for a gamma variable with ex = 8.1.005 o o 20 40 60 80 100 120 X Figure 4.9.1.11.025 I I I I I I I I I _I_ ~ I I 0...- -------~ ~ I I I I - . Probability density function for a gamma variable. £(X) = L xP(x) . the corresponding definition is £(X) = f'. ______ 1-.02 I __ t __.~ ---1--------+---------. (4. lL)'f(x) d. is defined by H'" m. too.63 THE EXPECTATION OF A CONTlNtJOUS RANDOM VARIABLE This.4) The moments of a distribution are defined in two ways.11. the expectation of a function g(X) is defined by = E[g(X)] L+: g(x)f(x) dx .11. and is often denoted by IL. . It is sometimes called the mean of the distribution.11.1. It is the topic of the next section. = E(X) 1 ?~ 0 (\ X e xlfJ /3"(0' .11. Let X have a gamma distribution with parameters a and /3.11.x . can be thought of as the long-term average or the ccnter of gravity of the probability distribution. (4.3) o More generally.6) The second central moment is the variance.1. or the rth central moment. Find E(X). We write (4. = J _~ (x . 1"0 xe E(X) == -1 () xiii u L dx W = e x/o dx =0.2) o Example 4.11.11.I)! The numerator has the form of the gamma function with a instead of (r . (4.5) The rth moment about the mean. Example 4.2. Find the expectation of a variable with the exponential distribution. The rth moment about the origin is E(X'). so that (4. (4. 12. The positive square root of the variance. denoted by VeX) or by defined by VeX) = I~"oo (x - JJ-)lf(x) dx . It is often easier to use the following formula: _ 2 I .1.12.64 CONTINUOUS VARIABLES 4. THE VARIANCE The variance of a continuous random variable. Find the variance of a random variable that has a gamma distribution with parameters a and {3.2.12. (T2. Following the method of example 4.<16 We saw in example 4.11. Find the variance of an exponential variable.13. = 20 2 • Then (4.12. I J ('" m2 = -0 Jo 2 -xiO x e = 2 L'" xe -. is called the standard deviation.11.2) Example 4. That is. we calculate Then (4.m z .3) o Example 4.12.12.JJ.2.• (4. a function that generates the moments. is (4. as the name suggests. MOMENT·GENERATING FUNCTIONS A useful technique for finding the moments of a distribution is to take derivatives of the moment-generating function. .12.4) o 4.1 that E(X) dx dx = 0. fT.1) It corresponds to the moment of inertia of a distribution of mass. 13. + (Ot). It follows that we can obtain the rth moment about the origin by taking the rth derivative of M(t) with respect to t and evaluating it at t = O.1) and expand it as a series in powers of t. The moment-generating function is M(t) :::: 1 (0 "0 Jc) e" dx -. .2. Suppose that X has a uniform distribution on the interval (0.-I (eI I ) .. the coefficient of t' in the series is m.. X.g.1) 8t 8t (8t)2 2 3! =-+--+ .9t)] dx 1 = 1- Of .+ -4!. Engineers will notice the similarity between the moment-generating function and the Laplace transform.. Find the rth moment about the origin.13. and we can see immediately that the rth moment is 1 (~ m.1) We recall that elx can be expanded as the infinite series IX e == 1 + tx (IX)2 (tx)~ (IX)3 + -2. When we evaluate the integral of equation (4.f.+ -3!. Example 4.+ . = 0)0 xr dx = Or r+1 . o .. For the exponential distribution M(t) =. This is a simple example.lr!.X/II dx o Jo = ~ L'" exp[ -xO 0.6S MOMENT-GENERATING FUNCTIONS The moment-generating function (or m.1. .13..13. is defined as +OC MAt) = E(e IX ) = J_~ elXf(x) dx. (4.~ (~ elfe. (r+l)!' and the coefficient of (Ir! is or m 'r-r+1' Example 4.. We have f(x) = 1/8.) of a random variable.8). but it illustrates the basic idea. + . y) dx dy .J3/)] (1 . We will see that the sum of 11 indcpcndent exponential variables is a gamma variable with u = nand J3 = H.. BIVARIATE DISTRIBUTIONS We discussed bivariate discrete distributions in section 3. .14. y) (ii) f ~ +~ f'~ _0.J3 t )-'" .13. if we are investigating a random variable and wc find that its m.f.. Summations arc replaced by integrals. c (4. y).66 CONTINUOUS VARIABLES This is the geometric series 1 + «(It) + ((It)2 + ((It)3 + .1) The density function satisfies thcse two requirements: (i) cf>(x. we use the moment-generating function in an inverse way.X(lJ3. o In latcr chapters. and so m~=8rr!. of random variables.14.g.1)!J3"] ) = o r 1x x"'-lexe dx L" X. y . 4. We have a bivariate density function cf>(x.3.14. the m.f.2) cf>(x. Suppose that we denote two continuous random variables by X and Y. or the averages. M(t) For the gamma distribution. Example 4. The probability that both (l < X < band c < Y < d is given by the doublc integral b i Lei a cf>(x..< _00 0 for all x.-I cxp[.g. y) dx ely = I. This is useful when we wish to talk about the sums. is = [(a -l)!J3"T' = 1(0' . For example.15..2/) -" we can say that the variable has a gamma distribution with parameters a = 3 and J3 = 2. is (L . The same ideas carry forward to continuous variables. The probability density is spread over an area rather than along a line. . (4. 14. if Y> E( Y) whenever X> £(X).15. Y) = E {( X . Suppose that the joint density function for X and Y is c/>(x.1) can be rewritten as Cov(X. (4. Here we obtain the marginal density of X by integrating out the other variable. X and Y.£( X)][ Y . x is a given value. In the discrete case. foe l/J(x.. 4.5) In this function. is obtained by dividing the joint density by the marginal density of X. the covariance measures how the variables vary together. (4.O<y<l .E( Y)I) . For purposes of calculation. equation (4. E( Y)] is never negative. (4. The conditional density is not defined when lex) = o.15. y) == f". In the simplest casc. h( ylx) is a function of y alone. and so the covariance is positive. y) and zero elsewhere. the product [X . y) f(x) .1) The definition holds whether the variables arc discrete or continuous. is a bivariate generalization of the concept of variance..15. Example 4. and if Y < E( Y) whenever X < E( X). (4.14.2) As the name implies. we obtained the marginal probability of one variable by summing over all the values taken by the other. Y) = E(XY) - E(X)E(Y) . (4. Y: = f(x) r: l/J(x. y) dx dy . =x +y • O<x<l.1.67 COVARIANCE The distribution function is defined by P(X < x.15. It must be zero when both variables arc at -00 and one when they arc at +00.£( X)Jf Y . y) dy . COVARIANCE The covariance of two random variables. It is defined by Cov(X.14. evaluated at the value x: h(ylx) = l/J(x. it is no longer a variable.15. . Y < y) = F(x.3) It has to be a nondecreasing function of both x and y.4) The conditional density function for Y. given that X = x. Y) 7 7 = ~ . This fact does not make X and Y independent. is obtained by substituting x = 1/3 in equation (4.16. This gives h( y Ix =!)=2(J+3y) 3 5' The conditional density of Y when x = 1/2 is (2y + 1) 12. Le. the joint probability function is the product of the two marginals.15.5).68 CONTINUOUS VARIABLES The marginal density of X is (I 1 f(x) = Jo (x + y) dy = x + 2 ' which is similar to the marginal density of Y. It follows that V(X) 5 = 12 - and E(XY) = ( 7 12 JIJI () () )2 = 11 144 = V(Y) . x x 0 1) 7 + -2 dx = -12 = E(Y) ' and Jot 2 £(X ) = 2( x x+ 21) dx = 5 12 . 1 xy(x + y) dx dy = -3 ' whence Cov(X. . which is the same as the marginal distribution of Y. given that X == 1/3. The conditional density of Y. 0 4. INDEPENDENT VARIABLES In section 3.( 12)( 12) = - 1!4 . For independence. we have just seen that this is not true for x == 113. we need h(ylx) = f(y) for all values of x.14.. we said that two discrete random variables are independent if P'i == p. Then F(X) - = I I (..p J for each i and each j . ( + 1. Thus.'" 1-+: xycfJ(x. The lifetime of a certain component has an exponential distribution. their covariance is zero. Suppose that X takes the values -1.1. and +1. = 0. The marginal probabilities for Yare P-0 - - I 3 and P-1 == ~ .1. y) == f(x)g( y) . and = (-1)( + 1)( !) + (O)(O)(!) + (+ 1)( + l)(! ) = 0 . what percentage of the components will last for more than 400 hours? 4. + 1).16. we have E(XY) = 1-+.'" xyf(x)g(y) dx dy = f.1. in exercise 4.15. The variables are not independent because Pou ¥ Po-P-u.16.2..0). This is illustrated in example 4.16.1.1) holds. If (J = 1500 hours. If.0.. The sample space consists of the three points (-1. each with probability 1/3. y) dx dy == f:. (0. zero covariance may not imply independence.+: xf(x) dx I-+". When the condition of equation (4. It follows from equation (4.16. and that Y == X2.~ yg( y) dy = E(X)E(Y). Example 4. if X and Yare independent. + 1) with equal probabilities. When the variables are not normal. even though the 0 EXERCISES 4. Pou Po-P-o = 0)0) = ~.1 ) cb(x. Y) variables are not independent. However.2) that Cov(X. The two variables are independent if. and only if. a buyer wishes to have a 95% probability that a . The converse is true when both the variables are normally distributed. E(X) == 0 and E(XY) = ~. (4.'" 1-+".69 EXERCISES The corresponding statement for continuous random variables is this: Let f(x) and g( y) be the marginal density functions of X and Y respectively. 6.10. An engineer puts 100 lamps on test.70 CONTINUOUS VARIABLES component will last for 5000 hours. What percentage is still operating at the end of the warranty period? 4. 710 of the Jamps have failed. What fraction lasts for 10. Compute m3 and m 4 for the exponential distribution.000 hours? 4. what should the buyer specify the average life to be? 4. variance. Usc it to calculate the mean.000 hours? 4.9.5.7. Suppose that in exercise 4.) 4.3. A continuous variable has the density function f(x) = kx 2 1<x<3. They arc warranted for a life of only 2 years.4.2% of them last for SOOO hours. . and find the mean and variance. Calculate the constant. 63. What would the engineer estimate the value of o to be? 4. The life of a certain device has a Weibull distribution with {3 = 2. Suppose that the lifetime is exponentially distributed. Show that the moment-generating function for a Bernoulli variable is q + pet. but that we did know that 25% of the devices lasted until 90()O hours. Lifetimes of field windings for a generator have a Wei bull distribution with a = 12 years and {3 = 2. and m J • 4. we did not know {3. k.5.8.000 hours? 4. A device is designed to have a mean lifetime of 20. What is the median lifetime? What percentage fails in the first year (8760 hours)? What percentage lasts for 5 years? What should be the mean lifetime if we wish 90% of the devices to last for 40.8. What fraction lasts for 10. After 1000 hours of use.000 hours.11. (See exercise 4. Derive the following identities for obtaining central moments from moments about the origin: 4. k.m-I(1 . 0<y<2. E(X 2 ). Y).j --I 13 if its O<x<l. . +0): 1 lex) = 20 =0 -o<x<+O.15. Find the marginal densities of X and Y.3. and calculate E(X).14.12. and find E(X) and VeX). Calculate the constant. Derive the moment-generating function for it Poisson variable. Confirm that. It is stated in the text that the sum of two independent random variables that have uniform distributions on the interval (0. A variable has density function [(x) = k sin x O<X<1T.. 4.0. 4.I.)n--I dx x x = (m -1)!(" -1)! (m+ n. VeX). and VeX). has a uniform distribution on the interval (. have joint density cf>(x. 1) has the distribution of example 4. 4. X and Y. Calculate the constant. Calculate E(X).16. 13) Calculate the mean and variance of X. x).17.13.I( I - B(a. The Beta function is defined by B( In.71 EXERCISES 4. 4. and Cov(X.1. Two variables.. k. elsewhere. 4. n ) = LI() . y) == kx(x + y) O<x<l. A variable has a beta distribution with parameters a and density function is X [(x) = .1)1' . X. A random variable. The mean.1. however. It is. the median. John Copyright © 1990 by John Wiley & Sons. Actually.2. At about the same time. Inc CHAPTER FIVE The Normal Distribution 5.1) Its graph is the familiar bell curve shown in figure 5. is the cornerstone of statistical practice. or probability law. The curve is symmetric about the value z = 0. M. in his early nineteenth-century work with astronomical data.1. F. so there are. of course.Z2) v'21T exp 2 . THE NORMAL DENSITY FUNCTION The simplest form of the normal density function is f(z) = 1 (.2. 5. exp -2 (5.2. This distribution is sometimes called the Gaussian distribution after K. INTRODUCTION The normal distribution. Gauss.Statistical Methods in Engineering and Quality Assurance Peter W. The normal density function cannot be integrated by the ordinary methods of calculus. Abraham DeMoivre derived the normal distribution as early as 1718 as the limit of the binomial distribution as n becomes large.2) . arguments about priority between supporters of the two men. which is also the maximum value. (5. used it as a law for the propagation of errors. It is perhaps too extreme a simplification to say that it is called the normal distribution because it is the distribution that naturally occurring random variables normally have. and the mode of the distribution arc all at z = o. it is true that many random phenomena have probability distributions that are well approximated by the normal. who. Laplace published the same distribution in his book on probability theory.2. However. proved in advanced calculus texts that f +'" -oc 72 (/2) dt = v'21T . J._.(x) dx • where dz = dx /(T...-..2. We write X .N (J.J.. - -------0---- oL..1 ..... 0 . q.. The general form of the normal density is obtained by introducing the two parameters. (T Then.2.3 .l.t and that the scale is changed. >< 0.- ~ .J__L_L.1.l_J__'__L..2 8 0.(x) looks exactly like figure 5.tl] q.t and (T2. V(X) = (T .I.73 THE NORMAL DENSfIY FUNCflON - 0. writing dP for the clement of probability._l. and setting x-J.. The density function for X is 1 [-(x-J.... and variance. Z is said to have .: J(z) dz ::... It is the standard deviation of X.1.. or expectation..t z=--. (5. we have dP ::. E(X) = J._J -5 -3 3 -1 5 x Figure S...t..iii c 0.(x) = (Tv"fiT exp 2(T2 .-... - -------~--------I I I I ~ . We show later that the farameter (T is the scale factor....l_. Probability density function for a normal distribution.2.l__L_L~.l_~Iio. V(X) = (T .-.l_.l_J__'__l.3) The graph of q... except that it is centered at x = J. (T2) to denote that X is normalll distributed with mean.t.. 3. From this table. THE TABLES OF THE NORMAl. J.N(89.1.1. We see that 99% of the population lies within the 2. we see that the values of F(z) change from 0.NCO. we can. 0'2).9750 to 0. 96U' :::.L)/a . From the symmetry of the curve.9900 0. (5. 1). By writing X == J.L. we also have P(Z < -1. 1).9750.00). Entering the table at z == 1.9974 0. x :::.OO5U 0. We have seen that if X .2) If we enter the table at z = 2. 2.L + 1.00 2.96)=0. one observation in 800 on the high side. i.9544 0.9()00 0.N(J.0047 0.L ± 3U'.3.575 2. this translates to the following two statements that are each true with probability 0.e.9906 0. within a distance ±2CT of the mean. Probabilities for Special Values of z I.96 2. P(X> J. so that P(Z> 1.96).96. Several other useful values of z are listed in table 5..0). and one in 800 on the low side. 96.0013 . a standard normal variable will lie in the interval (-1.1.1.0250 and 0.96.L .0.96) == 0.96:S Z:s 1. Z> +1.025 and P(-1.0 < X < 98. Let X .96U') = P(X < (5. We write Z .95.0500 0.96.5).9500 0.3.74 THE NORMAL DISTRIBUTION a standard normal distribution or to be a standard normal variable.5).L + ZO'.3.N(O.0228 O.96. Example 5. find values of F(x). Find (a) P(x > 93.1584 0. The change is small enough for liS to take as a rule of thumb that 95% of the members of a normal population lies within the two sigma limits.60 J. and 2. Table 5. 36. we see that F(1. and (c) P(S6.025.3.6R32 0. Z < -1.5% of the time it will be in the upper tail.95: J. 96a .9554.O(l z P(Z > z) 0. given E(X) and VeX).6 sigma limits.1. DISTRIBUTION Table I (see Appendix) lists values of F(z).L + 1.96) = 0.5% of the time in the lower tail.L .645 l.3.0250 0.J. For 95% of the time.0228 and 0. then (X . 5. Only one observation in 400 lies outside the bounds set by J.(XI 1.96)=0.1) J.0 rather than 1.96a) == 0.025. (b) P(X < 81. + 1. When X = 93. E(X) = uE(Z) + p.1056. the variance.z) .5) = P(Z > 0. Z = (93. wc have fez) = f(.F( -0.5 .0.0. THE MOMENTS OF THE NORMAL DISTRIBUTION In this section. It follows that the third central moment of X is also zero. Z = -0. Our procedure is to calculate the moments for the standard normal distribution. (c) When X = 86.5. (b) When X=R1. and the third and fourth central moments of the normal distribution.5. Since the standard normal density function is symmetric about the line z = 0.4. P(Z< -1.89. we calculate thc expectation.6247.0=-1. We then argue that since X = uZ + p.8944 = 0.0.1 + 1'(0. A similar argument shows that all the odd moments of Z are zero.5) . when X = 98.5.5) = 1'( 1. the standard deviation is 6.75 THE MOMENTS OF THE NORMAL DISTRIBUTION (a) The mean of X is 89.F(1.0.75) = 1 .25. so that (-z)f(-z) = -(z)f(z) and foo zf(z) dz = - r'~ zf(z) dz. Z = + 1. and the rth central moment of X is (rr times the corresponding moment of Z. Hence.0. We need to calculate F( 1. . and E(X) = p.6915 = 0.5) = 0.2266.0.9332 . From symmetry.25) = 1 .75) = 1 F(0. Z=-7.7734 = 0.0) 16.5.25) = 1 ..25) = P(Z> + 1.. 5.75. and so P(X> 93.I + 0. E(Z) = L: zf(z) dz = f~ zf(z) liz + Lh z[(z) dz = 0.5/6.0 = 0.5) . Application of L'Hopital's rule shows that the first term on the right side is zero. The skewness is measured by g_.76 THE NORMAL DISTRIBUTION Since E(Z) = 0.3' c. THE MOMENT ·GENERATING FUNCTION The results of the previous section can also be obtained from the momentgenerating function. which is M*(t) = M(t) exp(--.6. Thus.z exp -. V(Z) = 1 and A similar argument shows that the fourth momcnt of Z is m 4 == 3. V(Z) = f~."" z2f(z) dz = r: z[zf(z)] dz .1) We actually want the generating function for the central moments.r 2t ) 2 .5.r • For a symmetric density function. m. 5. _00 _00 fez) dz . Thc second tcrm is unity. meaning that the curve f(x) is symmetric .t + c. SKEWNESS AND KURTOSIS Scale-free versions of the third and fourth moments are used as measures of the shape of density functions. and the 4 fourth moment of X is 3c. It can be shown that the moment-generating function for the normal distribution is 2 M(t) == exp ( fl. Differentiating M*(t) as many times as necessary and putting t = 0 gives us the desired central moments. By integrating by parts. (5. ] [ (2)]+00 + f+"" V(Z) = 21T . = ..t) == exp ( 2 2t ) (T 2 .r • 5.fl.5. They subtract 3 from the kurtosis and call the difference the excess. One of the most important statistics we use is the mean. it is possible for two densities to have the same mean and variance and still differ markedly in shape.8. and moment-generating functions M. Suppose that X has a uniform distribution over the interval (. The kurtosis of a density function is measured by This a measure of the peakedness of the density curve. Let Xl and X 2 be two independent random variables with density functions fl(X) and f2(X). In the next three sections. We now consider how the sum of two independent random variables behaves. which is their sum divided by fl. The even moments arc (]2 V(X) = 3 and If (J = 0.10. which is less than the 0 kurtosis of the more widely spread normal distribution. It is symmetric about the line X = E(X) = O. Although the standard deviation is a measure of the spread of a density curve. of a random sample of observations. we introduce the important central limit theorem. The kurtosis of the normal distribution is m 4 (Z) = 3. The normal distribution has tails going off to infinity. Theorem 5.1 . The odd moments arc zero.8. They clearly do not have the same shape.6. SUMS OF INDEPENDENT RANDOM VARIABLES In the next chapter. Let U = XI + Xl' Denote the momentgenerating function of U by M(t). 8). we begin our investigation of statistical procedures. the odd central moments are all zero. or average. The difference is manifested in the kurtosis. The graph of the uniform distribution is a rectangle. The kurtosis of the uniform distribution is 9/5 = 1. Example 5. respectively.SUMS OF INDEPENDENT RANDOM VARIABLES 77 about the line x = E(X). In section 5.7. 5. we discuss the behavior of sums of random variables.1. and so the skewness is zero. this density has the same mean and the same variance as the standard normal distribution.7. Some use the normal distribution as a reference distribution. respectively.(t) and M 2 (t). The result of this section is expressed as a theorem. The density is 1120. ) ""' L. + tX z )!\(X\)!2(X 2 ) dx\ dX 2 exp(tx\)!. Furthermore. (5.5) . we can show that if U is the sum of several independent random variables.) dX I fY~ exp(tx 2 )!2 x : dX 2 = MI(t)Mz(t) .(O)M.3) The generalizations of equations (5. = W.78 THE NORMAL DISTRIBUTION We wish to find E(tx[ + tx 2 ).3) are that the expectation of the slim of independent random variables is the sum of their expectations.J w.(O)Mz(O) + M[(O)M~(O) = £(X\) + qx2 ) • (5.7..(O) + M[(O)M~(O) = E(X.) and (5. we differentiate M(t) with respect to t and equate t to zero. To find E( U). (5. their joint density function is f~(XI)f~(X2)' and so M(t) = = r:' r: exp(tx. V(W) .7.l-t. if WI' W 2 . ". • arc constants. its moment-generating function is the product of the individual moment-generating functions.7.) w. E(U 2 ) = M~(O)M2(O) + 2M. ? (T~ • W is said to be a linear combination of the random variables..(x.) + 2E(X\)E(Xz ) + E(X. Because Xl and X 2 are independent.E(X.V(X.7.1) o More generally.2) and (5.7. .7. (5.7. . and then = 2: E(W) = 2. Then £(U) = M.4) and 2 W. and the variance of the sum is the sum of the variances. The details of the generalization are left to the reader.2) Similarly. Nor do we just set up a booth outside the engineering building or the mathematics department. but because = = + 1. There are numerous examples. and its precision (the reciprocal of the variance) improves as we take more observations. Basically.5) become E(W) = 2: w.X z .8.5).4) and (5. we have WI = + 1 and W 2 :::: -1. Every member of the population must have the same chance of being included in the sample. If W == Xl . especially in political polling.4) and (5. Substituting in equations (5. . we obtain E(i) == E(X) (5. 5. but had biases built into them.8. fraternities. because that will bias the sample against students who take their lunch in residence halls. (5. It is not easy to obtain a random sample.4) and . . Independence is essential.E(X) (5. (5.7.V(X) . Then the formulas of equations (5.7. we do not just go into the student union at lunch time.xn ' We can treat them as realizations of n independent random variables. and sororities. in which the pollsters reached wrong conclusions because their samples were not random.2) and YeW) = 2: If we substitute 1/ n for w.8. wi w. RANDOM SAMPLES A random sample of n observations from a population is a set of n independent observations. we see that the expectation of the difference is the difference between the individual expectations. hecause that will bias the sample against the arts students.. Suppose that we do have a random sample of n observations: xi' . in the formulas of the previous section. n These equations show that the sample mean provides a good estimate of E(X) inasmuch as. We say more about this in the next chapter.1) w. on the average.8.3) V(i) = VeX) . If we want a random sample of the student population. each of which has the same probability distribution.7.7. the variance of the difference is the swn of the individual variances.S. i = E(X). it is a "fair" sample that is typical of the population being sampled.79 RANDOM SAMPLES An important special case involves the difference between two random variables.. In particular. which is the logarithm of a normal moment-generating function. InIM(t») = J-Ll + crf 2 . How large is large enough depends on the shape of the parent distribution. 5.2) which has the same form as equation (S. we take equation (5. One approach is to look at the logarithm of M(t) as a power series and to note that the coefficient of t' contains a multiplier.2). The proof of the central limit theorem takes us further into mathematical theory than we want to go in this book. the higher-order terms tend to zero.I). converges to the form of equation (5.J-L.10. The importance of the central limit theorem is that we can quite reasonably act as if the mean of a random sample of observations is normally distrihuted. We have shown that if W is a linear combination of independent normal variables. It follows from the results of section 5. the distribution of the sample mean approaches normality for large enough n. the sample mean has a normal distribution.2 cr. then W too is a normal variable. the distribution of the mean is well approximated by the normal when n is as few as six. We lean heavily on this in our applications.9.9. the moment-generating function of wX is M(wt). What if the variables do not have a normal distribution'? The central limit theorem tells us that for all practical purposes. are all normal variables.9. THE CENTRAL LIMIT THEOREM We saw in the last section that the mean of a sample of independent normal variables itself has a normal distribution. .9.1) If the X.1) and add the exponents to get M(t) " = exp ( L. as given in equation (S.t + " L. and we arc left with r.C) 2 ' (5.80 THE NORMAL DISTRIBUTION 5.J " w.7 that the moment-generating function of W is (S. no matter what the parent distribution of the population may be.5. M(t).J w. a larger sample is needed. For a uniform distribution.9. (1/ n As n increases. For a skewed distribution such as the exponential distribution. It consists essentially of showing that as n increases.S.1). LINEAR COMBINATIONS OF NORMAL VARIABLES If X is a random variable with moment-generating function M(l). 21)" 1 except that the factor I/O' = (1 .11. and we can use this property to derive the density function.4) Fortunately. we see that W has a gamma distribution with parameters nl2 and 2.z2(l - eXPL 2 2t)] dz.10. Table IV (see Appendix) gives tabulated values of the chi-square distribution for various values of degrees of fr~edom.11.z + tz 2 dz ~~ f I '" = \721T . we shall have no occasion in this book to use this formula or to quote it again! . The random variable W is said to have a chi-square distribution with 11 degrees of freedom..2t)1 /2 is missing.81 THE CHI-SQUARE DISTRIBUTION 5.11.f.:::o. the value of the integral is M(t)=(\ -2t)-1!2. The distribution of W becomes important when we investigate the behavior of the variance of a sample from a normal distribution in the next chapter. It follows that the moment-generating function of W is (5. Hence. The moment-generating function of Z2 is M(t) = E[exp(z2t )] = -1- v'21T 1 (2 ) f+~ exp .3 that the moment-generating function for a gamma variable with parameters a and f3 is (5.3) with Mw(t).3) Comparing equation (5.2) We recall from example 4. THE CHI-SQUARE DISTRIBUTION Let ZI' Z2"'" Z" be independent standard normal variables and let W = E Z.11.11.1) This has the same form as the integral for a normal distribution with zero expectation and variance (1 . (5. We can now write the density function of W as w. (5. The chi-square distribution is a special case of the gamma distribution that was introduced in section 4. '" 2 r . The expression "degrees of freedom" is usually abbreviated as d.13.11. of chi-square with 2n d.f. Example 5.' The moment-generating function of 2YI6 is obtained by replacing It is 1 by 2t10.80.f. We showed in example 4. xn be exponential variables with the same parameter.g. THE LOGNORMAL DISTRIBUTION The lognormal distribution is another useful probability model for the lifetimes of electronic devices..1. What is the probability that the supply of ten lamps will last for a year (8760 hours)? P(Y> 8760) = p( ~~ > 25... 2YI6 has a chi-square distribution with 2n d.03) . . Let Y be the sum of n independent exponential variables with the same fJ.0 is 0. We sum up the result of the previous paragraph as follows: Let xI' . . There is a 20% probability that the engineer's ration of lamps will last for a year. The model assumes that In( Y) is normally distributed with mean IL and variance u 2• The probability density function for Y is _ 1 1 [' [In(y) .f.. 0 5.e x p y uv'2-iT . 0. An engineer has a supply of ten lamps whose lifetimes are exponentially distributed with an expected life of 700 hours. and let Y= E Xl' Then. The moment-generating function of Y is 1 My(t) = (l-fJ/).11. As soon as one lamp fails. a new one is plugged in. Entering the chi-square table with 20 d.12.. Let Y denote the lifetime of a device.01 .IL]2] ji( Y) . we see that the probability of a value less than 25. (1-2/)'" which we recognize as the m.. 2u 2 .2 that the momentgenerating function of an exponential variable is 1 M(t) = 1..82 THE NORMAL DISTRIBUTION An important consequence of the mathematical result that we have just obtained is that we can establish a connection between the exponential and chi-square distributions.13.f. 1. The average value of the eighteenth-oruer statistic from a random sample of 20 observations from a standard normal population is 1.1 lists a random sample of 20 observations from a standard normal distribution and their normal scores. From the normal tables.13.L is the expected In(life).13.13. Example 5.12. The median of the lognormal distribution is y = exp( J. A similar result can be obtained by using normal scores. only 4% of the devices fail 60o before 10. then the distribution is not normal.13.000 hours. 0. 0. +(0) along the horizontal axis.605 and z = 4. This is the familiar ogive curve.000 hours? In( 100) = 4.00 0.L = 6.1. The lifetime of an electronic device (in hundreds of hours) has a lognormal distribution with J.2. It is called the eighteenth-order statistic for the sample. only two observations exceed it. and u is the variance of the In(life).0 and u:. hence. The 11rst observation. THE CUMULATIVE NORMAL DISTRIBUTION CURVE AND NORMAL SCORES The graph of the cumulative normal distribution is shown in figure 5.L) • The expectation is E(Y) = exp( J. in order of size. 0 5.800. its normal score is -1. F( -1.800 = -1 744 .59.4 = 60. the 6 avcrage lifetime is 100e .605 - 6. It plots the cumulative probability along the vertical axis and X( -00. It is shown as figure 5. Similarly. Normal probability paper changes the scale on the vertical axis and converts the ogive into a straight line.15. That is the normal score of the first observation in the sample. We can turn this around and argue that if we plot the cumulative distribution curve for some distribution on normal probability paper and the plot is not a straight line. The median lifetime is lOOe = 40. -1. or lazy S.200 hours.13. the fourth observation.13. .744) = 4%.L + ~2) . What percentage of the devices fails before 10. is the eighteenth member of the sample.THE CUMULATIVE NORMAL DISTRIBUTION CURVE AND NORMAL SCORES 83 2 where J.. The plot of the normal scores (Y) against the observations (X) should be a straight line. is the third-order statistic. We sce it again in chapters seven and eleven.300 hours. Table 5. and they do not.45 -0.52 -{UO -0. '--~---------T-------- ·5 ·3 -1 5 3 x Figure 5.13 -0.45 . Cumulative normal distribution function.96 ·-0.99 -0.19 -0.19 -1. Hi 0..92 0.13.13.74 -1.2 shows a sample of 20 observations from an exponential population with 8 = 100 and their normal scores.46 0. table 5.84 THE NORMAL DISTRIBUTION .06 0.13 0.32 -0.80 ··1. It is clearly not a straight line! Table 5. .3.0. - I I I 0. On the other hand. I 0.74 0.. 0.40 0..13 -0.13.92 -U.59 -0.87 -0.31 -1. Twenty Normal Observations (Po = 0.21 U' = 1.1.06 0.40 0.8 I -------~---------+-------.1.36 -0.59 -0. The plot of normal scores against the observations is shown in figure 5.52 0.-I . I I .87 1.31 1.97 -0.20 0.0) -0.11 -1.13. One would not expect the order statistics for an exponential distribution to behave like order statistics from a normal population..07 -1.2 .33 1. I I I - .. I I i 0..4.18 Normal Scores for the Sample 1.15 0.59 0. 00 -0.10 -1. 300 .. Normal score plot for a normal sample.40 Ohservalions fo'igure 5.70 1.0 - ** ** * * -1.2.70 0.13..40 0.2 * * * ~r-------~------~------~-------+------~----C11 -2.2 * * * ~r-------4--------+--------~------4--------+----C33 o 60 120 180 240 Ohscrvaliolls Figure 5.2 * . * 0.2 * ** * * ** *'" 0. Scores * * 1.85 THE CUMULATIVE NORMAL DISTRIBUTION CURVE AND NORMAL SCORES Scores * 1.3.13.0 2 * * * ** '" * ** * * -1. Normal score plot for an exponential sample. (b) Y < 74.5.4.8 < Y < 77. The diameters of the rods arc normally distributed with JL .3.0.87 -0. A manufacturer makes rods and sockets.8 cm and (T = 5 mm. What is the probability that the rod will go into the socket? 5.5.: 1. An oil company has a contract to sell grease in 500-g containers.13. the diameters of the sockets arc normal with JL :::= 1.59 69 93 Normal Scores -0.92 0.31 -0.19 -0. (c) Y < 69.5 < x < 20. The quantities of grease that the filling machine puts in the containers are normally distributed with the mean at the designated value and (T . A machine makes metal rods.87 -0.1.13 1.2. The lengths are normally distributed with JL = 19.06 . Find the probabilities of the following events: (a) Y> 75.4 are metal cans.2.40 0.2. 5. The amount of product made per hour by a chemical plant is normally distributed with JL = 72. To what value should the filling machine be set if the company docs not wish to have more than 2% of the containers rejected as below specified weight? 5.74 1.31 0.1.19 -1. An engineer picks a rod and a socket at random. The containers in exercise 5. Twenty Exponential Observations (8 24 248 130 76 2 56 86 86 17 1 60 53 =100) 146 80 10 3]4 49 90 -0. The specifications call for the length of a rod to be in the interval 19. Suppose that the lifetimes are normally distributed with JL = 1200 and (T = 400.92 -1.0 kg and (T = 2.05 em and (T. Their empty weights arc normally distributed with JL = 90 g and (T . Ten of those devices are put on test and their lifetimes are recorded.45 0. : 8 g.40 -0.1 < Y < 69.5 mm.0 cm and (T.1.3. The inspector weighs the cans of grease and expects them to weigh at least 590 g.59 0.6.: 2 mm.13 0. and (e) 68.06 1. (d) 75. A device is designed to have an average lifetime of 1200 hours.5. To what nominal weight should the filling machine be set so that only 2% of the cans of grease fail inspection? 5.86 THE NORMAL DISTRIBUTION Table 5.: 1. : 25 g.5 kg. What percentage of the rods meets specifications? 5.74 0.45 EXERCISES 5. Let Y denote the amount. . both of which have circular cross sections. 700 and u = 0. This verifies that as the number of degrees of freedom increases. and (c) 1320 < Y < 1400. 5. g3 tends to zero and value of the kurtosis converges to 3. What percentage of the devices operates longer than 10.000 miles? 5.6. g4' for the exponential distribution. 5. Calculate the skewness.90? 5.9. Repeat exercise 5. as n increases. Show that as parameter a increases. Calculate the skewness.6 assuming that the lifetimes have exponential distributions with () = 1200. Hint: You should use the following formulas (see exercise 4. equation (5. The life of a device measured in hours has a lognormal distribution with IL = 8.-3IL 4 1114 - 5. the shape of the chi-squared curve becomes closer to the nonnal curve. The moment-generating function of a binomial variable is M(l) = (q + pet)" .4.000 hours? What are the median lifetime and the average lifetime? 5. g3' and the kurtosis.9): m3 = m~ - 3m~1L + 21L3 . It also confinns that we are justified in using the central limit theorem to claim that.10.87 EXERCISES Let Y denote the average lifetime of the devices that were tested. the average of n exponential variables tends to behave like a nonnal variable. Find the probabilities that (a) Y> 1300. g4' for the gamma distribution.65 and (T = 0.5. The lifetimes of locomotive controls (in thousands of miles) have a lognonnal distribution with IL = 5.8. g3' and the kurtosis.12. 5.11.411l31L ' +6 m21L I 2 . What is the median lifetime? What percentage of the controls lasts more than 500. (b) Y < 1150. Derive the moment-generating function for the normal distribution. Calculate the third moment about the mean: m3 = npq(q - p).7. . What are the first and third quartiles of the average lifetime? How many devices would you have to test if you wanted P( Y > 1100) z= 0.1).13.m 4. . In each experiment. we take a sample of n observations. In this chapter. an inspection is made. We wish to estimate from the data the percentage of defectives being produced. A geometer would regard that vector as a point in an n-dimensional sample space.Statistical Methods in Engineering and Quality Assurance Peter W. and five can be regarded as a short course in probability theory. Each component is tested and classified as either defective or nondefective.X"' Alternatively. 88 .1. A sample is taken from each batch of components. and normal distributions. In each experiment. The data are a set of n numbers that can be regarded as the realizations of n independent random variables. we can write them in a column and denote the column by the vector x. Before· they leave the plant. or statistical investigations. designed to be the foundation for the statistical applications that are the main purpose of this book. J1.1. and (1 in the normal. They can be written as XI' X 2 .2. THE THREE EXPERIMENTS 6. we select an appropriate mathematical model: the experiments are chosen to illustrate the binomial. INTRODUCTION Chapters three.2. exponential. we turn our attention to applications by considering three basic experiments. John Copyright © 1990 by John Wiley & Sons. Experiment One We have a process for producing electronic components. 0 in the exponential. or distribution. M. These models involve unknown parameters that have to be estimated from the data: p in the binomial. four. x J " " . 6. each of which comes from the same population. We consider both point and interval estimates for the parameters. Inc CHAPTERSIX Some Standard Statistical Procedures 6. and make our first venture into testing hypotheses. 3.3. This does not say that the estimate will be exactly equal to p for every sample. They are put on test and their lifetimes are recorded. the estimate that we have chosen is correct. or recipe for obtaining an estimate. Experiment Two A sample of n light bulbs is taken from a production line. with probability of success p that is the same for all n trials. An estimator. We wish to find the average lifetime of the lamps being produced on the line that day.2. Let X he the total numher of defectives in the sample.89 THE SECOND EXPERIMENT 6. that has the property that it gives. A reasonable estimate is X p= --. a hypergeometric model would be correct. day in.2. It does say that if we follow this procedure. On the average. the binomial model is adequate.4. day out.1) .) We have to estimate p from the data.4. Then X has a binomial distribution with parameters n = 100 and p. with the batch size 50 times the sample size. on the average. XI is the (6. the correct value is said to be an unbiased estimator. Experiment Three We have a process for producing resistors with a nominal resistance of 90 ohms. so that E(P) = npln = p. which is unknown. n A (6. An appropriate model for XI is that of a Bernoulli trial. 6.3. 6. The data vector is a sequence of 100 zeros or ones. We wish to estimate the average resistance of resistors being produced by the process that day. THE FIRST EXPERIMENT Suppose that the batch that was being inspected had 5000 components and that the sample size was n = 100. (It can be argued that since we arc not talking about sampling with replacement. An appropriate model for exponential model with density function X 2:0. THE SECOND EXPERIMENT Let XI be the lifetime of the ith bulb. but. 6. we will get the correct value. on the average. XI denotes the state of the ith component tested. Unfortunately. It takes the value one if the component is defective and zero if the component is nondefective.1) We recall from chapter three that E(X) = np.2. the resistance of the resistors varies from item to item. X. In the last chapter.9 gives no indication of how precise that estimate might be. (J=--=i. The simple statement that our estimate is. we discussed the probability distribution of i at some length.5. we have estimated a parameter.L. 2: x.2) 6. 6.5. 6. n (6. the average resistance of a large batch of resistors. J. the average lifetime of the light bulbs.L. say.7.1) We discuss estimating the variance from the data later in this chapter. . (6. ( 2). We recall that in the normal case. to be estimated from the data. we .4. THE BEHAVIOR OF THE SAMPLE AVERAGE In each of the three experiments. we can use the central limit theorem and act as if i were normally distributed with expectation p and variance pq / n in the first experiment and with expectation (J and variance 0 2 / n in the second experiment.6. A reasonable estimate . If we were reasonably sure that the true value of the parameter would lie within 0.L and (F2.N( J. CONFIDENCE INTERVALS FOR THE NORMAL DISTRIBUTION The estimate. by the sample average: the average number of defectives per component. Let us assume that the resista~ces are normally distributed. i. 77. with E(i) = E(X) and V(i) = V(X)/n. i itself has a normal distribution.L. In the other two experiments. or value.1 of our estimate. p. 8.L z::::-- u/Vii is a standard normal variable. Since E(X) = J. it is reasonable to estimate it by il = i. and so X-J. of E(X) is a point estimate in the sense that it is a single point. THE THIRD EXPERIMENT Let X denote the resistance of the ith item. Now we have two parameters. or J.90 SOME STANDARD STATISTICAL PROCEDURES We want to estimate the parameter is (J from the data. 1). Neyman and E. They were introduced by J. day after day.L would be outside the interval on the high side.250. if we thought that the true value was quite likely to be anywhere in the interval .500/2 = 0.965 Z 5 + 1.5% of the cases. We start by obtaining a 95% confidence interval for the mean of a normal distribution.5%. we should be less comfortable. (6.96. On the other hand.7.1): .96. The average length of the four rods in the sample is 29.i ± 10.0.7. we develop a class of interval estimates known as confidence intervals. obtain 90%. 77.L .0. Assuming that the lengths are normally distributed with standard deviation 0.575 cm. Example 6. Let Z be a standard normal variable. are examples of interval estimates. x-J. in the long run.L . S. We have 95% confidence in the interval because it was deduced from a statement that is correct with probability 0.x -1. we were to take samples of size n from this normal population and compute the interval given by equation (6.7. The following statement is true with probability 0. in the remaining 2. J. Pearson. 95% of the intervals would indeed contain J.1) Equation (6. J. 96 (T _ 1. The two estimates. If. 95%. and multiplying by -1.CONFIDENCE INTERVALS FOR THE NORMAL VISTRIBUTION 91 might feel that we had a good estimator. In this section.L S x+ Vii .0 cm in length.96(T Vii 5J. Hence. A sample of four rods is taken from a process that produces rods about 30.9 ± 10.95: -1.7. We have (T/m = 0.L would be outside the interval on the low side.965 a/Vii Then and _ x- 1.1 and 77.96.96 5 aIv7i S + 1.L.95. J.500 em. The 95% confidence interval is given by equation (6.1) is a 95% confidence interval for J.7. 5 +1.9 ± 0.1.\.L. and 99% confidence intervals for the process mean. [n 2. 3) For 99% confidence. Shorter intervals can also be obtained by taking larger samples.575 + 1.L"$ 29.250cm? We wish to have 0.• i ± 0.822 COl.1.2. In some problems. how large a sample should the engineer take to have a 95% confidence interval no longer than 0.7.36 .645 for 1.e .219. 96 ( 0. z ~ -1. a shorter interval can be obtained with a decrease in confidence. we substitute 1. the length of the interval increases to 1.92 SOME STANDARD STATISTICAL PROCEDURES 29.500 cm. The 90% interval is shorter. (6. A sample of at least 16 observations is needed.96(0.575: 28.288 cm.92 . (6.085:-5 J. D Example 6. Then .L"$ 30.4) The length of the 95% interval is 0.2) For the 90% interval.96 ( Vii :s 0.250)"$ J. we substitute 2.250) 29.645.96(0.250 so that Vii '? n'? 0.1) is a two-sided interval.96 in e4uation (6. It sets both upper and lower bounds to the value of E(X). D The interval dell ned by equation (6.7.7.065. 0.7. (6. On the other hand.7. In example 6.7.L 50 30.L :s 29.7. 15.98 cm.250 = 3.1. The cost of increased confidence is a longer interval. i. if we are not content with an interval in which we have only 95% confidence and call for 99% confidencc. one-sided intervals are more appropriate.95.1): 29.575 .931"$ J.500) 1.164 :s J.986 .500) 1. We can start with this statement: With probability 0. so be it. 2.6) Intervals like those of equation (6.57.82. 645 CT Ii$X+-Yri .645.15.95.645CT Vii .58. The choice between using a one-sided or a two-sided interval depends on the application.7. 2.98. 2. Find an upper 95% confidence limit for the amount of butenes present. A purchaser of acid might specify that the vendor deliver acid whose strength is at least 60%. + 1. 2. We wish to be sure that the amount of contaminant present does not exceed a specified level. In case of doubt.745 and (T = 0. an impurity. The value of z that corresponds to a probability of 5% in one tail is z"'" 1. use two-sided intervals.. if the vendor wants to send him 65% strength at the same price.5) would be used if the engineer wants to be confident that Ii exceeds some lower bound and is not concerned that the value of Ii might be too high. We arc not concerned about taking action if the water is purer than required. we have i == 2.7. [J . 2. Intervals Ii ke those of equation (6. A good working rule is to use two-sided intervals unless the situation clearly calls for a one-sided interval. Ii < 2.5) Alternatively.73. in a crude oil shipment. 2. For this data.645.7. If it docs. 2. we take some action. Z $ whence _ 1. We say more about this when we discuss hypothesis testing in the next chapter. The data are 2.7.X- 1.15)' vlo ' i.68.84. 2. Sometimes it is obvious that the situation calls for a one-sided interval.!B . (6.6) are used if the engineer wants assurance that Ii is not too high.92.7.645 (0.c. Ten measurements are made of the percentage of butenes.59.74. (6. One example might be water quality. The upper confidence interval is given by Ii __ <x + 1. X could be the percentage of some undesirable contaminant. Example 6. 2.3.93 CONFIDENCE INTERVALS FOR THE NORMAL DISTRIBUTION and _ Ii?:. we can start with this statement: With probability 0. 28.80.8)/HlO '5 P '50. and. p = 20/100 = 0.2411000:$ p:$ 0.xln for q. Then equation (6.20 .24 11000 .1.7.94 SOME STANDARD STATISTICAL PROCEDURES 6.1.96V(0.96V pqln .3 has 20 defectives and 80 nondefectives. If p = 0. this leads to the interval x x .60 .8. Following equation (6. 0.1) There is an obvious problem.8. Find a 95% confidence interval for the fraction of defectives in the whole batch. the required sample size increases. The best solution to this difficulty is to substitute our point estimates for the unknown quantities: p = xln for p.57:5 P :5 0.20.1. following the development in section 6.5 nearly 100 years later. This was the first version of the central limit theorem. we substitute prior estimates of p and q. They showed that for large enough values of II. the binomial distribution approached the normal distribution in the limit as 11 became large.4) .+ 1. De Moivre showed that for a fair coin. the confidence interval would have been 0. the number of successes in n trials. q = 0..3) If the sample size had been 1000 with 600 defectives.2) Example 6. 0. CONFIDENCE INTERVALS FOR THE BINOMIAL DISTRIBUTION As early as 1718. The ends of the interval involve pq. Suppose that the sample of 100 electronic components in the experiment in section 6.96(0'()4) 0. How large is "large enough"? A useful rule of thumb is to choose 11 large enough for np > 5. Laplace generalized De Moivre's results to values of p other than p = 0.60 + 1.8. We may assume that X is normally distributed with E(X) = np and VeX) = npq.8. and if = 1 .J. and we do not know p. the normal approximation will be adequate for a sample of only ten observations. (6.96V pqln '5 p '5 .8. and nq > 5.8.20 + 1. the distribution of X. (6.96Y pqln .1) becomes ~ .S.5.96v'0. This means that we can obtain a 95% confidence intcrval for p by writing z= np -x v'npq. n n (6. In these formulas.8.5.12:5 p :5 0. As (J moves farther away from 0.96\1'0.63 .2). is well approximated by a normal distribution. n n (6.1.2)(0.96Y/Jqln '5 p '5 ~ + 1. 5. That is a more lengthy procedure.96ypqln or p - p = ± 1.1).20 for becomes I03. this estimate would give the interval . Squaring both sides.1.289. In this example.20. whence p = 0.21 and pq = 0.5.8. where p is only 0. One is to take the upper and lower inequalities as quadratics and solve them for p.8. It is reasonable for quite a broad range of values.21.24. not on fi = 0. but it is moved to the right. The end points of the interval can be written as p = fi ± 1. The interval has approximately the same length as the interval that we obtained earlier in equation (6. when p ::= 0. 2 2 A2 P . o The other approach.8.84r/ . we have pq = 0. so that the equation + 4 =0 .95 CONFIDENCE INTERVALS FOR THE BINOMIAL DISTRIBUTION There arc two other ways of handling the difficulty with the ends of the interval in equation (6.(2nfi + 3.25. Even with p = n.133 or p::= 0. so that this procedure gives a confidence interval that is only 9% longer than that from equation (6. and to substitute this value for pq in equation (6.7.3 or 0.20.458. we substitute 100 for nand 0.43.3). This procedure is clearly conservative.pp + P A ::= 3·R4p(p-fi) n • When the terms arc rearranged. It is centered on 0.84)p2 .841' p. When p = 0. 96y pq / n .8.4 or 0.84)p + niJ 2 ::= O. pq = 0.6. and the degree of improvement in the estimate docs not really justify the effort. this equation becomes the following quadratic in 1': (n + :'\. this is a point against the method.1).8. is to note that the maximum value that pq can take is 0. which is useful when p is closer to 0.1). For example 6. 6.10$ P ~0.5) One hopes that this approximation will not be relevant to industrial situations in which p represents the percent defectives! It is a useful approximation for political polling.8. 0.17 • The opposition vote was split between two candidates. and few of the bottom teams have failed to win at least 30%.30. Since 1910. we recall that V(X) = 8 2• The formula for the 95% . The equation then simplifies to (6. replace 1. which is the model in the third experiment. The percentages of the votes obtained by the winning candidates in the presidential races from 1960-1984 are shown in table 6.57 50. which was introduced in chapter five. Finally.8.9.96 SOME STANDARD STATISTICAL PROCEDURES Table 6. this leads to a quick method of obtaining approximate confidence intervals.1). which is longer than the other two intervals. no pennant winner in the National League. Applying the central limit theorem to the exponential distribution.0 and pq by 1.5.34 43. says that for most densities. Presidential Election Results Year Candidate 1960 1964 1968 1972 1976 1980 1984 Kennedy Johnson Nixon Nixon Carter Reagan Reagan Percentage of Popular Vote 50.1. since it is rare for one candidate in a two-person race to receive more than 70% of the votes.09 61. the distribution of i approaches the normal distribution as n increases. section 6. The central limit theorem.96 by 2. and only four in the American League.I. CONFIDENCE INTERVALS FOR THE EXPONENTIAL DISTRIBUTION The binomial distribution is not the only distribution whose mean approaches normality for large n.56* 60. In equation (6.S. has won as many as 70% of the games that they played in the season.99* 59.91 50. Another example is found in baseball.8. L. 968 _ 1. If we know J.Vii sO:5x+ Vii .6:5 0 :s 301.10. and the -(1 +vn.L 2 + J.1) n If we do not know J. (6. an obvious estimator for the variance is the average of the square of the deviations of the observations.L)2] = E[X2 . i.L2 = £(X2) 2J. We substitute the point estimate confidence interval becomes x for 0 in each of the cnd points. change the denominator to n .e .: E(X2) -.L2 . x. 0 6.2J.10.L)2.97 ESTIMATING THE VARIANCE interval is _ I.1. Suppose that the average of n = 100 observations is The 95% confidence interval is = 252.4 .804) s 0 s 252(1.J. 1.LZ] ::.J.9.2) . we can substitute the point estimate. This change is justified in the next few paragraphs. wc deduce that V(X) = (}"2 = E[(X .2) Example 6.2J.196) 202.9.10. ESTIMATING THE VARIANCE We recall that V(X) = £(X .J.Lx + J. From the definition of the variance.%) -( l .1. A preferable method for small samples appears in section 6.96) sO:sx x vn i (6.1..13. We must.9.1 ) (6.L2 = E(X2) .L. 960 x. 252(0.LE(X) + J. whence (6. however. Although S has 11 terms.5) S is called the sum of squares of deviations (of the observations from their average).. the estimate of the variance is said to have fl . and S = 16 + 9 + 1 + 1 + 1 = 28 . The first" . It should be noted that the derivation of S2 in equation (6.2 - L x" 0 I 2x-L. Then .5) does not depend on the data being normally distributed. Example 6.J " ( X. . Estimate V(X). " + X-2) X. r. and it is easy to calculate the deviations by subtracting x from the observed values. Xi X = "L. The deviations arc -4. but the last deviation is dictated by the restriction that the sum is zero. +3.lO. 4 . and + l. 6. + I1X-2 -2 nx . A sample of five observations is 1.98 SOME STANDARD STATISTICAL PROCEDURES and (6. For that reason. For this sample. which leads to an estimate o S" S = .10.(). they are not independent because the deviations add up to zero.( S) =(n·· 1)0'2. i = 5.1 of them may take any values. = n( JL 2+ 0' 2) .3) Let ~> .n (2 (F2) JL +-. . - -)2" = L. -1. + I.10.1. (6.. We estimate the standard deviation of X by the square root of the estimated variance.I degrees of freedom. lOA) and so we estimate V(X) by 2 S S=--l' n- (6.. = 2 X.J (].10. 4.: 7 . 8. and 6.X- - X.J = L. 99 ESTIMATING THE VARIANCE That calculation worked smoothly because n is small and x is an integer. If x is not a "convenient" number, the squares of the deviations will not be so easy to calculate, and the method used before may introduce roundoff error. In practice, it is preferable to compute S by the formula S '\.' 2 =,L, X, - (6.10.6) n In this example, S .:=.. 1 + 64 + 36 + 16 + 36 - (25)2 T = 2R . o Estimate VeX) from the following sample of six obser- Example 6.10.2. vations: 17.021, 17.02g, 17.026, 17.024, 17'()26, 17.022. The estimate can be ohtained hy following the procedures of the previous example. It is easier to subtract 17.020 from each observation first; recall from equation (3.10.3) that VeX - b) = VeX). This is only a shift of the origin and does not change the spread of the data, which now become 0.001, 0.007, 0.006, O. ()()4 , 0.006, 0.002 . Multiplying by 1000 changes the data to 1, 7, 6, 4, 6, 2, which is an easier set of numhers to handle. Formally, we have made the transformation Z = 1000(X - 17.(20) with V(Z) = 10 6 V(X) . We estimate V(Z) by 2 50';: = (26f I + 49 + 36 + 16 + 36 + 4 - --6- = 30.33 if; = 6.07 . The estimate of VeX) is 6.07 x 10 6 o 100 SOME STANDARD STATISTICAL PROCEDURES 6.11. STUDENT'S t STATISTIC In section 6.7, we derived confidence intervals for the mean of a normal distribution when the variance is known. The intervals were obtained from the standard normal variable X-/-L z == u/v'fi . Usually, the standard deviation is unknown and has to be estimated from the data. We substitute s for u in the equation for z. The resulting statistic is denoted by x -. /-L (6.11.1) (=-- s/v'fi . This statistic docs not have a standard normal distribution because the known variance has been replaced by its estimate, which is itself a random variable. The properties of the t statistic were first investigated by W. S. Gosset, who was one of the first industrial statisticians. He worked as a chemist for the Guinness Brewing Company. Since Guinness would not allow him to publish his work, it appeared under the pen name of "Student." Thereafter, Gosset was known to statisticians as Student and his statistic was called Student's t. The graph of the density function of t looks much like the standard normal curve, except that it is more spread out. To obtain confidence intervals, we repeat the earlier derivation starting with t instead of z, and replacing 1.96 by t*, the corresponding tabulated value of t in Table II (see Appendix). The value of t* depends not only on the level of confidence desired, but also on the number of d.f. in the estimate of the variance. With infinite d.f., the density of t is the same as the standard normal. Example 6.11.1. In the third experiment mentioned at the beginning of this chapter, a sample of ten resistors was taken from the production line. The resistances of the items in ohms were 114.2, 91.9, 107.5, 89.1, 87.2, 87.6, 95.8, 98.4, 94.6, 85.4. Obtain a 95% confidence interval for the average resistance of the items being produced. A dotplot of the data appears as figure 6.11.1 and a boxplot as figure 6.11.2. • • • •• •• • • • -+!--------rl-------;I--------+I--------rl-------;I---- Cl '14.') 90.0 %.0 102.0 108.0 Figure 6.11.1. Dotplot for exampk 6.11.1. 114.0 CONFIDENCE INTERVALS FOR THE VARIANCE OF A NORMAL POPULATION --1 101 + -rl-------+I-------+I-------+I-------+I-------+I----Cl 84.0 90.0 96.0 102.0 108.0 114.0 Figure 6.11.2. Boxplot for example 6.11.1. Since there is no indication that a one-sided interval is needed, it is appropriate to use a two-sided interval. The estimates of the mean and standard deviation are x==95.17 and s ==9.36, respectively. The value of /* for a 95% interval based on 10 - I read from the table as 2.262. The confidence interval is Ci 9_. = 9 d.f. is 17 - (2.262)(9.36) < . 9S 17 (2.262)(9.36) -fLs .. + ViO ViO 88.47 S fL S 101.87 . Nowadays, one rarely has to carry out these tedious calculations by hand. The printout for obtaining this interval using Minitab is given in the following. The data were entered in column one of Minitab's work sheet. The command TINT 95 CI called for a 95% t interval for the data in Cl. SET in cJ 114.2 91.9 TINT 95 cJ N=10 107.5 MEAN = 89.1 87.2 87.6 95.8 98.4 95.170 ST.DEV. = 94.6 85.4 9.36 A 95.00 PERCENT C.1. FOR MU IS (88.4697, 101.8703) o 6.12. CONFIDENCE INTERVALS FOR THE VARIANCE OF A NORMAL POPULATION We showed in chapter five that the sum of the squares of independent standard normal variables has a chi-square distribution. Hence, has a chi-square distribution with n d.f. 102 SOME STANDARD STATISTICAL PROCEDURES 2 It can be shown that SIif also has a chi-square distribution, but with n - 1 d.f. This property can be used to obtain confidence intervals for the variance as follows. Example 6./2.1. Obtain a two-sided 95% confidence interval for the variance of the resistances in example 6.11. I. In this example, S = 789.141 and there are n - 1=9 d.f. The distribution of chi-square is not symmetric. We read from Table III (see Appendix) in the row for 9 d.f. that the 2.5% value for chi-square is 2.700 and the 97.5% value is 19.023, i.e., with probability 0.025 chi-square will be less than 2.700. Then, with probability 0.95, we have S 2.700$ -2 S 19.023. (T Inverting this inequality leads to the desired confidence interval S 2 S ---<if'<-19.023 - 2.700 . Substituting 789.141 for S gives 41.48 $ if2 $ 292.27 and 6.44::: if::: 17.10. Our point estimate of the variance is S2 = 789.141/9 = 87.68. The confidence interval is not symmetric about this value. One should also note that the interval is quite wide. It stretches from 47% of the point estimate to 333%. Admittedly, there are only ten data points, but this docs indicate that we need considerably more data if we want to get a really reliable estimate of variance. 0 6.13. CHI-SQUARE AND EXPONENTIAL VARIABLES We pointed out in sections 4.10 and 5.11 that the exponential and chi-square distributions are both special cases of the gamma family of distributions. We show in this section how the chi-square tables can be used to obtain confidence intervals for the parameter of an exponential distribution. The method depends on the fact that if we have a random sample of n exponential variahles, Y = 2 E x,l9 = 2nil9 will have a chi-square distribution with 2n d.f. CHI-SQUARE AND EXPONENTIAL VARIABLES 103 Example 6.13.1. Suppose that only n = 10 lamps had been tested in the second experiment, and that i = 252, as before. Obtain a two-sided 95% confidence interval for H. In this example, 2 E XI = 5040. We read values of chi-square with 20 d.f. and obtain, with probability 0.95, 9.591 ~ 5040 -0--:<;; 34.170, whence 147.5 ~ 0::5 525.5. [J The method that we used in section 6.9 gives a different inverval, 95.8::5 H s 408.2. For small n, the method of this section is preferable because it is hased on the exact distribution of the statistic and not an approximation to it. For larger values of n, the chi-square distribution approaches normality, and the two methods give almost the same answers. This method can also he used to obtain one-sided confidence intervals (upper and lower bounds). Example 6.13.1 (coni.). with prohability 0.95, To ohtain one-sided intervals, we argue that 5040 --o <314 ., whence 0> 160.5 , or that 5040 -()- > 10.85 whence 0< 464.5. o (04 SOME STANDARD STATISTICAL PROCEDURES 6.14. ANOTHER METHOD OF ESTIMATING THE STANDARD DEVIATION In setting up quality control charts, it has been customary to use a different method of estimating the standard deviation based on the sample range. The range is the difference between the maximum and minimum observations in the sample. The advantage of this method is that the range is much easier to calculate than i; modem advances in computing have made that advantage less important. The estimate, 52, that was obtained in section 6.10 is more efficient than the estimate that is derived from the range. However, for the small samples that are normally used in x-bar charts (n = 4 or 5), the differences in efficiency are trivial, and little is lost in using the simpler range method. The validity of the range method is very dependent on the assumption that the samples come from a normal population. The range estimate is obtained by multiplying the sample range by an appropriate constant. For example, it can be shown that for samples of four observations from a normal distribution, the expected value of R is 2.060'. The corresponding estimate of 0' is RI2.06. The multiplier 2.06 is denoted in quality control work by d 2 • Table 6.14.1 gives values of d 2 for some values of n. The standard deviation is then estimated by • 0'= R (6.14.1) d' 2 The range of the first five observations in example 6.11.1 is R = 114.2 - 87.2 = 27.0. The value of d 2 for n = 5 is 2.33. The estimate of the standard deviation from that subsample by the range method is _ 27.0 0' = 2.33 = 11.6 ; the estimate from the other subsample, the last five observations, is 98.4 - 85.4 = 5 58 .. 2.33 We cannot say which of these estimates or the previous estimate is closer to the "true" value for the popUlation that provided this sample. They are each estimates of the unknown standard deviation. It is obvious that the highest resistance, 114.7. made the first estimate larger than the second. The Table 6.14.1. Expected Values of the Range, d z 2 1.13 3 1.69 4 2.06 5 2.33 6 2.53 7 2.70 lOS UNBIASED ESTIMATORS optimist should resist the temptation to calculate several estimates and choose the smallest one! 6.15. UNBIASED ESTIMATORS Equation (5.8.3) shows that i is an unbiased estimate of E(X), which means that if we use i to estimate E(X), we will, on the average, have the correct answer. Before giving a formal definition of the property of unbiasedness, we define the term statistic. Definition. Let x be a data vector. A statistic, t(x), is a function of the data, i.e., a number calculated from the data that does not involve any unknown parameters; i is a statistic. 0 Definition. A statistic, t(x), is said to be an unbiased estimate, or estimator, of a parameter, say, l/J, if E[t(x)J = l/J . (6.15.1) o Unbiasedness is an attractive property for an estimator to possess, but it is not adequate just to have an unbiased estimate. In the first experiment, section 6.3, E(X]) = p. This says that if we throwaway all the data except the observation in the first component, and say that p = ] if that component is defective and p = 0 if the component is okay, we will have an unbiased estimate: unbiased but also unwise! How might one choose between estimators? Which is the best "recipe" for estimating cf>? One course of action is to use unbiased estimators wherever possible, and, when faced with two unbiased estimators, pick the one with the smaller variance. It is a good principle. In many problems, it is possible to find an unbiased estimator of l/J that we know to have a smaller variance than any possible competitor. Such an estimator is called the MVUE, minimum-variance unbiased estimator. There is an interesting mathematical theory behind the search for MVU estimators, but to investigate it would take us beyond the scope of this book. It suffices at this stage to say that under most circumstances, the sample mean, i, is the MVU estimator of E(X). Definition. An estimator of l/J that is a linear combination of the observations is called a linear estimator. It can be written as '-" a x - L." aix; . (6.15.2) o 3) By equation (5.5). by equation L a.7. We are considering estimators of the form W(x) Since W (5. in which case. In the hinomial experiment. we equated the number of successes X to E(X) = np and solved for the unknown p. = 1. . Show that the sample mean is the best linear unbiased estimator of £(X). we discuss two standard procedures for finding estimators: the method of moments and the method of maximum likelihood. .15.15.4). 0 6.x. V(W) = [a~V(X). a" are equal. The unbiased linear estimator of cf> that has the smallest variance is sometimes called the BLUE. = lin for all i.1. an unbiased estimator. but we gave little indication of why they were chosen in particular. hence. Example 6. MOMENT ESTIMATORS In the previous sections. That was the rationale for the estimates that were used in the three initial experiments. and so we must minimize [a~ subject to the restriction of equation (6.3). It consists of equating sample moments to the expected values of the corresponding population moments and solving for the unknown parameters.106 SOME STANDARD STATISTICAL PROCEDURES Linear estimators constitute an important class of estimators. (6. The method of moments goes back a long way in the history of statistics. In the normal and exponential examples. E( W) = £(X). The estimators that we chose were reasonable. Direct application of the method of Lagrange multipliers shows that this occurs when all the coefficients. Are there perhaps better estimators available? In this section and the next.E(X) = E(X) . The sample mean is a linear estimator of E(X). This idea is best illustrated by a proof of a result that common sense would regard as obvious.15. and La. best linear unbiased estimator. They arc used extensively when we fit equations by the method of least squares.7. we have discussed estimates of parameters. we equated the first sample moment i to E(X) to estimate IL and 8. IS =L a.16. a. The first two sample moments are 2: x.17 and so we write it == i = 95. and x=95. 10. .1. we equate the first two sample moments to their expectations and solve the simultaneous equations. This estimate is Sin rather than SI(n . E(X) = 012.107 MOMENT ESTIMATORS Example 6.17 and whence . A quality control engineer samples items from a production line until five defective items are found.. which is not compatible with the largest observation.8). m. o Example 6. Then x = 4 and 8 = 8. The probability density function is f(x) = 110 =0 OsxsO x> () .1).2 = 78. and so the estimate of 0 is () = 2x. We denote the estimate of the variance obtained by the method moments _7 bY (I-. Suppose that we have five observations: 1.16. Estimate the unknown parameter O. o This can sometimes give an unreasonable answer.16) with r = 5. Then E(X) "'.91 .1. and so 80= ~ and p 5 p = 80 = 0'()625 . X. inspected has a negative binomial distribution (section 3.3. 2~ 3. it is biased on the low side.rip.24 . If there are two parameters to be estimated. The number of items. Estimate the percentage of defectives in the process.16. = 10 = 9136. and 10.2. A sample of n observations is taken from a uniform distribution on the interval (0.16. Estimate the variance of the resistances in example 6.11. 4. The fifth defective was the eightieth item inspected. o Example 6. .)-' = '8 In(L) = -n 10(0) - whence. MAXIMUM-LIKELIHOOD ESTIMATES Another standard method obtaining estimates of parameters is the method of maximum likelihood. Fisher. .17. and 8 =x. Suppose that we have a continuous variable and only one unknown parameter. denoted by cpo The density function can be written as f{x. and P(x). It is often more convenient to maximize the natural logarithm. The idea is that we look at the data and pick the value of the parameter(s) that. We have and L(x. 0) (1)" exp (_ n.108 SOME STANDARD STATISTICAL PROCEDURES 6. which we can write as P(x.).1. The maximum-likelihood estimate of cp is the value that maxImIzes L(x. X. cp) to emphasize that it is a function of both x and cpo Definition. Example 6. which was developed by R. differentiating. In(L) rather than L itself. in retrospect. maximizes the likelihood of that particular sample.17.. Consider a sample of observations x = (x l' X2' likelihood of the sample is defined as the product . A. we have a probability mass function cp). Find the maximum-likelihood estimate of the parameter in an exponential distribution. The o In the case of a discrete distribution.. IlX 8" ' . This value is usually obtained by differentiation. cp). 0 which appears as an exercise. Find the maximum-likelihood estimator of uniform distribution..17. One is the gamma distribution. I. Usually..3. There are exceptions. See example 6.. The likelihood function is (I in the L(X. The maximum is obtained by taking (J to be as small as we can... it finds a minimum when (I = +00. this estimate is biased on the low side. Example 6.=i and 2 S. .( "2 In(21T) - n In(cT) - 2: (x. the two methods..e. consistent with the data..8)=(~r Differentiating L does not find a maximum.)2] 1 L = «(TV21i/ The likelihood function is CXp - 2172 • Then In(L) n) = . the maximum-likelihood estimate is 10.109 MAXIMUM-LIKELIHOOD ESTIMATES Example 6. The normal distribution. Another is the uniform distribution. Since the largest observation is usually less than (J and never 0 exceeds it.16.3.2.3.. by setting (I equal to the largest observation. In the data set that follows example 6. [2: (x. . moments and maximum likelihood..)2 20'2 Differentiating.16. . agree on the same estimator. 17 = - n Notice that we would have obtained the same estimates if we had written v = u 2 and differentiated with respect to v rather than u.17. we have and Solving the simultaneous equations gives the estimates p. 3.2. omitting the eighth wafer. a discussion of thcm would take us beyond the scope of this book. How many independent measurements must be taken for an engineer to know the octane numher to within 0.20 10.5. 6.863 I.5 7. Obtain a two-sided 95% confidence interval for the mean speed.3. EXERCISES 6.2 6.80 11.50 14. 6.6 7. A machine produces ball bearings.30 Make a dot plot of the data.40 11.1.563 0. They are.5 7. Obtain a two-sided 95% confidence interval for the sheet resistance.6 7.10 10.90 14.80 14.30 10.3 7.30 14. Find a two-sided 95% confidence· interval for the average diameter of the bearings heing produced.20 13. however.4.50 10. The resistances arc 7. The mean grain sizes in a sample of six thin films of Al-l %Si on Si measured in /Lm arc 0. treated in most textbooks for a general course in mathematical statistics. . A sample of 25 devices is tested for speed.50 14.9 7.61 mm. \02 1.10 15.673 0.50 10.10 10.70 10.10 10. however.4 7.5 7.4.22 mm. Make a dotplot of the data. Their speeds in MHz are 10. 6.4 7.60 11. 10. The standard deviation of a measurement of the octane number of a gasoline is 0. the diameters are normally distributed with if = 0.6.5 7. Obtain a one-sided 95% confidence interval that gives an upper bound on the mean grain size. Repeat exercise 6.30 10. A sample of 20 bearings has an average diameter of 12.4 7.3 7.30.110 SOME STANDARD STATISTICAL PROCEDURES The method of maximum likelihood has several very attractive theoretical properties.60 10.50 14.7 7.3.934 0.1 numher with 90% confidence? 6.3 7. 6.10 14. Sixteen wafers are tested for sheet resistance.061 .50 10. The speeds for five such samples in MHz are (samples arc rows) 16. Twenty devices arc placed on life test. The average lifetime of the devices tested is 1832 hours.2 14.11. (b) the chi-square distribution. Use the normal approximation to obtain a two-sided 90% confidence interval for the percent defective in the batch. vcrify that the value of d 2 = 2.0 19. 6.06 by taking several samples of size 4. Obtain estimates of the standard deviation. The lifetimes are assumed to be exponentially distributed. A quality control inspector takes four devices off the production line at the end of each shift and tests them for speed.0 13. computing their ranges.0 22.8.10.12. Obtain a two-sided 95% confidence interval for the percent defective in the batch. If you have a software package that draws random samples from a normal distribution.2. CT. A sample of 47 items from a large batch contains 18 defectives. A sample of 50 parts from a large batch contains six defectives.2 16. and (b) using the more usual method of computing S2 == ~ (y .)2/3 for each sample and taking the square root of the average of these estimates of (. 6.9.7.>.3.3 12. Obtain a two-sided 90% confidence interval for the variance of the sheet resistances from the sample reported in exercise 6. A television newscaster reports the result of a national poll in which ](X)O people split 556 in favor of a question and 444 against.8 17.5 13.13.5.5 14. and showing that the average range is close to 2.6% in favor has a margin of error ±3%.2 12.111 EXERCISES 6.5 15. Obtain a two-sided 95% confidcnce interval for the average lifetime of a device using: (a) the central limit theorem.2 18. 6.4 10.4 17. (a) using the sample ranges averaged over the five samples.0 14.0 17.5 L3. 6.6 15. He says that the figure 55.06. You can do this with Minitab in the following way: .0 17. How is that figure calculated? 6. 6. 20 -O<x<+O.00.!. 6.16. =0 elsewhere. Estimate the fraction defective. An engineer tests devices until ten defectives are found. A random sample of 20 observations has 1: y = 16.14. rmax c1-c4 c5 rmin c1-c4 c6 let c7 = c5-c6 describe c7 6. Use the method of moments to estimate parameters a and p.. 6. A sample of six observations is -3. A variable. and +8. Use the method of moments to obtain estimates of parameters a and p. . Usc the method of moments to find estimates of a and p from a sample of n independent observations from a gamma distribution..17. normal 0 1. -9. The moments of this distribution were calculated in exercise 4. f(x) = . +5. +10. 6. .15. has a uniform distribution on the interval (-0.15. X. -2. A sample of 100 observations from a gamma distribution has I: y = 492 and E l = 1024.0 and I: l = 13. The beta distribution was introduced in exercise 4.18. 6. +8).12. Use the method of moments to obtain an estimate of 0. The tenth defective is the 267th device tested.112 SOME STANDARD STATISTICAL PROCEDURES random 500 cl-c4. Hypothesis testing is also used in comparative experiments. p. INTRODUCTION Scientists make decisions on the basis of data. if it is too high and falls outside the acceptance region. accept or reject the hypotheses.1. In quality control work. If one of the points falls outside the control lines. using the binomial distribution. Testing for attributes. In these two examples. If there are too many defectives. M.Statistical Methods in Engineering and Quality Assurance Peter W. engineers test hypotheses and make statistical decisions without actually using those expressions. he decides to take action as if the process were out of control and starts looking for an assignable cause for the bad point. If the point falls within the control lines. In the control chart. The engineer is testing the hypothesis that the percent defectives in the lot is at some predetermined acceptable value. after analyzing the data. When he plots a point on the chart. the hypothesis was made about the parameter. the engineer looked at a random variable that was taken to be (normally) distributed with a certain expectation. John Copyright © 1990 by John Wiley & Sons. the engineer accepts the hypothesis and decides that the process is in control. JL(). if it falls in the acceptance region. A quality control inspector takes a random sample of items from a lot. otherwise. the engineer rejects the hypothesis and takes the appropriate action. This a fundamental part of the scientific method. If the point falls outside that acceptance region. They postulate (set up) hypotheses. of a binomial variable. the engineer rejects the hypothesis. An engineer looks at a quality control chart on which the averages of the daily samples are plotted. In the second example. The number of defectives is the test statistic. he rejects the lot. he is testing the hypothesis that the process mean is at its usual value. is discussed further in chapters 11 and 12 under the heading of acceptance sampling. collect data. he accepts it. JL. and. The hypothesis was that JL was equal to some particular value. the engineer accepts the hypothesis. Inc CHAPTERSEVEN Hypothesis Testing 7. the hypothesis being tested involved a statistical model. A chemical 113 . The data are compatible with that hypothesis. One morning. measure the resistance of each. we develop the theory of testing hypotheses about the mean of a normal distribution in the framework of an example. The engineer is testing the hypothesis that the means for the two reagents are equal against the alternative that they are not. The chemical engineer in the laboratory could set up a confidence interval for the difference between the two means. There is a strong connection between confidence intervals and hypothesis testing. If the interval included zero. of an item is a normal random variable with expectation fL and standard deviation R. For the latter. The inspector docs not want a confidence interval for the fraction defective. Our probability model for the process is that the resistance. Our problem is to decide whether the average resistance of the items in the batch is 200 ohms. The engineer may conclude that one reagent is clearly better than the other or else that there is not an important difference between them so far as the quality of the product is concerned. This is the classical hypothesis-testing situation. We wish to test the hypothesis We decide (arbitrarily for the present) to take a random sample of n = 16 resistors. AN EXAMPLE In the next eight sections. the emphasis was on what are called decision rules. our experience with this supplier has been that their resistors have a standard deviation of 8 ohms (a number chosen to make the arithmetic simple).114 HYPOTHESIS TESTING engineer may have a choice of two alkaline reagents in a process. A few runs using sodium hydroxide and a few using potassium hydroxide are made and their averages arc compared. the choice might be made on cost or other grounds. the engineer could accept the hypothesis that the means were equal. The confidence interval and the hypothesis test arc mathematically the same. He wants a rule for saying yes or no. In this case. We wish to determine whether they meet the specification of 200 ohms. 7. We do not expect that each resistor will measure exactly 200 ohms. In the previous examples. the hypothesis of equality would be rejected. X. and calculate the average resist- . If the interval did not include zero. The information on the size of the difference between the means is important to future decision making.2. The quality control engineer and the inspector both need rules to apply to take appropriate actions. a large batch is delivered. We can live with that amount of variability. the engineer should calculate the interval and report that as well. Suppose that we purchase 2oo-ohm resistors from a supplier. many bad batches would get through. On the other hand. Rejecting the hypothesis means deciding that the average resistance is not 200 ohms. There is strong evidence that the hypothesis is false. presumably. reject 92% of the batches for which fL was actually equal to 200. accepting the hypothesis does not really mean agreeing that the hypothesis is correct. one risk can only be reduced at the expense of increasing the other risk.THE DECISIONS liS ance. If we accept the hypothesis. and the middle ground-not proven. The prisoner is declared to be guilty only if there is a preponderance of evidence that causes the jury to reject the hypothesis of innocence. We know that it is an unbiased estimator of fL and that its standard deviation is 8/V/i = 2. refusing to buy that batch. . since P(i ~ 200. the hypothesis of innocence is accepted. and we buy it. we decide that the batch of resistors meets our specification. The choice of a suitable acceptance region involves balancing these risks in some way. For a given sample size. The interval 150 < i < 250 would be the acceptance region. We can accept the hypothesis or we can reject it.8 < i < 200. How do we define close? Look first at two extreme possibilities. 7. Otherwise. and reject if i fell outside.8) == P(z ~ 0. The two intervals i ~ 250 and i:5 150 would constitute the rejection (or critical) region. Hypothesis testing is like a judicial process. Accepting the hypothesis means that there is not strong evidence to reject it. The type IJ error is to accept the hypothesis when it is false. This will be our test statistic. On the other hand. the prisoner may actually be guilty. respectively. How do we decide'! If i is close to the hypothesized value of 200. and.10) == 0. THE DECISIONS There are two decisions that can be made on the basis of the test data.46. we argue that the hypothesis is false and reject it. We can make two errors in the decision process. we should. The type I error is to reject the hypothesis when it is true. i. Scottish criminal law used to allow three possible verdicts: guilty. We reject the hypothesis if i is "far" from 200. but the evidence is not strong enough for conviction. we argue that the sample is compatible with the hypothesis.2) == P(i:5 198. Neither we nor the supplier would doubt that the batches we rejected were indeed substandard. When the acceptance limits are wide. We might decide to accept the hypothesis if i fell within the range 150 < i < 250. The probabilities of making these errors are conventionally denoted by alpha and beta.2. if the acceptance interval were 199. but. Those limits are unreasonably restrictive. If i is not close to 200. not guilty. we are strongly persuaded by data with i outside those limits that the hypothesis is false.3. We start by assuming that the prisoner is innocent. with acceptance limits so far apart. Both risks can be reduced by increasing the sample size at the price of increased inspection costs. Now we have a decision rule. Note that the null hypothesis gets the benefit of the doubt-the cases in which the old Scottish juries would have returned a verdict of not proven. or the critical region. An engineer testing with a = 5% is arguing that a value of i.4. is strong evidence that the null hypothesis is false. NULL AND ALTERNATE HYPOTHESES We actually consider two hypotheses.. We mean by the acceptance region the set of values of the test statistic that result in our accepting the hypothesis. In this example. a two-sided 95% . The other hypothesis. Values of the test statistic that are in the critical region arc said to be (statistically) significant.0 + 1.0. we argue that if HI) is true. we accept Ii.1<:i<203.0 . The hypothesis that we have mentioned. which is so far from the hypothesized value that it could only have happened less than one time in twenty. therefore. rejecting the batch may mean several possible courses of action. we reject H". and average the 16 readings. The null hypothesis is to be rejected if the average resistance is either much less than 200 or much greater than 200.5. To calculate the acceptance region with a == 5% tor the resistor example.96(2) < i < 200. from scrapping the whole batch to testing 100% of them. The acceptance region that we have just obtained looks like a confidence interval. In this example. The traditional choices of level for alpha arc 5% and 1%. reject the batch. test them. If :i :s 196 or if i 2: 204. or test procedure: take 16 resistors at random from a batch. and so it is defined by 200. In the example of the resistors. ACCEPTANCE AND REJECTION REGIONS The usual procedure in designing a test to choose the test statistic and the desired level of alpha.116 HYPOTHESIS TESTING 7.1. to contain :i 95% of the time. If 196 < :i < 204..0 and standard deviation 2. or HI' is called the alternate hypothesis. Iill' is called the null hypothesis. then i has a normal distribution with mean 200. the alternate hypothesis is It is the two-sided alternative. The acceptance region is. there is a very strong connection. denoted by H a . The rejection region is chosen so that i will fall inside it only 5% of the time.91. 7. When we reject Ho.96(2) 196. and then to calculate the acceptance region. The other values of the statistic constitute the rejection region. accept the batch. when we accept Ho. the buyer agrees to buy that load of gasoline." In terms of hypothesis testing. The vendor may look at the problem differently.1 < i < 203. or some other figure.3. if i .1. 196.0. This is a one-sided test. The vendor would prefer a situation in which the customer had to buy the batch of gasoline unless it was established that the batch was below specification. unless.0 (or ::590. i > 90. A tank truck of gasoline arrives at the buyer's garage.0 or higher.e.0) and When the null hypothesis is rejected.t = 90. i.3.92< 200 and i + 3. Several samples of the gasoline are drawn and their octane numbers are measured. i. The truckload will only be accepted if i exceeds 90.0 by some "safety factor.t is i ..92 . and the buyer will buy the batch only if the data convince him that the gasoline is not below specification.t = 200.6. The interval contains the hypothesized value.92 < J.9. this would be very unlikely because few buyers have such test facilities. which is identical with the acceptance region..) The buyer wants to have gasoline with an octane number of 90. the buyer would take for hypotheses Ho: J. The rejection region that the buyer chooses is the upper tail of the distribution of i. the set of values defined by i> c.r . and only if. J. where c is a value calculated for this test." so the buyer will use a decision rule such as "buy the load if. but does want protection against gasoline with an ON less than 90. 7.e. (In the real world.92>200.e.117 ONE-SIDED TESTS confiden~e interval for J. i. The buyer does not care if the oil company supplies higher-octane gasoline than specified.t < i + 3.. ONE-SIDED TESTS Suppose that a company that runs a large fleet of cars contracts to purchase gasoline in large quantities from a major oil company. 3. we reject the null hypothesis. each party is looking after its own interests. different parties may choose different alternatives. If in the example of the resistors. We wish to test Ho: iJ- = 60. we have improved the power of the test to spot departures from specification on the high side.0) and They are looking at the situation from two different sides. we would like to increase that figure. If the average yield with the new process is significantly greater than 60.. The current process converts 60% of the raw material to finished product. i >203.0. The choice between one and two sides rests with the person designing the test. the alternative had been Ji. This means that we decide to carry out further investigation into the new process at the pilot plant level.0 against Ha: IL > 60.7. but in a particular situation.5 would call for accepting Ho in the two-tailed case and for rejecting it in the other.== 90.118 HYPOTHESIS TESTING was significantly less than 90. In the same situation.L > 200. The cost that we pay is that we can no longer spot departures on the low side. A good rule of thumb is to usc a two-sided alternative unless a convincing case can be made for either of the singlesided alternatives.645(2) . This is illustrated in the following example.: f. A value i = 203. or only if i is too low. the critical region would have been i > 200 + 1. or if i is either too high or too low? The correct choice is usually clear from the context of the problem. We make some runs with their new process and calculate the average yield. The cost of a type I error is a few weeks' work on the pilot plant. 7.0. The vendor would prefer the hypotheses Ho: iJ. we may not care about that. The cost of a type II error is . Suppose that we arc considering rcplacing our present process for making a chemical product by another process. THE CHOICE OF ALPHA The choice of alpha depends on the consequences of making a type I (or a type II) error.0. The chemists in the laboratory come up with a possible new process. Notice that the upper limit in the one-tailed test is lower than the limit in the two-tailed test. We can have either one-sided or two-sided alternatives to the null hypothesis. This example illustrates two points. By choosing the critical region appropriate to the one-sided alternative. The essential question is this: Do we want to reject flu only if i is too high.0 with a = 5%.0 (or 2::90. This brings us back to the subject that was only briefly touched upon in section 6. Thus. i is a mapping of that point on to the real linc. The new drug will be approved only if the patients who receive it have significantly fewer attacks per person than the control group. and that i takes values (at least in theory) anywhere on the real line from -00 to +0:0. The last example in this section concerns drug testing. At this stage we can afford to take a fairly large value of alpha (and a correspondingly small value of beta). It might be the difference in the numbers of attacks per patient in the two groups. The second virtue is that wc know how the statistic behaves. At the next stage. that is a matter of opinion. With luck. We can rephrase the development of the test in the following way. which calls for a low value of alpha.. there is the test. The others are given the placebo-pills that do not contain any drug (although the recipients are unaware of this). One group is given a course of pills made from the new drug. It is a good estimator of IL. The real line is then divided into two regions.15: i is the minimumvariance unbiased estimator of the parameter that is being investigated. we will be allowed enough time and research budget to achieve that also.119 THE Z-TEST the loss of an exciting new process that could make the company a lot of money. The usual procedure is to select two groups of patients. it is the best estimator. or reduction. Unless the scientists think that there is some danger that the new drug will actually make the patients worse. that property was used to obtain the acceptance limits. 7. Now we had better be sure that we do not erect a costly monument to our misjudgment.TI':ST There are two important advantages in the choice of i as the test statistic in the problems of the previous sections. THE z. It is normally distributed. Rejecting the null hypothesis means approving the new drug for use. by most standards. We can regard i as a summary. the consequences of an incorrect decision are different. acceptance and rejection. Suppose that a pharmaceuticat company proposes a new drug that allegedly alleviates the symptoms of some illness. As for beta. Indeed. The null hypothesis is that the new drug is of no use-the two means are equal. We could just as well use any equivalent mapping. but now the consequence of rejecting the null hypothesis is that the company will choose the new process and build a new plant at a cost of several million dollars. of the set of n data points into a single value. a small beta would be nice. a very conservative test. We do . The responses of the patients are noted and a test statistic is calculated. The hypotheses tested after the pilot plant runs are the same. The mathematician sees the sample as a point in ndimensional space. the test has a one-tailed alternative.S. That is a very serious step. too. We know that i is the statistic that we would use to estimate /J. We want alpha to be small. So we may as well focus on a function of i that we can handle. A plot with p. Then (3 = Fz ( -0.3 . The alpha risk is 5%. the rejection region is i > 203.0 = -1. is the type II error when p.0. we have 203.9.9. the beta risk is 20%. Thus. We can use any function of i that has a one-to-one correspondence. i is normally distributed with mean 205. = 200 against the alternative. and the mean resistance is not 200.0.{3.0? In this case. = iJ-o.-c) crVii ' 1 . z has a standard normal distribution.0.7 = -O. This is the rationale for the z-test. but has some other value? What. = 205.120 HYPOTHESIS TESTING not have to use i.0.) = F (p. The OC curve is the plot of (c-p. It has the advantages. It measures the ability of the test to discriminate between p. Instead of focusing on i directly. = 204. p. we focus on x . Similar calculations show that when p. for example. we accept the null hypothesis if Izl < 1. of being more general in its statement and of pointing out more clearly the connection betwecn tests and confidence intervals. if.205. and the corresponding z statistic is (i . and the value JLo in the null hypothesis. along the horizontal axis and the power along the vertical axis is called the Operating Characteristic (OC curve) for the test.0 and standard deviation 2.3.20. THE TYPE II ERROR The one-sided test that was developed earlier for resistors tests p.Fz crVii z . The power of the test is defined as 1 . When the null hypothesis is true.96. The OC curve for this example is shown in figure 7.855) = 0. at least pedagogically.1. The z-test is no different from the test using i itself as the statistic. 7. Beta is the risk of failing to reject the null hypothesis when we should do so. we wish to have a two-sidcd test with a = 5%. = 203. What if the null hypothesis is false.855CT(i). > 200. {3 = 36%. for example. We write FAz) for the cumulative distribution of the standard normal distribution. Measuring in units of the standard deviation of i.205) /2.iJ-o z = CT/v7i ' where the null hypothesis is Ho: p. the value of beta is seen to depend on the value of p. With CT = 8 and n = 16. Thus.. and when p. {3 = 56%. 0 with standard deviation 8/VTi.(1. -_.8 I I I .0 ohms._____ _ I I I I I I I I :z2' ~ .2 _______ IL _______ _ -------+---------~---------~---------~I I I r I 1 I .9. i is normally distributed about a mean 200. . we have c = 200. - I I I I I I I 0. It is then determined by only two points. n. i is normally distributed with mean 203.. Two numbers have to be specified: c and the sample size. Suppose also that thc engineer wants to have a 90% power of rejecting the null hypothesis if the resistance is as high as 203..0 and the same standard deviation.-1- ~ __ _ _ •• __ • __ 1_ _ _ _ _ •• _ _ _ I I I .0 . (7. where the rejection region is i> c.0 ohms.28)( :n) . Suppose that the engineer is prepared to take a 5% alpha risk of accepting a batch with resistance as low as 200.0. It is the cumulative normal curve.0 + (1.9.645)( ~) . the graph is a straight line. To satisfy the requirement that a = 5%.1) If J. ___ _ 0.4 I r I I 2 I I I r 0. When the null hypothesis is true. When it is plotted on normal probability paper...1. (7.121 THE TYPE II ERROR r I J.l.6 -------~---------+---------~---------~----I . The power requirement leads to c = 203.~ I I I I "0 ~ ~ I I 0.~ 0. OC curve.2) . How should the test be designed? The engineer will choose a one-sided test with the critical region defined by i> c.L = 203. r 0 190 193 196 199 202 205 208 Mean Figure 7. or ogive.9. which leads to the next consideration. 28.0. the estimate is obtained from a sample of n observations. Let za and z/3 denote the values of z that correspond to our choices of alpha and beta (in the example.36.2. If ex = 5% with the two-sided alternative. The calculated value 1.1) t= . and so the null hypothesis is accepted.11 with confidence intervals. which is the tabulated value for fl . THE I-TEST If we do not know the variance of the hypothesized normal population.36/VlO = 1. Then the sample size required is given by (7. of the variance. The normal test statistic z is replaced by Student's t: x .7.0 9. is to< = 2.17 with estimated standard deviation s = 9..1 d.9..262.2).<.17 -.L and the desired value of alpha. we have to estimate it from the data. In example 6.122 HYPOTHESIS TESTING Solving equations (7.11. If the observed value falls in the critical region.5% in each tail. We did this earlier in section 6.90. Example 7.746 is less than 2. they were 1.- s/vTl . a sample of ten resistors was taken and their resistances were measured. and the appropriate tabulated value for ( with nine d. and so there are fl . The t statistic has thc same number of degrees of freedom as the estimate. respectively).10. The value of t obtained from the sample is compared to t*. the null hypothesis is rejected.1) and (7.f.262. The derivation of a general formula is left as an exercise for the reader.1.645 and 1. Suppose that we wish to test Ifu: IL = 90.10.ILl' (7. we get n = 61 and c=201. The test statistic is / = 95.9. 0 ..f.9. there is to be probability 2.1.0 against If. The average resistance was i = 95.10.I d.: IL "" 90. and let the difference between the means in the two hypotheses be 5.3) 7. . In this case.746. 47:5 IL :5 101.68.0.961 98.364 2.48:s The point estimate is / =- 87. (T2 :s 292.9 107.000 VS MU N . the engineer had chosen to assign alpha a value of O. We can construct tests in a similar way. Examples 6.1 and 6. That is not good enough. for some peculiar reason.1 both deal with the sample of ten resistors.11. in some problems.4 T P VALUE 1. it contains the hypothesized value 90.6 85.141. which is much closer to the calculated value of 1.0. the data are compatible with the null hypothesis. would be significant. There is much more information here.11. Calculating and printing the P value clearly avoids having to look up (* in the table.123 TESTS ABOUT THE VARIANCE The engineer could just report the bald fact that the t value was not significant. 90. The observed difference between i and IL would have been declared significant if.27 . we should have had t* = 1. We can see from the confidence interval that a larger sample with a correspondingly lower value of t* might have led to rejection of the null hypothesis.000 N MEAN STDEV SE MEAN Cll 10 95.E. The 95% confidence interval obtained earlier is 88.75 and infinity and can now be easily calculated by modern computers. . and we calculated a 95% confidence interval for the variance 41.4 94. The sum of squares of deviations is S = 789. and.746.170 9.8 MTB>end MTB > ttest 90 ell TEST OF MU = 90. 1. As the sample size increases.2 91. 7.75 0.833. we might accept the null hypothesis at a = 5% and reject it if we chose a = 10%.2 87. It is the area under the density curve for the { distribution between 1. The interval gives us a range of values of IL.6 95.12. The Minitab printout for the {-test is as follows: MTB > set in ell MTB> 114. TESTS ABOUT 1'81-: VARIANCE In section 6. therefore. but only barely. /* decreases. Clearly. one cannot conclude that IL =F 90. we used the chi-square distribution to construct confidence intervals for the variance of a normal population.87 .75. Thus.11 The P value is the smallest value of alpha at which the observed value of t. II or higher.5 89.12.1 87. We can also go back to the tables and note that had we taken alpha to be 10%. p. For a test with ex = 0.153 < 16. based on 31 observations. of which 20 were defective.92. we cannot say. we compare this value to the 95% value. 87. If Ho is correct. the null hypothesis would have been rejected. S/u 2.5 to the tabulated 95% value. with alpha set at 5%. 16. In that case. only nine d.05.f. the statistic 2 LXI -(J.00.92.8. This is illustrated in example 7. The same approximation can be used for testing hypotheses about the percentage.77 with 30 d.l' the statistic S/60 = 13. we obtained a 95% confidence interval for the parameter of an exponential distribution.13.124 HYPOTHESIS TESTING Suppose that we wish to test the hypothesis against Ha: u 1 > 60. from the tables. therefore. We obtained a confidence interval for the percent defective in the population by using a normal approximation to the binomial distribution. that the hypothesis is false.f. 31. We can rewrite the test statistic. . we recall that 2 ~ x. Since 13. we considered a sample of 100 components.1.88 is nearly 50% higher than the hypothesized value. as (n .13. Even though the point estimate of 87.13.f. In example 6.84. TESTS FOR THE PERCENT DE}t'ECTIVE In example 6.4. There were ten lamps in the sample. we accept the null hypothesis. To test the hypothesis Ho: 8 = 160 against Ho: (J > 160. The reason is that there was not enough data.f.1 )S2/U 2. Suppose that we had the same estimate. and their average lifetime was 252 hours. 7.153 has a chi-square distribution with nine d. THE EXPONENTIAL DISTRIBUTION Chi-square statistics can also be used in tests ahout exponential distrihutions.68. we compare the calculated value 3].68)/60 = 43. 7.1. Under lI. With ex = 5%. Then the test statistic would have been 30(87.= 20(252) 160 = 31. and reje<-'t the null hypothesis.12.iO has a chi-square distribution with 2n dJ.1. which is greater than the tabulated value of 43.5 has a chi-square distribution with 20 d. 1. We argue that under the null hypothesis. In this section.np -5 z = Vnpq = 4. A sample of 100 components has 20 defectives. X.1 (cont. 0 7.125 CHI-SQUARE AND GOODNESS OF FIT Example 7.13.to.25 against the alternative HI: P .E Defective Nondefective 20 25 80 -5 +5 75 . the number of defectives. [n 1900.25.75. This is shown in table 7. The deviations in the bin numbers are thus 20 .14.25 either. we discuss these goodness-of-fit tests. We consider a model with two "bins": defective and nondefective.14.75 = +5. has a normal distribution with expectation np = 25 and variance npq = 18.SQUARE AND GOODNESS OF FlT We have been using the chi-square distribution as the distribution of the sum of squares of standard normal variables.25 = -5 and 80 .1 Observed (0) Expected (E) Deviation (dev) = U .1. and the evidence is not strong enough for us to reject the null hypothesis. It has led us to tests about the variance of a normal distribution and the expectation of an exponential variable. Test at a = 0.14. However. beginning with another look at the binomial test of the previous section. Table 7.155 . Example 7. respectively. em. the history of chi-square goes back earlier. It would not be strong enough for the one-tailed test with the alternative HI: p < 0. Karl Pearson introduced chi-square tests in the context of binomial and multinomial distributions.05 the null hypothesis: Ho: P = 0.33. the expected numbers are 25 and 75. respectively (notice that the deviations sum to zero). The test statistic is x . The chi-square test statistic is [n this example. respectively.13.). U 25 25 = 25 + 75 = 1.33 = 1. The observed numbers in the bins are 20 and 80. L.99. 18 failed at the first stage and were rejected. but the value of U would have been tripled. We note again that the sum of the deviations is zero. 20 failed at the second stage.15.84. accept the null hypothesis that p = 0. because the two deviations arc forced to sum to zero. the derivation appears as an exercise. U = 225 225 75 + 225 = 4. Had there been 300 components with 60 defectives. the percent defective would have he en the same. and we compare U to the tabulated value of chi-square.. The frequencies were .00 o which exceeds the critical value.2. we listed the frequencies of the digits 1 through I) in a sample of 320 digits. 7. with 65% passing both stages.20.15.155)2 and the tabulated value of chi-square is 3. lJ Example 7. There is only one d.84. two d.84=(1. pz = 0.84. we reject the null hypothesis.33 to 3.33 = (1. Example 7.96. Suppose that an electronic component undergoes a two-stage testing procedure before being put in service. and the others passed.15.f.f.15. and since 1. The extension is illustrated by two more examples.12. therefore. and P3 = 0.155 to L. U = 1. Stage 1 Stage 2 Pass o 18 E 24 -6 20 32 -12 104 +18 O-E 122 36 144 324 U=24+ 32-+104=9. for chi-square with one d. the z-tcst and the chi-square test are mathematically the same.99. 5. In tahle 3. It is essential that the test be applied using the actual numbers in the bins and not the percentages. This equivalence is always true for the binomial test.25. 3.96)2.2.126 HYPOTHESIS TESTING We compare the value of U to the tabulated value. Out of 160 components tested. Since U > 5.84 is equivalent to comparing L. Comparing 1. In the example.1. There are. It had heen expected that 15% would fail at the first stage and 20% at the second.65 against the alternative that those arc not the correct values. We wish to test the hypothesis Ho: PI = 0.33 < 3. 3. MUL TlNOMIAL GOODNESS OF FIT This test can be extended to the multinomial distribution with several bins. 170.7. Test at £l' = 5% the hypothesis that the octane number is 79.260. The deviations are.17 Test the hypothesis that the actual average thickness by their method is indeed 2.2.170.3.80 12.0 }. the expected number of occurrences is 32.190. how many measurements should be made if {3 is to be 10% when the octane number is 8O.2 The standard deviation of an octane determination is 0.0 79. In exercise 7. 0 EXERCISES 7.0? .50 11. Is the data compatible with the assumption that each digit occurs \0% of the time in the parent population? For each digit.L = 8.25 vs.92. A research group is trying to produce silicon-on-sapphire films with a thickness of 0.5. 7. A sample of nine of their films gives the following thicknesses: 0. the alternative }. and £l' = 5%. Five measurements of the octane number of a gasoline are 80.3.90 5% the hypothesis 11.30 11.2 }.00 Test at 10.£ 8.250.00 }.7 against the alternative that it exceeds 79. which is 16.90 8.Lm against the alternative that it is not. 8 -3 -3 4 o 2 -2 -6 -I and U = 144/32 = 4.90 (~ = 7.60 9.25.2 80.1. 7.f.127 EXERCISES frequency o 1 2 40 36 29 3 33 4 5 29 34 6 32 7 8 30 31 9 26 A quick glance might suggest that there arc too many zeros and not enough nines.200.20 11.8 79.Lm.3. 7.270. That is considerably less than the tabulated value of chi-square with nine d.L 0. A sample of ten devices was tested for speed. therefore.6 80.4. The frequencies of the devices in MHz were 10.11 0. L = 13.50 against the alternative Ha: p > 0. 7. test the hypothesis Ho: tive I. 7. Test Ho: p = 0.3. In a samplc of 80 freshmen engineering students.31. how many observations should we take in order for a 50% increase in the sample variance. Test. A sample of 50 components from vendor A has a mean time between failures (MTBF) = 402 hours. and the rest are particolored.50.0 at a = 10%. Test the hypothesis that the average MTBF is the same for both vendors against the alternative that it is not. A company buys components from two vendors.L > 13. When they bloom. What is the value of f3 if 0 = 3000? 7. 703 are Democrats. S2. . 32 were exempt from the first semester of calculus.10.0. Is this sample compatible with the claims of the nursery that white.11.6. and red flowers ought to occur in the ratios 9: 6 : 1? 7. Following exercise 7. particolored.L = 12. the fraction of students exempt was p = 0.L ~ 12. A sample of 18 items are put on test.7. Assuming that the lifetimes are exponentially distributed. In previous years. 81 are white.8. The average lifetime is 2563 hours.13.0 against the alternative hypothesis (T2 > 4. In this an unusually good year? 7.12.5 against the alterna- I. A samplc of seven items from a normal distribution has a sample variance S2 = 6. Is an apparent 50% increase in the variance significant if it is based on only seven observations? 7. HYPOTHESIS TESTING I. The company gardener plants 120 flowers in a large bed outside the main entrance to the factory. We wish to test the variance of a normal distribution. A poll is conducted on the eve of a closely contested congressional race. test the hypothesis Ho: tive I. using a = 5%. a sample of 50 from vendor 8 has MTBF = 451 hours.9.128 7. Out of 1300 voters polled.7.2. Let p denote the fraction of Democrats in the electorate. over the hypothesized variance to be significant? 7. at a = 5%. test the hypothesis Ho: (J = 2400 against the alternative Ha: () > 2400.00 against the alterna- 7.5. the hypothesis (T2 = 4.5. 20 are red.04. In exercise 6. In exercise 6. 129 . ILl . XI' x 2 ' . estimating and testing hypotheses about the mean. first. IL. . comes from a normal population with unknown mean ILl and known variance The second sample. or to test the hypothesis Ho: ILl = 1L2. we discussed some common statistical procedures in which the emphasis was on drawing conclusions about a population from a single sample. when their variances arc known.•. u 2. 8. which does not assume normality. . Since the first sample has m independent observations. INTRODUCTION In chapters six and seven. Is the yield of our plant the same whether we use hydrochloric or nitric acid as reagent? Is the yield of wafers the same with two different etching regimes? Do we get less variability in our product whcn wc lower the operating temperature? We begin by comparing the means of two normal populations. M. i is normally distributed with expectation ILl and variance u7lm. John Copyright © 1990 by John Wiley & Sons.1. . This is followed by Wilcoxon's nonparametric test. which are similar problems. when the variances have to be estimated from the data. consisting of Y 1> Y 2' . tTi.. We turn now to comparative experiments in which we compare two populations and ask whether they are the same or whether they differ.X m .Statistical Methods in Engineering and Quality Assurance Peter W. of a normal distribution from a sample of n obscrvations. The first. for example. u. and variance. later. The chapter ends with a discussion of comparing two binomial populations and the use of chi-square for analyzing 2 x 2 tables. Inc CHAPTER EIGHT Comparative Experiments 8. . We estimate ILl by i. We considered. Y n' comcs from a normal population with unknown mean IL2 and known variance We wish either to estimate the difference. COMPARING TWO NORMAL MEANS WITH KNOWN VARIANCE Suppose that we have two random samples.2.1L2. Then we discuss comparing the variances of two normal populations and comparing the means of two exponential populations. and. H3 + 1.1 and y= 72./-Lz < i - Y + 1.9 69. we obtain the following two-sided 95% confidence intcrval for 11-.10 -72./-Lz: i - Y- 1.93 . we replace i by or .2.8 70. 96uV 11m + lin < 11-1 .2.2. -.h ' estimate 0 f 11-.79 3. o .7 74.1 75.4 68.2. = 11-2 is 1.1.6 72.ln.2.1.2 74.93.2 70.N(J.913 = 2. If we assume that the variance of A yields is 6. .3) and x-y -====== (8.ln < 11-1 . (8. Example S.2.0 70.11-2 < -l. . A or B.1.2.1) Similarly. a z-tcst for the hypothesis 110: 11-1 .2. The z statistic for the hypothesis that 11-.96Y uilm + IT. is an unbiased . = u 2.1) and (8. If.4) z== ITv'T1fn""+ lin .5 74.2 72. Y- equations (8. Twelve runs with each of the reagents are made and the following yields are observed: A(x): B(y): 71.2) take the simpler forms 1. a 95% confidence interval for the difference between the population means is 71. in equation (6.5 75. The two sample means are i = 71.11-2 Wit variance u.3 74.96\1'6/12 + 4/12 < 11-1 .1 74.8 71.1 72.04 .96uv'TTm + 1In (8.3 66.7 68.96Y u~/m + u./-Lz < -0. ») .3 70.130 COMPARATIVE EXPERIMENTS Similarly.ln).ILl < i - Y + 1.L2 .83 z = 0. which is barely significant.1).R 72.2) If u~ i - = u. u. can be used.y and ulYfi by Yu~/m + u.7.0.21 m + IT 221 n.00. A dntplot of both samples on the same scale (see Figure 8.ln.62 < 11-1 .11-2 = 0 is obtained by setting (8.1 72. A chemical engineer has a process in which either of two reagents.0.6 73. d = i -)1.1) suggests that the yields with B might be higher. Their difference. and the variance for B is 4.1 73. we calculate estimates from the two samples: = 6.f.0 •• • I 68.541.1 d. m + (n ) 2 -~ (8. . The pooled estimate of the common variance is S2 = 5.1.131 UNKNOWN VARIANCES •• I 66. (8.2 d.3. Dolplol for t!xample 1l.592. the test statistic becomes x-y / ' ' lI'' 'm=''=+=::l=::/='' (=s-Y (T in (8.2. By substituting s for equation (8. We deal first with the case in which it is reasonable to assume that the .f.3) We illustrate this by returning to example 8.2. i - Y- t*sVl 1m + I In < ILl .3).2. where t* is the tabulated value for t with m + n . two vanances are equaI to (T 2• Let and Then 5 l /(T2 has a chi-square distribution with m .1.513 and = 4. the confidence interval is obtained by substituting s for (T and t* for 1.L In a similar way.0 I 70. s. UNKNOWN VARIANCES 111e situation when the variances are unknown has to be separated into two cases.1.2.f.2 d.4).0 • •• A •• I 76.0 • •• • • • • • • I 72.1 ) + n-2 is an unbiased estimate of (T2 with m + 11 .I d.2.1L2 < i - Y- t*sVl 1m + lin .). Example S.2. If we do not assume that the variances are known.3. 8.96 in equation (8.1)5.3. (m . and 5 2 1(Tl has a chi-square distribution with n .J)(T2..0 •• • • I 74.2) This statistic has Student's t distribution with m + 11 . and so .t'. s.2 d.0 B Figure 8. and E(5J = (n .1 (cont.3. ).4. There does exist an approximation. which is less. UNEQUAL VARIANCES When it is not reasonable to assume that the two variances are equal.90. and so the data are compatible with the hypothesis Ho: 1-'1 = 1-'2' The (-test statistic is t =- 1. but now it is routinely available in many software packages. and so the two-sided 95% confidence interval for the difference is calculated by equation (8.ln < ILl .1-'2 < 0. in which case a 95% confidence interval is given by . We can certainly estimate the two variances by 2 81 m -] S =-1 and and we can then write the test statistic as (8. the situation becomes difficult mathematically and is known to statisticians as the Behrens-Fisher problem.074. you are not going to get into serious trouble by acting as if equation (8.JLz <.99 . Ll 8. in absolute value. than the tabulated value for alpha = 5%.3. the Minitab program gives the as i' = -1. value of I' For this data set.96Ysi/m + s.1-'2 < -1.i - Y- 1.82 < 1-'1 .16.83 + 1. this approximation was awkward to use.3): --1. 2.9] . and so we do not reject the hypothesis of equality.83 0.4.f.2) Example B. This interval contains zero.4. J) has a normal distribution. If m and n are greater than about 20. before the days of computers.96Ysi/m + '\'.074.2.074VS.1 ) The mathematical problem is that this statistic does not have Student's distribution and the actual distribution cannot be derived in a tractable form.917 = -1.3. (8.S41/6 < 1-'1 .2.132 COMPARATIVE EXPERIMENTS The 5% value for / with 22 d. is (* = 2.i - Y + 1.4.In . .1 (cont.83 -. for that stage and that stage alone.) = 2u z. there are 12 wafers receiving treatment A and 12 receiving treatment B. . In the earlier treatment of the reagent data. and ell' ez.). we had a simple model: and where i-Ll and P-z are the unknown constants whose difference we are trying to measure. one-half received reagent A and the other received reagent B.1 as if the 24 observations were all independent of each other. we take another look at the data of example 8.. the other sources of variation have canceled out. are independent random errors. + ez.f. The value of t' in this case is not much different from the t value of the previous section. Now we add the batch term and write and y. The engineer. which we assume to have the same variance (]'2. u 2 and u. The difference. Other expressions are also used. ) = V(Y. and the appropriate analysis is called the paired t-test.1 (cont.4 d. where f3. x. and the differences: The data have been rewritten to show the pairs .) = u. the variance between batches. V(x . To illustrate this. In semiconductor work. denotes the contribution from being in the ith batch. The test procedure consists. of subtracting Yi from x. This is called a paired experiment. Example 8. free of the batch effect. THE PAIRED t·TEST We treated the data in example 8. took 12 batches of raw material and divided each into two halves. . of wafers is split into two halves for one step of the processing.2.P-z. is The batch component has vanished and V(d. in each pair and carrying out a one-sample t-test on the differences.2. 0 8. which had canceled out. = i-Lz + (3." perhaps 24. split-lot experiments are run in which a "lot. + (]'2. Then the difference between the two observations made on the same batch was a measure of i-Ll .5.. Suppose that the engineer was aware that there were two sources of variation in the process: random noise and also variability between batches of raw material.1. The variance of a single observation has two components.2.Y.133 THE PAIRED t-TEST which should be compared to the tabulated value of t with approximately 22. therefore. therefore. the largest has rank m + n. they add up to (m + n)(m + n + 1)/2.e.9 -2.5 72.7 75.1). Wilcoxon's test can be as good or better than the I-test.0 Pair A B d 7 73.2 -3. Whitney. The first step in this procedure is to arrange all m + n observations in increasing order of size.2 -2. The test statistic is the sum of the ranks of either the X or the Y observations. a test that does not depend on assuming that the populations arc normally distributed. Fewer degrees of freedom means that a higher value of /* must be overcome to establish si'gnificance.8 1 5 71. the increase in precision from eliminating it will more than compensate for that increase in the critical value of I.0 0. the Mann-Whitney version is more popular among social scientists.8 -1.1 8 66. i. If the X median is appreciably lower than the Y median..9 9 72.5 -2. the average rank of the X variables should be about the same as the Y average.53 with 11 d.7 70.4 74. WILCOXON'S TWO-SAMPLE TEST During World War IT.f. we give each of the tied observations the average of their ranks. When there are departures from normality.6 74. developed a similar test that bears their names and is mathematically equivalent. From now on. One chooses whichever . At about the same time. we are testing the equality of the population medians rather than their means.2 4 72. Formally. R.2 2 68. However.2 72.6 0.1 0. The smallest observation has rank one. an industrial statistician. but if the variance between batches is considerable. Mann and D. which is clearly significant.134 COMPARATIVE EXPERIMENTS Pair A B d 71. Notice that the number of degrees of freedom has dropped to 11 (12 .1 75. the twosample I-test is the "best" test when the conditions of normality and equal variance are met.1 72.1 12 68.3 69. developed a nonparametric test.6. If there are tics.3 70.3 -2.8 10 11 74. and so on.8 74.6 3 74. H.7 ··-1. Frank Wilcoxon. Theoretically. If the medians are the same.1 70. we work with the ranks of the observations.1 -3. o 8. the X obscrvations will tcnd to he clustered to the left and the Y observations on the right.1 73. . working at Ohio State University. the superiority of the (-test over Wilcoxon's test is not great even then.6 The one-sample {-test on the differences gives {= 4.6 6 70.8 -1. 9 Y . Tables of the critical values of Ware available for various values of m and fl. each X observation is given a score equal to the number of Y observations that exceeds it. three red and three white. the sum of the Y ranks has E( W) = n(m + n + 1) 12 and the same variance. the highest 15.4 X 8 70. We can now formulate a simple test for the hypothesis that the two populations have the same median.2. The theory of the test can be derived by thinking of a box with three red balls and three white balls. 10 11 12 9 10 11. We are merely applying that approximation to the orderings of m red balls and n white balls as they come out of the box! This is illustrated by going back to the original data of example 8.1 X 7 70.1 and ignoring the remarks about pairing that were made in section R. a = 1/20 + 1/20 = 0.1 (conI. and only if.05.5. usually the sum that has the fewer observations to bc added. we can act as if the sum of the X ranks is approximately normal with E(W)=m(m+n+l)/2 and V(W)= mn(m + n + 1) 12. namely. When m and n are ten or more.2 Y 5 70.13S WILCOXON'S TWO-SAMPLE TEST is more convenient.5 X X X The 24 ohservations in ascending order are 4 69. reject the hypothesis if. Example S. We remove the balls from the box one at a time and note the order of the colors.). They are XXXyyy XYXYXY YXXXYY YXYYXX XXYXYY XYXYYX YXXYXY YYXXXY XXYYXY XYYXXY YXXYYX YYXXYX XXYYYX XYYXYX YXYXXY YYXYXX XYXXYY XYYYXX YXYXYX YYYXXX The sums of the X ranks for the 20 sequences are W = 6 7 898 13 12 13 14 9 10 15. Consider the simple case where m = n = 3. In the Mann-Whitney test. For this test. 1 66.2. Note that we are not making any normality assumptions about the actual data. II 12 The lowest value is 6. and the test statistic is the sum of those scores. We suppose only that we have 12 observations on A and 12 on H.e. if X has the three lowest observations or if X has the three highest observations.3 3 68. i.0 Y 6 70.3 2 68. and so there are 20 possible orderings (ignoring ties) of X and Y. W = 6 or W = 20. There arc 20 possible arrangements of six balls.. each of those scores has probability 1/20. 1 Y 72. the table of critical values consists of the three numbers 7. 10 is the 1% value.8 Y Observations 11.) Under the hypothesis of equal medians.5 Y X Y 75. and 13 in order are all equal. we subtract 1 from D. 13 is the 0. is the sum of the two overlaps. 10. but it is easy to use and the table of critical values can be carried in your head.2 X 11 (72. we reject the hypothesis of equal medians at a 5%. (11 + 12 + 13)/3 = 12. "not only quick but compact.7.3 X Y Y 22 74.6 X 10 71. Tukey produced a test that is. in his words. They each receive the same rank. Suppose that the smallest observation is from the X population and the largest from the Ypopulation." Duckworth wanted a quick·and-easy test for comparing the medians of two populations that did not need any tables and could be used for a wide range of m and n.8 23 24 74.8 16 73.1 74.150 == -1 67 V 300 . 12.136 COMPARATIVE EXPERIMENTS 9 71.2 20 21 74. THE DUCKWORTH TEST In the first issue of Technometrics. If either 3 + 4nl3 s m s 2n. . (It is a good idea to check your calculations by adding the Y ranks. the two should add to (24 x 25)/2 == 300. or vice versa. and 13. The test only works if the largest and smallest values of the observations eome from different populations. o 8.6 X Y Y 14 72. The test statistic is Z =: 121.1) X 18 19 74.. that total is 179. too. W is normally distributed with IL = 150 and (1'2 = 300. It is preferable that the two samples have approximately the same number of observations. The sum of the X ranks is 1 + 2 + 3 + 6 + 7 + 9 + 10 + 12 + 14 + 17 + 18 + 22 = 121 .7 75. John Tukey (1959) presented a simple test that he had constructed to meet "Duckworth's specifications." This test is not as powerful as some of the more elaborate procedures. the number of X observations that are smaller than the smallest Y plus the number of Y observations that are larger than the largest X. D. If D 2!! 7. or at least that neither mIn or nlm should exceed 2.1 X 17 73.7 15 72. The test statistic. Under these circumstances.1 % value.1 Y 12 13 72. The use of the table is illustrated in this example. . o 8.8. 15) = 40 = 2.8. we note that there are three X observations below all the Y's and one Yabove all the X's.137 COMI'ARING VARIANCES Example S. .1) If F is close to unity./s~ of equation (8.'1 (8. The numerator is (Sl/O'~)/(m . Using the Duckworth test for the data of example 8. 15 d. and so F is the ratio of two independent chi-square variables. which is not significant at a =.f. To derive the distribution mathematically.. 10 d. a complete table would occupy many pages. Because there are two parameters involved.1 (cont. and F reduces to the simple variance ratio s.4.1). = 80.1. which implies that only values of F greater than unity need to be given. We compute the variance ratio with the larger estimate in the numerator: 80 F(1O. In section 8.8.1).). 5%.2.f. si Example S. we conclude that the variances are different. some hooks economize by letting be the larger of the two estimates and s. The F statistic will sometimes be denoted by F( 4>\. in the denominator. In the present situation. .l. the smaller.f. we start with the ratio JO' 21 s. each divided by its degrees of freedom. Suppose that two samples have the following estimates of the variance s. and s~ = 40.'/O'i S 2 F= -.t. The distribution of F depends both on the number of d.0. How do we test the hypothesis o'~ = The test statistic is the variance ratio 0'.0.0.1). in the numerator and the number of d.S.-. The total is D = 4. If F is large or small. the denominator is (S2/(T~)/(n . 4>2)' Table IV (see Appendix) lists critical values of the F statistic. which we can denote by 4>1 and 4>2' respectively. we obtained estimates s~ and of the variances. we hypothesize that the two variances are equal.2. we accept the hypothesis that the variances are equal. COMPARING VARIANCES s. For a 95% interval. that difference is not significant.138 COMPARATIVE EXPERIMENTS The table gives the value of F that corresponds to the given percentage in the upper tail. good estimates of variance need plenty of data. we want to compare F with the value that would be exceeded by chance 5% of the time (not a/2).L can be a long way off. 0 8.9.54 . In fact. 15) = 2.a)% confidence interval by writing then. Since we have already taken care to put the larger estimate in the numerator. CONFIDENCE INTERVALS FOR THE VARIANCE RATIO We saw in the previous section that the statistic that actually has the F distribution is We can.025. The appropriate value is found in the table under a = 0.1 ) One more difficulty remains. construct a (1 . we get Unfortunately. The confidence interval that we are seeking is thus (8. the value of F that appears on the right-hand side of equation (8..9. and so we have to switch the subscripts throughout.1) is found in the table under a = 0.9. dividing through by sils. It can be obtained by using the following equation: (8. It is perhaps discouraging to see that even though the one estimated variance is twice the other. this is upside down. therefore.9.2) . and estimates based only on a few d. The value on the left side is not in the tables.05 in the tenth column and the fifteenth row: F*(lO. . respectively./2(2n. _m) .1) to the tabulated values of F(2m. First. We can. 5) 3.06 < 2 .8.a)% confidence interval for (J1/82 can be obtained by the methods of the previous section as ily £. therefore. The second. We can then compare (JI and 82 in two ways. The first sample..10. Yi> Y2'" . (·_1) IT.2n) distribution.. and take the ratio. their average lifetime is 178 hours. (xly )F..2. Twenty components from supplier A are put on test.).10.1. A (I ~.52) ( ..2) V2 Example 8.f. we can act as if x and yare both approximately normally distributed with variances Xl and y2. (T~ < (2. 30 components from supplier B have an average lifetime of 362 hours. XI' x 2 ' ••• . (Tz 8.139 COMPARING EXPONENTIAL DISTRIBUTIONS Example 8. n _81 / <-. There is an alternative way that has the virtue of not being an approximation.1 (cont. . We assume that . 2n). test the null hypothesis (J\ = ()2 by comparing the test statistic Ho: (8. COMPARING EXPONENTIAL DISTRIBUTIONS Suppose that we have samples from two exponential distributions.00.Yn' comes from a population with expectation (J2' We estimate (JI by x and (Jz by y. o (TI 0.. the variance ratio: We can now obtain a 95% confidence interval for .. _ _ ? II -. Method 1..10. (8. we have an F statistic: has the F(2m. If we divide each of them by their d.xm • comes from an exponential population with parameter (JI. and then follow the procedure of section 8.82<"2 <8.' ral2 (2 2) m. respectively. We recall that 2 ~ x/81 and 2 ~ y/82 have chi-square distributions with 2m and 2n degrees of freedom.5)(3.10. J78 . n We recall that the variance of the number of defectives in a sample of size n is npq..1.2 z = and _ V(y) (362)2 = 3 ( ) =4368.33 . Suppose that the first sample contained x defectives out of m parts and that the second contained y defectives out of n parts. and at a = 5%.02 < -184 + 151.1.362) .335 < 0\ . 60) < 02 < 362 [FO. Under the null hypothesis that PI = P2 = p. A random sample of 200 parts from supplier A contains 30 defectives. we would decide that the reliabilities differed.2.2 .2 + 4368. Arc the two suppliers producing at the same rate of defectives? We can act as if the fraction defectives in the two samples are normally distributed and construct a test and a confidence interval along the lines of section 8.02 < .96\1'1584. to be compared to FO.64. 40) The confidence interval is = 1.11.1 < 0\ .05(60. ~ =Y -.28< 8\ 8" <0. a random sample of 100 parts from supplier B contains six defectives. 178)( 1 ) 01 (178) ( 362 Fu u2s(40. Method 2 362 F= 178 =2.140 COMPARATIVE EXPERIMENTS _ V(x) (178)2 = --W = 1584.362 = 2 38 v'V(i) + V( y) . COMPARING BINOMIAL POPULATIONS A manufacturer receives supplies of a certain part from two suppliers and wishes to compare them. the common value p is estimated by pooling the two samples: .40)] 0.88. A 95% confidence interval for the difference between average lifetimes is (178 . the estimates of the binomial parameters are X P\ =m A and A p.03.. 025 (60. o 2 8. we no longer have to assume that PI We take the best estimates of the variances: = Pz.15 p.CHI-SQUARE AND 2X2 141 TABLES X +y P=m+n· A In constructing the z statistic. n For the data at the beginning of this section.96Y(V(xlm) + V(yln» <PI .0. For the confidence interval. we take V(xlm) whence = pqlm and V(y/n) xlm .0006375. n = 100. and the (0. x = 30.11.12)(0.15 .1.12.96Y(V(xlm) + V(yln» .11. = 200.02 < PI . def.0.06 z = \1(0.1) .1. P2q2/n interval is = 0. the table would be Supplier nondef.]6.1. (8.l m n -1.OO12 < PI . m Example 8. A B 170 30 94 6 .88)(1/200 + 111(0) = 2.96v'0. PI = 0.P2 < ~ .12.06.yIn z = Vpq(llm + 1 In) = pqln.l .000564.:: pA2 .:: 0.15 . CHI-SQUARE AND 2 x2 D TABLES A commonly used alternative to the z-test of the previous section is to set up the data in a 2 x 2 table (a table with two rows and two columns) and make a chi-square test.pz < 0. n and A two-sided 95% interval is given by ~.:: 36/300 = 0. Pooling the samples.26. and P2 = 0. y = 6.09 + O'(17 0. V(y).06) . S. For the confidence interval.pz <0. For this data set. Plqllm. m 0. there is only one degree of freedom. the probability that it would be a nondefective is 264/300 = 0..2 54.1 55. Similar calculations show that the expected values in the other cells are 88.3 34.6 46.bc)/N.11 = z· In the general case. A B 170 30 200 94 6 100 264 36 N = 300 If the probability of a nondefective were independent of supplier.o. = -6. Twenty wafers were sent to a vendor for a boron transplant. nondef.4 48.142 COMPARATIVE EXPERIMENTS We can add the marginal totals and see immediately that if a part were to be chosen at random from the combined sample. .5 48. The deviations all have the same absolute value.7 55. -6. and the probability that it carne from supplicr A is 200/300.4 57.1. The yields were 54.9 52. the expected number of nondefectives from A would be 300(0.7 63. h. with deviations +6.3 Control 59. and 24. The chi-square statistic may be calculated by the formula: u= --:-----:-:-:--N.C)-:2---:-:--c-:(a + b)(c + d)(a + c)(b + d) . Twenty received the standard deposition.5 31.4 . . 176 + 88 + 12 + 24 = 5. we have for that cell 0= 170. After the treatment. 12.. wc ean denote the cell frequencies by a.4 44..88.. respectively.9 46. The value of the chi-square statistic is U = 36 36 36 3 6 .2 45.-_h-. Thus. +6.88)(2/3) = 176.7 31.(a_d:-:--..3 66. def. all 40 wafers were processed together and the probe yields were measured. c.1 62. 0.7 43.E E = 176. and d: a c (/+c b d b+d a+ b c+ d N=a+h+c+d The deviation in each of the four cells is ±(ad . EXERCISES 8. For vendor B.143 EXERCISES 54. a layer of gold.7 12. Devices from two manufacturers were said to have the same nominal speed (frequency of operation). s = 1.5 12.6 Boron 51.7 9.8 13.7 13.9 14.1 74.1 B: 16.6 17. The specification calls for (I to be 325 hours or longer.5 16. During wafer preparation.7 12.3 Was the boron implantation effective? 8.3 12.7 11.1 B. 8.4 12.6 9.7 13.2 13.6 15.8 75.3.8 49. the following results were obtained: For vendor A.6 49.2.1 14. is sometimes sputter-deposited onto the back of the wafer.8 84.8 12.0 13.0 13. 8.4.2 13. The mean times between failures (MTBF) of an electronic component have an exponential distribution. s = 2.3 57.9 17.5 52. Wilcoxon's test.3 13. A summary of their results is: Control average mpg = 31.3 16.7 48.2 82.8 13.6 12.2 12.9 12. 32 out of 45 units tested were void-free. 8. Twenty-two devices from manufacturer A and 25 from manufacturer B were tested at room temperatures. The engineers in the research group take tcn cars and run them on the company's regular brand and another tcn cars that they run on the regular gasoline with additive. and Duckworth's test.5 10.2.0 16.2 13.5.9 17.9 13.1 34.4 Is the new additive effective in increasing the mpg? Is there greater variability in mpg with the new additive? .6 Do the devices from the two manufacturers have the same average speed? Make dotplots of both samples on the same scale.5 56.4 18. Test the hypothesis Ho: 0 = 325 against the alternative HI: (:J < 325.7 13.4 55.3 52.8 14. Eighty failures were investigated and the mean time to failure was 282 hours. Their observed frequencies in MHz were A: 16.8 Additive average mpg = 32.5 84.6 16. In a comparative test of two vendors.3 54.7 12. An oil company is considering introducing a new additive in its gasoline that it hopes will increase the mileage per gallon.0 13.7 13.7 52.5. 3000 Angstroms thick. Use three methods for this exercise: the t-test. Test the hypothesis that the percentages of void-free units for the two vendors are equal. 39 units out of 76 were void-free. Denote the MTBF by (I.0 14.1) 12.4 13.9 17.4 49.2 61. we cannot construct a process that will continue forever to produce absolute clones. INTRODUCTION A great deal of the industrial production in this country comes from production line work-a method of production that calls for carrying out a repetitive process again and again and again. The latter cost more. you would do well to have a mix that is robust to the fluctuations in ovens from house to house. We arc going to assume in the next few chapters that we havc a production process. No matter how hard we try. M. that are identical with one another and each conforming exactly to the standards that we had in mind. if you are selling cake mixes to be made into cakes in the huyers' own kitchens. although we can reduce it. Indeed. or within a tenth of a degree. both in initial capital investment and in maintenance. we can eliminate almost all the noise. and out of the other will come a steady succession of objects. and there is variability in the raw material. Some of the noise we cannot eliminate. We also assume that we have some measure of the quality of each widget. all we have to do is to put in raw material at one end. It is convenient to say that we arc producing widgets. called the product. It would be nice to think that once a production line has been set up. out of which comes a product. These measures may be discrete. each production process has built into it a certain amount of natural variability. Components in the production machinery wear out. Unfortunately. which we can call noise. life is not like that. if we try hard enough. In making a cake on a production line in a bakery. We can construct ovens that maintain a desired temperature within 5 degrees. we cannot persuade the hens to lay eggs that are absolutely uniform in size or the cows to produce cream that does not vary in the percentage of fat from day to day. Perhaps. Whether we like it or not. If the cake recipe is robust to temperature changes. as in chapter 144 . John Copyright © 1990 by John Wiley & Sons.1. Inc CHAPTER NINE Quality Control Charts 9.Statistical Methods in Engineering and Quality Assurance Peter W. which is statistical jargon for saying that a deviation of a few degrees in the oven temperature will not seriously affect the palatability of the product. it will not be cost efficient to invest in the highest-costing ovens. grade 3 is sold at a discount to college residence halls. and still does. or five. widgets off the line at regular intervals and measure their lengths.2. that situation is described as sampling and classification by attributes. the research arm of American Telephone and Telegraph Company. In 1931. we discuss quality control charts for continuous variables. and grade 4 is inedible. What Shewhart did was to introduce a simple graphic procedure for monitoring the output of the process that enables the manager to tell whether the process is in a state of control. that arc difficult to quantify. grade 2 goes to the grocery stores. there is a stir in the shop. If the process has gone out of control. That is a more complicated situation than we want to discuss at this time. . QUALITY CONTROL CHARTS Statistical quality control charts were introduced about 60 years ago by Dr. we could grade each cake on a scale of 1 through 4. we clearly have the attribute case. We plot the average length of each sample on one chart and the range of each sample on the other chart. it is the subject of the next two chapters. On the other hand. 9. the manager gets a signal to take action. at least. In this chapter.QUALITY CONTROL CHARTS 145 three. a scientist at Bell Labs. At that stage. which seems to improve things. Perhaps they introduce inspection of the outgoing product so that they can. Perhaps they make some ad hoc adjustments to the machinery. such as palatability. cut down on the number of unacceptable widgets that leave the premises. is that people set up the machinery and start production. We can get into arguments when we try to measure characteristics. defective or nondefective. They continue to produce widgets as long as the items look satisfactory and there is not a preponderance of complaints from the clients. The clients in this sense may be the actual end users or they may be the people in another plant some miles away who are using the widgets in the construction of some more complex pieces of equipment. Walter Shewhart. conforming or nonconforming. In the jargon of quality control. These charts are then used to tell us whether or not the process is in control. Production continues until it dawns on someone that they are now producing unacceptable widgets. The hasic idea is that we take a sample of four. If we either accept each cake or reject it. he wrote a book entitled Economic Control of Quality of Manufactured Product. Grade 1 cakes we sell to the best restaurants at the premium price that they deserve. Then each widget is classified as either acceptable or not acceptable. "tweaking" this knob or that. The measure of quality is something like the weight or length of the widget. It is often rather haphazard. What happened in those days in too many shops. We then prepare two charts. along the vertical axis. i. It could have happened by chance. We stop production and. With each sample. Is it moving along steadily? 9. when we look closely. For the present. we say that the process is under control. In the United Kingdom. This cause. How do we make adjustments to get the process to meet specifications? That is the topic of process improvement (off-line experimentation). we take a random sample of n widgets from the production line. The lower line is 3uivn below it. With a normal distribution of noise. From the point of view of hypothesis testing. we seek an assignable cause. We plot the sample averages. If the probability . may turn out to be something serious that calls for corrective action such as replacing a part of the machinery or an operator who has not hct:n properly trained.09uiv'fi above and below the center line. but only about one time in 400. in the jargon. usually denoted by i. Some engineers call each sample a rational subgroup. the probability that a point falls outside those limits is only one in 400. or sample number. or at our estimate of that distance. The average length of our widgets may not be what the customer wants. only one time in twenty will a variable differ from its mean by more than two standard deviations. Under the null hypothesis of control. The upper control line is at a distance 3u/vn above the center line. The area outside the outer control lines is the critical area. If a point lies either above the upper line or below the lower line. or the variability may be too great. THE x-BAR CHART At regular intervals. that the capability of the process is not what we would wish. we have a null hypothesis that if the process is under control. A process is said to be in (statistical) control if it is performing within the limits of its capability. which is to be discussed in the final chapters. It may only have been a passing phenomenon that is easily explained and is not expected to recur. when found. The center line corresponds to the process mean. we say that the process is out of control. with time.3. We can now give a definition of control in this context. Commonly used values of n are 4 or 5. there is a certain amount of regularity. The other is that in a random process.146 QUALITY CONTROL CHARTS Shewhart's ideas depend upon two statistical concepts. it is more usual to place the outside lines at a distance 3. we are concerned with the stability of our process. we carry out a test. that the product being manufactured and the customer's specifications are some way apart. The first we have mentioned earlier: a certain amount of random noise occurs naturally in any production process. As long as the plotted points fall between the two outside lines. The piece of paper on which the sample means are plotted has three horizontal lines. is normally distributed with expectation /J.and variance u 2in. the random variable. along the horizontal axis. such as every hour. We may find. or rejection region. or once a shift. SETTING THE CONTROL LINES When we speak of i having expectation j.693 1. Everyone is involved. and a simple action. Its simplicity is one of its great virtues.4. One of the workers is assigned to make the entries.534 (l. Everyone can focus on the problems of quality.14.4l-l3 0.975 2. Then we can multiply that estimate of the standard deviation by 3/vn. or decision. The_range.06 for II = 4 and 2.326 0.704 O. We recall that d 2 = 2. (T21 n).000 .12R 1.955 2. we are talking about the mean and variance of the process as it actually is and not about the specifications. A pilot set of about 20 rational subgroups usually suffices to get started.L. We have to estimate those two parameters from a pilot set of data taken from actual production.L and variance (1'2In. We now have a simple graphical procedure that shows how the process is behaving. Their grand average is denoted by R.577 0.880 1. Table 9. as in section 6.33 for n = 5. the procedure is a little different.4. In practice. R" of each subgroup is recorded.059 0. 1t can be divided by d 2 .147 SElTING THE CONTROL LINES distribution of i is really N( j.1. and fixes the center line of the chart. The upper and lower control lines are then set at Values of d 2 and A z are given in table 9. It provides our estimate of 1-1-.933 2. 9. Factors for the x-Bar Chart n 2 3 4 5 6 7 d2 1. That figure is the alpha risk. Everyone. workers as well as management. 1.729 0.1. The standard deviation has traditionally been estimated from the ranges.4. to obtain an estimate of the standard deviation.023 0.4II) O.992 2. The average of the individual i is denoted by { and called x-bar-bar. We define another multiplier. The two steps are combined into one. the probability is 0. can see how their process is doing. The chart can be kept on the shop floor.I)J 1 A2 Eff.0025 that we will mistakenly stop production and that there will be no assignable cause to find. rule. is not an unbiased estimate of the standard deviation. but the concept of range as a measure of variability is easier to understand than the concept of standard deviation.> is plotted along the vertical axis. It can be argued that this advantage is less important today than it was 25 years ago. The center line is set at R. The rationale for computing multipliers D3 and D4 is now shown for the case of n = 4.4) -I. we want everybody to be able to relate to it. where c 4 is a multiplier whose values are given later in table 9. every modern engineer carries a hand calculator that can compute /.076 3. Denote the range by w.12. The chart is meant to give everybody a picture of what is happening. one could obtain an unbiased estimate of the variance from the squared deviations in the usual way.12. one computes S2 = L (x.~ are the topic of section 9. Not only can the worker on the floor easily calculate the range of a sample of four observations. The simplicity should not. Charts based on .1. The primary advantage of using the range to estimate u is its ease of calculation. s.574 2.1J04 1. be dismissed.282 2. Table 9.924 . The asymptotic value of 11 c4 for larger n is t + (4n .1.114 2. there are three control lines. however.428w .1. E(w) = 2. Multipliers for the Range Chart n 2 3 4 5 6 7 o o o o o 0. Values of D3 and D4 are shown in table 9. Again.12. R. The sample range. - i)2 n -1 The square root.148 QUALITY CONTROL CHARTS Alternatively. the average from the pilot set. The range estimate for u has very high efficiency by comparison to s/c 4 • Those efficiencies are also given in table 9.059u = d 2 u and that Yew) = O. The upper and lower lines are obtained by multiplying R by D4 (upper line) or D3 (lower). and s/c 4 in an instant.267 2. A similar chart of the sample range keeps track of changes in the variance.5.1. For each sample. This is a diagnostic tool.5. THE R CHART The x-bar chart keeps track of changes in the process mean over time.5. The unbiased estimate is s/c 4 .183w 2 and sd(w) = 0.775Ru 2• Substituting w/d z for u in the latter expression gives Yew) = O. It can be shown that if the observations come from a normal population. 9. the upper and lower control lines are set at 10.6. To compute 0 3 . the process is certainly not under control. which is not true. we set the upper control line at 0 4 = W + 3sd(w) = w + 3(0. and so we replace it by zero. and a process that produces outliers has problems. Had that value been. which has a sample mean 11.30.13 + (0.13 .284w.1 and 9. However. If the variability of the process is out of control.149 AN EXAMPLE where sd(w) denotes the standard deviation of w. The R chart is obviously sensitive to outliers. whether you like it or not.50. The data are in table 9. Remember that. Following the same procedure as for the x-bar chart. even when the observations are normal. this is the rationale that has traditionally been used. The charts of the pilot set arc in figures 9. The process is clearly rolling along steadily.80. instead. we .2. 9.13. The distribution of w is certainly not symmetric. It is possible that the search for an assignable cause will find a reason that could justify the engineer in throwing out that point and recalculating the estimate of (T without that sample. AN I<:XAMPLE The following pilot set of 20 subgroups of four observations each was used to set up x-bar and R charts for a process. It can be argued that the engineer should look at the R chart before looking at the x-bar chart.82) = 11.1. The difference between the factor 2.282 is due to roundoff in the calculations.82) = 8.284 and the actual value of 0 4 = 2. the average range is 1. The grand average i for the 20 groups is 10.428)w . but this quantity is negative.729)( 1. we should take 03 =W - 3(0. There is a tacit assumption here that w ± 3sd( w) are the correct limits and that w is normally distributed. The only point that comes close to an outer control line is the seventh.6.82. A single bad observation can make the range of that sample so large as to exceed the upper control line. For the x-bar chart.6.45 and 10. it is an observation.729)(1. That should only be done if the engineer is very sure that it is justified.(0. as manifested by the R chart.428)w = 2.6. even though inflated ranges may make the control limits on the x-bar chart far enough apart to include all the points. 11. 3 I::---t------ til .1..8 I -------~ I ______ - I I I I I I S ~ 1 -.8 10..2 11.7 1..13 9.90 11.28 10.4 1..0 8.6 10.7 9.3 11.9 11..7 10.3 9.6 8.9 9.7 10.5 8..7 1..80572 . _.5 10.85 10...1...50 11.9 0.6 8.1 10..83 9._'_"___' 4 8 12 Sample number Figure 9..8 H>..4 9. I I I I ' I I I I I I I ..45 2.8 o I -------+---------~I I I I I I I I I I ~.9 10.1325 10.150 QUALITY CONTROL CHARTS Table 9.60 10. l .1 2.6..5 10.---------~---------~-----I I I I 1 I I I I .00 10.._ ' _ _ ' _ .2 11.2 10. t i R 10.5 11..1 11..8 1..8 7.2 1O.8 f I .6.2 2.60 R 2..7 11. 11.70 9.2 1.4 11...2 1t 12 13 14 15 16 17 18 19 20 11..0 1..I 9.9 to.4593 I I 1 I I I I I I I I I I I ~---------4----------~t I I .9 10.4 to.. l .30 10..7 11..6 9.43 9.6 10..2 11.2 9.0 8.1 10...8 8.......20 9.15 10.9 11.6 3.5 9.5 11..4 11..3 9.10 9.. 16 20 8..8 I I I I I I I 9.63 10..3 10..5 10..5 10.5 10.7 10...5 8.5 9.3 10...7 9. _.9 2. °nlC x-har chart.1 9._ ' _ _ ' _ .---+-+:+----t=I 10.5 2.6 to.0 8.8 8. l ..6 10.6 11.6 10.00 9.8 9.9 11.0 x 9..0 1.3 8.9 1..7 1. t I I I ~r:---------+---y-----------------+:-::-..1 9..0 2.8 8.7 9.3 10.0 10.8 0.2 8.3 I I I I I I I I -------+------ 10.1 9._ ' _ _ ' _ .50 10.3 1.1 11..4 9...4 to.53 10...9 9.6 10.. _. I •••••••••••••••••••••••• 1•••••••••••••••••••••••• 1•••••••••••••••••••••••• 1•••••••••••••••••••••••• 1•••••••••••••••••••••••• 11..6 9.7 10..9 9. Twenty Samples 2 3 4 5 6 7 8 9 10 to..5 11...1 9. .7. One might suspect the following scenario.. would have noted that it was above the upper control line and proceeded to look for an assignahle cause.7. It could. The upper limit for the range chart is taken at (2.82) = 4. ····· . All the points fall between the limits.15 .········.1 is.--f-~r-Ir---\---I 1... The process has an unfortunate tendency for the mean to increase.82 I I I I I I I -------'t-----I I I I -. -------~---------~---------~---------~------ 4. in this case.I ---------T---I I I I I I I I I I ----'1'------- I I I I I I I I I I I Sample number Figure 9. the (supposed) problem seems to have been temporary. for example. . on the other hand. the lower limit is zero.282)(1...-+----.6.. the operator . be due to a new operator who was a bit heavy handed until he acquired the knack. an example of a process that is manifestly out of control.. ·········~· ... It could occur by chance.. Certainly. 9... There are numerous points outside the limits on both the high and low sides.--..15356 3 2 I--+---rl--+--. The range chart. As soon as this is noticed.151 ANOTHER EXAMPLE I I I I 4 ························1························1························l······ .2. One outside point in a string of 20 points may not signify anything serious. ANOTHER EXAMPLE The process that is charted in figure 9.. ..... I I ...... tweaks one of the variables to bring the value of the mean back down......... I 1 I .. 11....... . ....... ...1. and other external sources of variation.... I 9... ...00364 20 Sample number Figure 9..... I 8 12 16 9.. but came............ 8.... I 1 .... 4....~--~--~10... ... We obtained it from the ranges of the individual samples..... It does not take into account variability bctwcen shifts.... I ....... who might adjust the machine in different ways when they start their work turn.. The process drifts upwards and is "adjusted." One piece of evidence to support this theory is that points above the upper line are immediately followed by points that arc markedly lower...6864 1 1 I I I I -of-I I 11...1 + I --------~- ------+---------+------- .. The x-bar chart..1 I ~- o 4 I I I .. the variance that we have measured will be relatively small-the variation between four items produced on a precision machine under virtually idcntical circumstances.. I .1......... It is a measure of the variability of four items made on the same machine on the same shift. In some cases... What if the outside points had not occurred in such a systematic pattern...........1 1 I I I I I ~~----~.~--~--~---.. ... Then another cycle starts. it could also include differences between batches of raw material.345 I 10..7. at random among the 20 points recorded? How might such a phenomenon be explained? It is important to recall how the estimate of the process variance was obtained...---+~~~----. . and usually overdoes it. But the variability bctween shifts could include differences between operators. seemingly..1 -------~----- I I I I I I I I 1 I 1 I 1 1 1 I I I I . .152 QUALITY CONTROL CHARTS 12. 1 .1 I ........ 65 10.. there was a value of i that exceeded the upper limit..6........90 11.8.L..I-.18 10...95 9.4~4i-...-L.1 and the first ten in table 9...J'--l.95 9...J.60 1UO 9.. The x-bar chart. 15 samples after the change had taken place.J'--l.25 lU18 11.8.80575 o 10 20 30 40 50 Sample number Figure 9.8 Ii..93 10.1.L. DETECTING SHIFTS IN THE MEAN Figure 9.I I I I I I I I I I I I I I I I I I I I I I --------~---------~-----I I I I I I I I I I 8..1-A--~----.2. There was an increase in IL of about one standard deviation.1.-L.......J'--l..3 I I I I + _________ ...-L...6.23 11...j 10. Values or i ror 30 More Samples 9... we discuss alternative stop rules and a criterion for assessing them: average run length..I-...23 10..28 10.00 10.25 10.J'--l.35 10.1...75 10.J 8...1 and 9...L.6.1 is an x-bar chart for the same process that produced figures 9.88 11...4593 11.J.. something happened to the process..13 10.65 10.---.8..L. Thirty more samples have been taken and added to the previous 20.18 10.63 10..S.1)..23 9.8 -+---------~------ I 9.. e -a5 10.58 11...153 DETECTING SHIFTS IN THE MEAN Table 9. After the thirtieth sample (the 20 in table 9...3 b--h....-4-*"...30 11..J...70 11. 11... .I-... Their averages are given in table 9.3 10.8. After a while... That happened at the forty-fifth sample..8 L-. Is that an unreasonable delay in detecting a change? In the sections that follow..30 11..L.J.60 10.-L.88 10.00 10.I-.S.1325 ell I 9.. The R chart has a run of four above the center line from 9 through 12. roughly half the points should be above the center line and half should be below it. beta is the probability that if there had been a shift in the process average. and call them inner control lines. AVERAGE RUN LENGTHS In hypothesis testing. These lines are used for the second rule. If our process is behaving as we expected. too many or too few. points. Two consecutive points. In the context of control charts. Since they should occur at random.1. Too few runs might indicate a slow cyclic trend in the process average. One point outside the three sigma lines. We have been assuming that we are dealing with a random normal process with mean JL and standard deviation u.154 QUALITY CONTROL CHARTS 9. we spoke of the alpha and hcta risks. The probability of a run of seven is 2(11128) = 0.9. 0 There are also rules that are based on the simple coin tossing model. One could have too many runs in the extreme case of two operators working alternate shifts if one operator consistently made the widgets too long and the other consistently made them too short-the sawtooth effect. or more. either both above the upper warning line or both below the lower warning line. This leads to two classes of rules based upon runs. this rule would have been triggered after the thirty-fifth sample. either all above the line or all below it. Our only stopping rule so far has been Rule J. either all above the center line or all below it.0a/vn above and below the center line.8. A stopping rule is a rule for making the decision that the process no longer satisfies the original conditions. Rule 2. Rule 3. 0 In figure 9. the situation is the same as a sequence of eoin tosses with heads for points above the line and tails for points below. 9:10. ALTERNATIVE STOPPING RULES The term stopping rule is used by statisticians working with stochastic (random) processes. there is a run of five points below the center line from sample 8 through sample 12. o Some engineers draw two more lines on their charts at a distance 2. There are also rules based on the number of runs. In the data of table 9. alpha is the probability that we (wrongly) trigger a stop rule whcn the process is still in control.6.1. We choose the . A run is a string of successive points. A run of seven. or warning lines.016. we would fail to detect it. the chance that a point will be outside the control lines is 2(0.8 1. The word run is used in a different sense from the previous section.0cTlvn below the upper line.8.6 0.3. In figure 9.L to /.()()27.L + ulvn. the number of samples until one exceeds the limits has a negative binomial distribution with p "" (1. Using rule I.2 310 0. Suppose that the mean shifts from /.4 200 0. the average run length when there is a zero shift in the mean is 370.02275 and the average run length (ARL) is 44. Here it refers to the number of samples that arc taken before thc rule is triggered. Its expectation is lip"" 370. As long as the process remains in control.9 1.9 2.4 l.0 93 72 56 44 35 28 22 18 15 Rule 2 799 349 238 165 117 84 62 46 35 27 22 17 14 12 12 10 9 10 7 8 7 6 6 . since the distance to the upper linc is now reduced to one standard deviation of i. Average Run Lengths to Detect a Shift d(u/Vii) in the Mean d Rule 1 0.3 1.5 1. on the average. For a shift of 2u/Vii. the process average changed between the thirtieth and thirty-first samples.7 O. The change was detected in the forty-fifth sample-a run of 15 samples.1. How long a run will it take.0 122 1.2 1. We assume that our estimates of the mean and the standard deviation arc correct. but the chance that a point will fall above the upper line is 0.1.5 161 0. The chance that a point will now fall below the lower line is virtually zero. Table 9.10. but is 4(Ti-v7i above the lower line. to detect a change in the mean? We consider rule 1 in some detail. If the samples are independent.6 1.00135).7 1. With control charts. Now the process is operating about a new center line that is only 2.1 1. while the standard deviation remains the same.R 0. the emphasis changes slightly to a different criterion-the average run number.ISS AVERAGE RUN LENGTHS test that has the smallest value of beta for a given alpha. the ARL is reduced to 6. 0..( 0. and so the standard deviation of sis 0. s. For small samples.2 66U C L = s + . where p is the probability of a point outside a warning line. where values of C 4 are given. .S that charts can be based.389u.11. as usual. for several values of the shift in the process mean. the calculations arc easily made. 9. C 4 "" 0. if we wish to detect small shifts in the mean within a reasonable average run length.1. we have used the R chart for the control of variability.10. s" are plotted along the vertical axis. Some values of c4 • B 3 • and H4 arc given in table 9. However.1. using rule I and rule 2.389 ) . but on the sample standard deviation. such as n = 4 or 11 = 5. the sample standard deviations. and little is lost by using the method that is simpler arithmetically. With modern computing facilities. J contains average run lengths. we can use larger samples. Table 9..12. s CHARTS So far in this chapter. In practice.12. the engineer sets the limits at UCL= Hi" and where the values of B3 and B4 are obtained from tables. from the N samples in the pilot set. We mentioned in section 9. 9. One can also make charts based on the sample variance in a similar way.156 QUAliTY CONTROL CHARTS The ARL for rule 2 can be calculated by setting r = 0 and m = 2 in formula (13. The formula is (1 + p)/p2.2 s.f/c 4 for (T and set the upper control limit at . We now substitute .9213 s = . It can be shown that the variance of s is given by the formula For n = 4.3.12. SETTING THE CONTROL LIMITS FOR AN s CHART In these charts. in which case the relative efficiency of the range decreases.4). This is shown later in table 9. three standard deviations above and below the center line. S = E sJ N. The center line is set at the average of the sample deviations.9213. The control lines arc drawn. the relative efficiency of the range estimate to s is close to unity. not on the range. 13.738 0.712 0.572 1.850 n.0854 1. it is the standard procedure.552 1.030 0.f. As in the R chart.911 0. -? u- Then. Constants for s Charts n c4 l/c 4 B3 84 Efficiency 2 3 4 5 6 7 8 9 0.716 1.9650 0.975 0.869 0.185 1.970 1.510 3.267 2.933 0.382 0. .781 0.8381.618 1. with 3 d.0168 1.07172 < 3i < 12.9400 0.518 1.534 1.766 0.568 2.12.1546u < s < 2.482 0. derived from the statement 0.9823 0.751 n.0423 1.0363 1.9845 0.!H4 0.9515 0.354 0.992 0.9862 0.9213 0.466 0.8862 0.0317 1.646 1.831 0.448 0.098 1. calculate better limits in this case because we know the exact probability distribution of s when the data are normal.9835 0.955 0.725 0.0148 1.9754 0. are 0.0194 1.321 0.8381 .2533 1. Nevertheless.000 9.700 10 11 12 13 14 15 16 17 18 19 20 1.890 0.503 1. A two-sided 99% confidence interval for s is.406 0. therefore. We now develop these limits for the case of samples of size 4.284 0..5% values of chi-square. ALTERNATIVE CONTROL LIMITS In the previous section.9854 0.679 1.497 0.797 0. the upper and lower 0. With n = 4. the control limits were calculated as the center line plus or minus three standard deviations.428 0.157 ALTERNATIVE CONTROL LIMITS Table 9. We can. The reader will recall from section 6. respectively.7979 0.9727 0.594 1. after taking square roots and simplifying.0157 1.882 1.0229 1.0281 1.9776 0.068u .761 1. this procedure tacitly.0638 1. 0.9594 0.0133 0 0 0 0 0.0140 1.12 the procedure for obtaining a confidence interval for the variance of a normal population.118 0.185 0.0210 1.0252 1. assumes that the distribution of s is normal. however.0180 1.9794 0.1284 1.9693 0.239 0. and wrongly.0510 1.9869 1.9810 0.490 0.266 2.07172 and 12.1. 9. thus. the x-bar chart is robust to moderate departures from the normality assumption. We consider the case in which the specification is two-sided and symmetric. Our emphasis in process control has been that the process should be roIIing along steadily. the principle of the R chart is sound.168.168. Instead of having only one chance in 400 of a point outside the limits under normality. How essential is this? It is not serious as long as the distribution is roughly normal. the risk of a false stoppage may increase twofold or more. in theory. Notice two things: the limits are tighter. and so the rule of thumh of drawing the control lines at plus or minus three standard deviations does not pass muster mathematically.15. The situation with the R and the usual s charts is different.5 < s < 2.Lo 5. The robustness of control charts under non normality has been investigated by Schilling and Nelson (1976). 9. and the lower limit is positive. One should not.14. and it still serves as a powerful (and simple) diagnostic tool.Lo + 5) .9213 and obtain 0. We saw in chapter five that the distribution of the sample mean i approaches normality as n increases. NONNORMALITY The derivation of the control charts depends on the assumption that the observations are normally distributed. Even if the actual alpha risk is a little cloudy. They are measures of process capability. the distributions of Rand s are not normal. The control limits are.~ . The measured characteristic. such as the length of the item. UCL= 2.2455 . one in 200 rather than om: in 400. overreact to this. . Even for sample sizes as small as n = 4 and n = 5. Even if the data are normal.2455 and LeL = 0.158 QUALITY CONTROL CHARTS We now replace u by slc 4 = 5/0. PROCESS CAPABILITY Nothing has been said yet in this chapter about specifications. however. closer to the center line th.m the usual limits. has to fall in the interval (J. J. be under control but be way off specification! Two criteria tic process control and specifications together. A process can. 73% of the items will fall in the interval (Ik . We define two new indices CPU = US~O'- Ik CPL = Ik . and Virtually every item produced will be above the LSL. If. CPU) = C/l. If the process is on target-if Ik is actually equal to the target value Iko-a process with Cp = 1. It was reported a few years ago that the minimum acceptable value of Cpk in some Japanese . the percentage of items within specifications will fall below 99. If Ik """ f. both these indices are equal to Cpo The second criterion is Cpk "" minimum (CPL. Ik = Iko + CT. and only 97.ko + 2CT. We denote the actual average value for our process by Ik (as represented by the center line on the control chart). for example.LSL 6CT where USL and LSL are the upper and lower specification limits. If. We now introduce another measure of process capability that includes the failure to be on target.27% will fall above the USL. the specification limits wiJI be LSL = Ik . however. only 84.0. Cp measures the potential cability of the process. where k = Ilko - 5 Ikl .13% will meet specifications.0 has (5 = 3u and is capable in the sense that 99. the process is not properly centered.73% of the items will be within specifications. Ik ¥ Iko' The process capability index.k). Ik + (5).73%. as CT decreases. If the process is correctly centered. respectiveIy. Cp increases.LSL . k =0 and Cpk = Cpo A desirable aim for a process is to have Cpk > 1. ell' is defined as the ratio of the specification width to the process spread: C = 2(5 6CT p lJSL . but 2.159 PROCESS CAPABILITY Iko is called the target value of the characteristic and 25 is the specification width.(5.4CT USL = Ik + 20' . so that the process mean is indeed at the target value. and 3CT If Ik = Iko = (USL + LSL)/2. if the process is off target. 128 = 2. The control lines are drawn at a distance ±3it 1. i.e. LCL= 8.43 16 9.28 10.50 4 10.1 had been reported as single observations without any information about CT. What can be done when the samples are individual observations. The average of the moving ranges is i = 0. which corresponds to a fraction 0.629) = 11. Suppose that the sample averages in table 9. This can be achieved by reducing the standard deviation until USL . I CHARTS Thus far we have discussed setting up x-bar charts when the sample size is greater than n = 1.50 19 11. denote the value of the ith individual.50 = 1. = IX'+l .00 20 9. Example 9. The sequence of the values of r..128. Let x.4 defective parts per million.16.63 17 10.90 7 11.60 15 10.60 and r2 The first two values of r.70 - 10. which represents about 3.42 = 10.16.66i from the center line.1-£01 = 1.5. The estimate of 0. either to use the sample range or the sample standard deviation.28 = 0. We have a set of twenty individual data points: 1 2 x 10.0 and Cpk = 1. and the center line is at x = 10.and reducing the deviation of the mean value from target to IlL . Then r.LSL = 120.x.l is the range of the ith pair.33.1.53 12 10.80. There are two choices for estimating the standard deviation. 9. Therefore. manufacturers are striving for processes with Cp = 2. Nowadays. in the semiconductor industry.20 10 9.00006 (6 in 100.70 - 9.13 + 2.85 18 to. is called the sequence of moving ranges. are rI = to. .is given by it d2 = it 1.70 11 i 9. 10 9. Denote its average by i. control limits are set at veL = 10.160 QUALrIY CONTROL CHARTS industries is 1.83 9.20 .6.66(0. n = I? Such a chart is called an I chart (individuals).517.15 14 10.45 13 10.46.13.629.30 8 to.00 9 3 9. The standard deviation is estimated from the moving ranges of pairs of adjacent observations.000) of nonconforming units.13 5 6 9. 84 25. 26..161 EXERCISES UCL'"' 11.45 and 8.23.27 25.23 26.19. 24.6.s • * * :Q * * * * * * MU = 10.83 25.14.48 24. using all three stopping rules.90 23.10 23. 22.l :::I .72 23. 25.52 22. 25.76. 22.1.94 24. 22.01 25.69 2S.77 20. 20.55 18.456 o 10 5 20 15 Observation number Figure 9.02 22. 26. It will be recalled that the control limits in the original x-bar chart were set at 11.16. 23. 22.41.S9 26.39 23.82 21. o EXERCISES 9.10 25. 23.S3.25 * <:.52 25.62.77 22.0S 21.76 25.80.11 25. 25.84 25.99. 23.54 23.70 26.06. 28.80 11.16.32 23. 26.76 23.09 20.26. 22.96 23.61 21.58 24. 28.57 25.70 28.51 23.82.83 28. . * * c.32.15 24. 23. each with three observations..57. 22.64 23.86 23. 24.78 25.80 25. The following data sct shows 30 samples. 22.76.3S.83.48 23.39 23.13 * • * * 8. Make an x-bar chart and an R chart for the data and mark any points that are out of control.86.50.95.60 22.00 * -6 . 2l.10.88. 24.13 22. 23.04 26.73. The first three samples are in the first row. The Minitab I chart for this data is shown in Figure 9.42.16. 26.02.22 25.17 25.75 LCL =8.12 23.37.91 2S.31 25.j :::I .l.61 27. 21.38 19. 27.1. 2S.71 23.22 23. 24.02.'0s: • 10.S4.51 22.1.29 21.10 24.75. I chart for the averages in table 9.24 25.55. 8 19. 16.1 17. 63.2 64.8 62.5. using all three stopping rules.5 18.4 65.0 60.7 un 17.2 65.162 QUALITY CONTROL CHARTS 9.6 65.4 17. 65.2 64. When the process was under control. Samples of three items are taken from the production line at the end of each shift for tcn days (30 samples).6 61.8 63. The first row contains the first three samples.5.3 18. 17.0 17.5.3 17. 67. 67.6 18.3 18. 65.4.4 18. 18.0 66.4 17. 17.0 67.3 15.5 19.0. 18.4 62.9.8 67.4.0 63.2 16.0.5 16.6.4 65.6.7 17.7 18. Do your conclusions about points being out of control change? 9.1 assuming that you were given standard values IL == 24.4. 64.2 61. the process average was 64.2 17. 65. 64.0 64.4 64.5 17.4 63.4 16. 59.4 20. 17.2 66.6 65.8 68.8.2.5 16.9.8 17.4 65.4 63.2 17.2 64.0 69. 64.4 66.9.6 18.0 62.8.2 18.8.8 64.0 and the standard deviation was 1.9 17.2 66.4 18.6 16.2. 65.8 63.1. and so on.6. 64.4 17.6 18.1 19.2 62.5 17. 18. 65.6 66.1 17.9 15.0 65.0 16. Make an x-bar chart and an R chart and mark any points that are out of control. 65.2 65.0 and (T = 1.4 62.8 16.0 65. 62.7 16.7. 18.8. 19.0 65.9 17.4 18.0 15.2 66.4 60.8. The measurements of the items are given below with the first three samples in the first row.6 18.4 62.3 18.9.7.6.6 64. 9.4 17.6.8.6 66.6 65.6 64.6 63. 20.8 20.0.0 65.2 19.7 18. 17. The first two samples are in the first row.8 62.3. 67.0 20.5 17. 16.0 65.9 17. Make an x-bar chart and an R chart of the data and note which samples are out of control.2 63.0. 62.7.8 62. 15.0 17.4.5 17.0.2. Repeat exercise 9.6 65.4.2 16.5.6.0 63. 67.5.2 64.5.0. 63.4 19.8 66. 64.5 17. 62.8 62. 18.8. 16. This data set contains 100 observations taken from a production line in 25 samplcs of 4 each.3 16.6 9.4 62. Uti IS.8. Make an x-bar chart and an R chart for the data and mark any points that are out of control.4 18.0. 66.0 67. 64.8 60.0 64.6 62. 64.6 63. .7 18. 17.8 18.8 62.0 66.6 67.2. This data set consists of 90 ohscrvations in thirty samples of three observations each.8 64.2 20. 17.5. 65. 18.6.8 64.4 64.6 63.8 18.8 62.6 64.0 65.0 IX.4 65. 18.0 66.2.4.2. 18.6 18.6 17.2.0 18.0 19.4 64. 47 39.34 .16 40.14 40.43 39.9.80 39.38 40.33 41.32 38.98 39.38 40.73 41.67 40.95 40.01 40.S. Make an s chart for the data of exercise 9.94 40.e.54 40.8.79 40. 9.29 39. 9.41 40..80 41.20 40.55 39. Assuming that £(s) = c4 0'.53 41.35 38.94 40.79 39.15 40.14 40. 9.38 40.77 39.02 40.10 38.15 41.1.33 40.95 40.44 40.7.163 EXERCISES 40.65 39.19 40.83 40.76 40.25 40.57 41. Prove that s is a biased estimator of 0'.18 39.07 40.66 39.13 40.98 39.92 40.10.1.84 39.90 41.4.48 39.92 40.56 40.95 40.57 40. 0'.55 40.69 40.84 39.97 40.31 38.46 37.14 39.56 39.93 38.68 41.60 40.52 39.C~)(T2.58 38.80 9. Make an s chart for the data in exercise 9.97 40.16 40.17 41.99 40.30 39. prove that Yes) = (1. 39.86 41.47 39.49 40.07 39. i.01 39.20 41.68 41.53 40.53 41.41 39.6.87 39.36 38.54 41.92 40. that E(s) # 9.86 39. Make an I chart for the data in table 9. 10. We arc sampling. BINOMIAL CHARTS We assume a process rolling along under control. and p is the proportion of nonconforming units in the batch.. In the first part of the chapter.3 that if the inspector takes a random sample of parts. or as nonconforming. Many engineers use the terms interchangeably.S. terminology more in line with international standards.Statistical Methods in Engineering and Quality Assurance Peter W. M. if the sample consists of only a small fraction of the batch. Inc CHAPTERTEN Control Charts for Attributes 10. we discussed control charts in which the measurements were continuous random variables. We recall from sections 3. We can associate with each item tested a random variable x that takes the value zero if the item is acceptable and one if it is nonconforming. Those are known as control charts for variables or control charts for measurements. However. Another term used is go. INTRODUCTION In chapter nine. in which case it is accepted. by attributes.. in which case it is rejected. the number of nonconforming units theoretically has a hypergcometric distribution. where n is the number of parts sampled. An item is examined and classified as either conforming. we can reasonably act a. Then we introduce another series of charts that use the Poisson distribution. producing parts of which a proportion p do not conform. if the number of nonconforming units has a binomial distribution with parameters nand p. Either the item has the attribute of acceptability or it does not. the words nonconforming and nonconformity have lately replaced in some industries the older words defective and defect. One can picture an inspector drawing a random sample from a large batch of components and counting the number of nonconforming units. An inspector draws a sample of size n" of 164 . we look at the situation in which no measurements are recorded.1. or testing.. In this chapter.. no-go: either the itcm is acceptable (go) or it is nonconforming (no-go).2. Nonconforming units replaces the use of defective as a noun. we present control charts based upon the binomial distribution. John Copyright © 1990 by John Wiley & Sons. In order to bring U.12 and 6. BINOMIAL CHARTS FOR A FIXED SAMPLE SIZE Figure to.) = npq = . quoted an Air Force report that a certain defense contractor had. an Associated Press report.1 i(n - (10.1 %. It is a random variable and its variance is pqln. a sample size of at least 100 is recommended. reduced its overall defect rate from 28% to 18. In the second.1. iln is our estimate of p.1) P. The fraction nonconforming for that sample is X. The Numbers Nonconfonning in Samples with n 150 = 6 4 8 1 5 10 9 3 8 3 8 4 4 6 5 6 12 5 4 3 4 8 9 5 4 . we will need samples of 1000 items! The value ofp varies from application to application.1) Vex. the average of all the x. or else one can plot Xi and obtain what is called an np chart.i) (10. This means that if the fraction nonconforming is thought to be about 5%. The center line is set at x-bar. Two examples arc enough to illustrate the general procedure. 1989.=.1 shows a control chart for an inspection scheme in which samples of n = 150 were taken from each batch. n The upper and lower control lines are drawn at veL = i + 3'. Some semiconductor manufacturers are working in some stages in the region of one defective unit per million. nj where Pi is an unbiased estimate of p. and think in terms of upper and lower control lines at a distance of three standard deviations from a center line as before.3. over a 12-month period.2. There are two equivalent methods of making a chart. 10. One can either plot Pi along the y-axis..3. The first has a fixed sample size.165 BINOMIAL CHARTS FOR A FIXED SAMPLE SIZE which Xj are nonconforming.1.3. is normally distributed.3. If n is large enough. printed in the Austin American Statesman on December 24.3. we can act as if P. When we get the fraction of nonconforming units down to 0. How large should n be? A commonly used rule of thumb is to take a prior estimate of P and choose n so that np > 5. the sample size varies. in which case it is called a p chart.. is estimated by i(n .3.005. The data are given in table 10.2) i) In Table 10. Then the variance of x. On the other hand. We will construct an np chart. (10. The process is in statistical control. ..... ··... .. ·. ·.I.. and LCL = i .3y i(n ... ·rI ... hut there are two close calls.....3) is negative..1...+--+--+------+t... ···· . ( 10..-~.......3.3._IIi_+_-~+~----+--... In the p chart... which is often the case.....3) If the value given by equation (10. The only difference between the np chart and the p chart is in the scale of the vertical axis..820 12 9 I 1 I -------~---------~------I J I -~---------~-----I I I I I I I I I I I I I I I I I I ..... -_-_-_-_-_-_~ 5..... The average number of nonconforming units per sample is 5........7%. the center line would be labeled 3...... In these two extreme cases... I I I I I I I I ....3......166 CONTROL CHARTS FOR ATTRIBUTES I I I I I .. All 25 points lie between the control lines. and the seventeenth only one.........76..84% and the VeL 8.i) In.........3. The center line is set at this value. ·· ..8. ·· 12... I I ---~---------+-----I I I I I t - J I I ~I --~-----I I I I I I I I I • I I I I I I I I I I I I I I I 6 1-~++__. the percentage of nonconforming units in the samples was as large as 8% and as small as 0.. .30 and so the lower line is set at zero. In this example..53%.. Binomial control chart.. we plot the percentage nonconforming.....3. LCL is set at zero..· . Equation (10.. which corresponds to approximately a 4% nonconformance rate.. The VCL is calculated by equation (10..3) gives a value of -1.2) as 12. The fourteenth sample had 12 nonconforming units.76 I I ----1"------I 3 I I I I I 5 10 15 20 Sample number fIgure 10... 1 is a p chart for a process to produce electrical parts.. VARIABLE SAMPLE SIZES Another strategy is to sample every item in the batch: 100% sampling. I .I : ~ ~ -~----- I 0. -------T----I I I . and 100% testing would leave the engineer with no product.. 100% sampling can be virtually..4.. or actually.. ---~ 0.""1.OO02 15 Sample number . with batches of about 2000 items subjected to 100% inspection..O_o.. I I I I I I I ... It would seem that this method ought to spot all the nonconforming units. but in practice it does not.':': '....... in some cases..4.. Especially when the nonconforming rate is low and the procedure is tedious.:::: '':::':'f:': '':::':'.09 - I ..... Figure 10.. I I I I I I --..1206 0... ..".:r::::' :':'.~~~lT=f=.or+...1004 -: I I .. Serious questions have been raised about the practical efficiency of 100% testing.6=. . ':':':::: '':::'::':: :':':::':':: :':''::~'::::''::::'':: :':':: '.4.. The important difference between this example and the preceding one is that the sample I I I I 0...: ::':::':'f:':'. Sometimes 100% sampling is clearly out of the question..1... I 0... If the method of testing a fuse is to exposc it to increased levels of current until it blows...... especially when testing is destructive. I I ~~~~~~~~~~~~~~~~~~~0.t.. 30 .. However. or of having a piece of automatic testing equipment installed on-line.'igure 10.167 VARIABLE SAMPLE SIZES 10.... . automated.. 0. --.::::.12 'e5 I I I I I I I I I I I I I I I I I I I I I I I I I .4=. The test can merely be a matter of plugging a component into a test machine and seeing if the light goes on.1 . the test destroys the item.~~N:'.f1' ~=-=". . inspectors tcnd to become inattentive. .. The p chart (unequal sample sizes)..11 ~ I - 5 VJ I I t I I I I I I :: I I I G) I i-------.. . .. are smaller batches. we must calculate the limits using the actual sizes.::: lOOp).7 26 2231 254 19 177 8. using the average batch size.3 Average batch size = 2015. The data are given in table 10.!n.1004 .4.2. Control lines based on the average batch size are adequate to get the general picture.8 10. Table 10.1) Again. .0 3 2319 199 8.9 14 2451 261 10. 16 and 29.8 11. This gives values 0.4 8 1904 172 9. In this data set. the lower line is set at zero if this formula gives a negative value.2 21 2047 192 9.4 11 bateh size nonconf.4 22 1918 176 9.4 R.1.4.2 R.1004. The batch size varied from day to day for various reasons. ISO 8.6 2 1854 204 11.9 24 2183 201 9. The center line is set at 0.6 4 2322 245 10. This variation makes the simpler np chart inappropriate. 1 1705 180 10.0 13 2488 295 11.6 5 1796 209 11.6 9 2171 182 8.3.4.4.7 15 1549 143 9. % nonconf.3). % nonconf. % nonconf.0802 and 0.168 CONTROL CHARTS FOR ATIRIBUTES size is not constant. For each batch. the largest batch had 2488 parts and the smallest 1549. % noneonf.1206.2 23 1774 194 10. A chart prepared under these conditions is shown in figure 10. 1819 213 11.4 7 1950 222 11. the average of the Pi (some engineers use P:.6 20 2276 205 9. 2153 249 11. but for the close cases.4 2086 242 11.2) and (10.0 batch size nonconf. 2015.6 12 1884 213 11. 100% Sampling of Batches of Components batch size noneonf. The usual procedure is to compute the values for the outer control lines from equations (10.1.4 10 2383 224 9. Point 13 is now out of control because it was the largest batch in the group. No points are out of control.6 21 batch size nonconf.q.9%. The percentages of nonconforming parts ranged from 8. The lowest points.3 17 1755 18 1633 154 9. We can make a chart that ignores the variation in sample size.2 16 1842 151 8. (10.2 6 2103 29 27 28 30 1983 15l:l4 1901 2374 234 171 156 197 11. although point 13 on the high side and points 16 and 29 on the low side come close.3. we calculate the limits p ± 3YP. The outer lines are no longer horizontal. p = 0.2% to 1t . ~: .. the search for an assignable cause may lead to a reason... I .. . . .'I---+t-'--l-h'--44-#:--. INTERPRETING OUTLYING SAMPLES In section 9. I I I ___ L 107 I I .-. .a--:~ 0. we made a passing reference to the question of handling outliers in the data set that is used to set up a control chart. I I . . I ! . This will lead the engineer to set the control lines for the x-bar chart too far apart and to be lulled into a sense of security when all the averages fall comfortably between the unrealistically liberal limits. I" . ____ --~ " " ~-- . :".'1 ' . such as a malfunctioning piece of equipment that is replaced. : .1004 97 ... I I I I .. . . t. ... The p chart (unequal sample sizes). t---'rl-.--t-------t-------- 117 I . I I I I ....: L. I I I I f L_____ I L___ • f _ IL ___ _ I I I I I I f I ..' ---.l :.. Nothing was said about samples in which R was too small. because with the usual sample sizes of 11 = 4 or 5.. ..' ..--. fT~. 1 ..5. ----i-------t--. ..1. . ll.· 1 :. I I . I . !:..4.. Similar problems arise with p and np charts.: :...H-Ht---t.L. L.' :. . .-. 87 ' '' ~----I I I I I I .: .:..- . 10. . . "'-: .5. the lower control limit on the R chart is zero... ~ : ~: n~~~~~~~~~~~~~~~~~~~ o 5 10 15 Sample number 20 25 30 I<'igure 10. .. .. h.. ___ --L ___ _ I I .t-: . If a batch has an unusually high fraction of defective parts. .. -----.'::l~'. . I I ... That point can then be dropped from the data set and the control limits can be recalculated.. The interest then was in R charts.169 INTERPRETING OUTLYING SAMPLES (X10·3) 127~-r~-r~-r~-r~-r~-r~-r~~~~~~ :....· L. An unreasonably large range can give the engineer an inflated estimate of CT..2. The np chart is shown in figure 10.00 +--------------------.1.0..LCL '" 6.+ . This difficulty also occurs with c-charts. The UCL is at np =.55. Suppose that the three points below the LCL arc dropped from the data set and that the control lines are recalculated from the remaining 22 points. The new center line is at the new average. The observations are shown in table 10.. 19. Numbers of l>efect~ in 2S Samples In 17 13 19 17 18 24 24 17 20 17 23 21 15 16 31 22 19 19 22 =150) 26 20 21 234 Points that fall below the LCL may present a different problem. 32.-• UCL = 29..060 * 0. They are the numbers of defective items in 25 samples of n = 150 items each.05.0 o 5 10 15 * * 20 Sample number Figure 10.. 25 .. The following data set illustrates the difficulty. The np chart for table 10. There are three points below the LCL-numhcrs 18.5. np = 20.5.5.170 CONTROL CHARTS FOR ATI'RIBUTES Table 10... and 20. Look how well we can do when we really try hard! It is just as important to search for an assignable cause for a low point as for a high point.5. The last sample is now below the UCL and is under control. There is a temptation to greet such points with applause. The last sample has 31 defectives and lies above the LCL.. ..1. J.1. it often happens that the low point is too good to be true...1... The average value (center line) is np = 18..94 • * • * +----------------------np'" 18.. Unfortunately. One of the standard reasons for low points is that they have been reported by an inexperienced inspector who fails to spot all the defects...5... 1635 0. THE ARCSINE TRANSI<'ORMA TION In the last section.4661 0. 0. doubts were cast about the importance of observations that fell below the LCL. one calculates p.3844.4661 0.4949 0. expressed in radians.6.5735 0.1395.3672 0. the transformed values are 0. LCL = 0. 0.3281 The center line is at 0.2838 0. One can argue that when the LeL is zero. The control limits are UCL = 0. x = 2 arcsin( yp).3281 0. the engineer loses power to detect genuine improvements in quality. the percent nonconforming. In applying the transformation.p). When the arcsine transformation.4027 0.3672 .5223 0.ft).4949 0.2838 0.3281 0. That should not be interpreted as meaning that all points below the line are false.6294 and There are no points outside the limits.3672 0.2838 0. The difference between thc two transformations is a matter of taste and is not important. This is one of the reasons why the arcsine chart has been proposed as an alternative to the p chart when the sample size is constant. It can hc shown (and it appears as an exercise) that the LCL will be set at zero unless np > 9( 1 .1.3281 0. in the sample and replaces it by x = 2 arcsin( yp) .4027 0. Another reason is that the arcsine transformation makes the plotted variable independent of p. The LCL for the np chart is set at the larger of zero and nfi . The variance of x is I! n. and the control limits are set at Nelson (1983a) suggests an alternative transformation: x = arcsinl v' x(n + 1)] + arcsin! V(x + 1)! (n + 1) I .3.4661 0.3281 0. Jt is also claimed that the transformed variable has a distribution that is closer to normal than does the binomial.4027 0.3Vnj}( 1 .171 THE ARCSJNE TRANSFORMATION 10.3672 0. where arcsin(u) is the angle whose sine is u. is applied to the data in table 10.4661 0.3281 . .172 CONTROL CHARTS FOR ATfRIBUTES 10... 1 1 62 r---+I-t-#-hM-f-~f--II~+f-\--I+-.. the yarn will be weak enough to break.. 34..I .. 79.... The c-chart..·..L to denote the parameter of the distribution.. · I I I I I I I I I I Sample number Figure 10. ... 1 1 1 1 -----l-------~-------iI I I I 1 .L = 5...1 and 10.312 . based on the Poisson model.-\--~ 42 ..7...I ..... The average number of nonconformities per day during that period was c = 56. I .7...95 ± 3(7.. Since the mean and variance of a Poisson variable are the same. 56..952 1 1 1 1 1 ------r---... If we take a sample of ten reels and count the total number of nonconformities..3.. five nonconformities per reel. The letter c is used instead of I...L = 50.7. ten reels were inspected in the course of each day.6. . ........ That is a nonconformity... r·I ........ At various places in a spool..I ........ that total will have a Poisson distribution with I..1•••••••••••••••••••••••••••• I 79..7. .... Figure 10...1 is a c-chart made from the first 21 points.. The charts shown in figures to. . the standard deviation of the process was taken as 7.I . on the average. there arc... ..1. The output of 50 days' work was inspected. The data arc given in table to..... Charts for the total number of nonconformities.. and the control lines were drawn at 56.2 are taken from a process similar to that described before.95.. I 1 82 1 I 1 I I -----~-------~--­ ... and the total number of nonconformities in the ten reels was recorded...·...55.. are called c-charts.1. The center line of the chart was set at that level.....·..... If the number of nonconformities per reel has a Poisson distribution with I.7....55) = 34....... ·· . c-CHARTS Suppose that we are looking at the output of a machine that makes yarn for knitting and weaving.·... 7...592 ... closer investigation revealed that the regular inspector was sick on those two days... Points 15 and 16 are below the lower line......312 Sample number Figure 10. 62 ..... The c·chart. 1. and the engineer noticed it immediately without the need for a chart... 1..7........ Investigation over the weekend found an assignable cause in a part of the machine.7.. As a result.... Throwing out those three points would have increased the average number of nonconformities to 59..1............. I I I I I 1 I ....952 F--Hr+if+i--\-+-++ifr'1----+ 1 42 34... 1 __________ 1 79..... A temporary repair was made and production resumed on Monday.. .. Unfortunately..2... Assignable cause-inadequately trained worker..... I I I I ... he missed quite a few weak spots and reported too few nonconformities.............. There are three points outside the lines.. That was not done because a new part Table 10....2..173 C-CHARTS 1 1 1 82 ________________ 1 1 1 1 +1 _________ ~---------L------ .......592 1 I I ' 1 I ~---------L------ 1 I -*--z'"rio-*~ 56....... Point 10 is too high..... Her duties were taken over temporarily by another employee who did not know exactly what to look for.... Number of Nonconformities in Ten Reels of Yarn 77 64 59 54 49 56 54 49 38 59 75 41 54 76 60 51 43 59 41 46 45 24 40 41 51 61 22 46 43 59 49 69 43 50 49 65 59 49 50 59 45 45 5S S6 53 85 67 43 42 S3 ........ It would be nice to conclude that the process had suddenly pulled itself together and that those points might set a standard for future performance.. .. 1....... This occurred on a Friday... Charts for demerits are similar to c-charts. During the last 29 days. and ten points. x s . one point. Let the numbers of occurrences of these defects be denoted by x . The replacement resulted in an immediate improvement. The total number of demerits is then Although Y is a linear combination of Poisson variables. but it still left room for further improvement. the average number of nonconformities per day dropped to 49. 10. This might be a more appropriate procedure in the evaluation of the overall performance of a defense contractor. arc merely cosmetic. five points. no doubt it would have heen eventually. Some defects. We can. + 25c s + lOOc lO • We then set the center line from the pilot data at y = i l + 5i s + lOXIO = Yo .174 CONTROL CHARTS FOR ATTRIBUTES arrived on the afternoon of the twenty-first day and was substituted next morning for the part that had been temporarily repaired. DEMERITS In constructing c-charts. A chart of the last 29 points shows a process under control at the new level of c.S. readily calculate its expectation and variance as and V(Y) = c. than computing an unweighted percentage figure. To handle this situation. and x 10' respectively. That represented a 10% reduction in the nonconforming rate. That may not be so. Figure 10.2. Suppose that a process has defects with three demerit levels.2. A more serious defect may carry a penalty of ten demerits. and so it does not. some companies use a system of demerits. The improvement in quality after the new part was installed is obvious. particularly the run of 13 consecutive points below the line. itself. and let them have Poisson distributions with parameters c l ' c 5 ' c lU ' respectively. to which reference was made in section 10. however.2 shows all 50 points plotted with the same control lines that were calculated from the first 21 points. A small defect may carry one demerit. it is not the simple sum of them.7. we have assumed that one nonconformity is no better or no worse than another. What justifies our concluding from the chart that there has been an improvement is the preponderance of points below the old center line. Jt is interesting to note that rule 1 of the previous chapter-seek an assignable cause if there is a point outside the outer control lines-was not triggered.. others are functional. have a Poisson distrihution. such as a scratch in the paint. 3. Make arcsine charts for the data in tabletO. which were made after the new part was installed. Make a c-chart for this set of data assuming that the average number of defects is 9. The batch sizes and the number of defectives are given in the following. Twenty samples of 200 items each were taken from a production line. 4 3 7 4 288 4 7 R 2 11 13 5 R 787 7 5 Is the sample with 13 defectives out of control? 10.2. Use I-t = 49. 11= 1953 70 1446 47 1441 55 1954 x= 2450 2448 1753 85 83 64 2549 90 1850 2154 90 1950 80 2340 82 64 1450 51 1554 57 1852 80 2038 89 2248 75 2450 1552 54 2449 120 2552 1746 49 1949 2049 84 2149 74 78 66 65 88 Batch 16 is clearly a bad batch.2. The average number of defectives is 73. Batch 2 has the fewest defectives.4.6.5? . Is it out of control? 10. The average batch size is 2013.1.7.5. Do your conclusions agree with those of section 1O. Twenty-five batches of devices were subjected to 100% inspection.5. The number of defectives in the samples are given in the following. Make a (Poisson) c-chart for the last 29 observations in table 10.EXERCISES and the control lines at EXERCISES 10. but it is next to the smallest batch.1. Make a p chart of the data. both with and without the three low points.5 and mark any points that are out of control: 13 9 10 l) X 17 14 12 18 12 12 9 15 10 10 11 14 13 19 15 8 11 10 14 8 10. Is the process under control at the new level of defects? 10. Make a p chart and a np chart for this data set. >4) The average number of defects per device is 0. .176 CONTROL CHARTS FOR ATTRIBUTES 10.e. and so the defects cannot be regarded as occurring at random. This corresponds to a process that produces "good" items with an average rate of one defect for every two items and is contaminated by about 10% "bad" items that have. they arc dreadful. Five hundred devices are inspected and the number of defects in each device is recorded.5 and 50 from a Poisson distribution with /L = 5.97.0. Let x. This data set was constructed with 450 observations from a Poisson distribution with /L = 0. denote the number of devices that have n i defects. A summary of the data is n= x= o 274 1 138 2 37 3 11 4 9 5 31 (i. when they are bad. they are indeed good.. When the devices are good. on the average.6. five defects per item. Show that the data are most unlikely to have come from a Poisson distribution. Statistical Methods in Engineering and Quality Assurance Peter W. M. John Copyright © 1990 by John Wiley & Sons, Inc CHAPTER ELEVEN Acceptance Sampling I 11.1. INTRODUCTION In developing the theory and practice of acceptance sampling, we introduce two people whose attitudes have traditionally been those of adversaries, the vendor and the buyer. The vendor manufactures widgets in a factory. When a batch of them is made (we henceforth use the jargon of acceptance sampling and use the word lot instead of batch), the lot is presented to the buyer, with the claim that it is of satisfactory quality. Before accepting the lot, the buyer calls for an inspection to determine whether the lot meets requirements. If the lot passes inspection, it is accepted. If the lot does not pass inspection, it is rejected. The reader will already have realized that we are going to talk about a special form of hypothesis testing. In practice, the role of buyer can be played by the actual purchaser or by the vendor's own quality control inspectors who stand between the production line and the factory gate. The emphasis in this chapter is on single sample plans. This is the simplest case. The inspector takes a random sample of n items from a lot and rejects the lot if the sample contains more than c defective items. The numbers 11 and c are predetermined. The discussion is, for the most part, confined to sampling by attributes. Chapter twelve begins with multiple sampling plans and sequential sampling. It ends with a description of the use of one of the major sampling schemes, MIL-STD-I05D, that is commonly used in defense work. The field of acceptance sampling covers a very broad area and we have space for only a fraction of the topics that could be covered. The reader who wishes to pursue the matter further is referred to the books by Schilling (1982) and by Duncan (1986). 11.2. THE ROLE 0.' ACCEPTANCE SAMPLING When a buyer is offered a lot, there are three choices: (i) accept the lot as satisfactory without any inspection; 177 178 ACCEPTANCE SAMPLING r (ii) inspect every item (100% inspection); (iii) use an acceptance sampling scheme in which only a fraction of the items are inspected. There are several arguments against 100% inspection. Among them arc the following: (i) if testing is destructive, e.g., if testing consists of twisting an item until it breaks, the buyer will be left with no items; (ii) testing introduces a chance of damaging good items; (iii) the cost of testing every item may be excessive; (iv) when the fraction of defective items is low, 100% inspection can become a tedious task, with the result that too many defects are missed. A desirable goal would be a manufacturing process in which the first option-zero defects-zero sampling-is valid. That is the aim of manufacturers who seek first-class quality and first-class competitiveness in world markets. This is already happening in situations where a major manufacturer has built up such a strong, close relationship with a supplier that the expenditure of time and money on acceptance sampling is no longer warranted. In such a case, nonconforming units are so rare that the occurrence of one will justify a very serious investigation to identify the error and to eliminate it. There is also a mathematical point. Whereas only a few years ago manufacturers talked about defective rates of 1% to 10%, we arc now seeing in the microelectronics industry processes that have a defect rate as low as 500 or even SO parts per million. If a manufacturer is only producing one nonconforming part in 10,000, there is only about a 1% chance that a sample of tOO parts will contain a defe·ctive part. If the nonconformity rate is one in 100,000, only about one such sample in 1000 will contain a defective. The situation described in the last two paragraphs is a worthy aim. However, many processes are far from that ideal. For that majority, acceptance sampling remains a major feature of the production system. It is there to give protection both to the consumer and to the producer. Schilling (1984) argues that acceptance sampling has other uses than merely deciding the fate of individual lots. He argues that "when applied to a steady flow of product from a supplier it can be made a dynamic element supplementing the process control system." It can be the buyer's quality control chart for the vendor's production. ll.3. SAMPLING BY ATTRIBUTES Each item in the sample is tested by an inspector who rates it as either defective or nondefective. The words conforming and nonconforming are SINGLE SAMPLING PLANS 179 often used instead of the older terms defective and nondefective. We stay with the older terminology, although we sometimes find it convenient to call items "good" or "bad." The quality of a lot is measured by the percentage of defective items: the lower the percentage defective, the higher the quality. Deciding whether a lot should be accepted or rejected is called lot sentencing. When a lot is rejected there are several choices: (i) The entire lot can be scrapped, which may incur a large financial loss. (ii) If the defects do not render them unusable, the items in the lot can be sold on an "as is" basis at a lower price. This can happen in the garment industry, where batches of shirts can be sold at discount outlets as "seconds." (iii) 100% sampling can be made of the items in the condemned lot and the defective items replaced by good items; the lot is then said to have been rectified. These choices are sometimes referred to collectively as lot disposition actions. The standard inspection system for sampling by attributes that is used by the U.S. Department of Defense is MIL-STD-lOSD. It is discussed in chapter twelve. 11.4. SINGLE SAMPLING PLANS We mentioned in section 11.1 that, in the simplest case, an inspector draws a random sample of n items from the lot and notes the number of defectives. He works according to a sampling plan that tells him how large the sample should be and gives an acceptance number, c. If the number of defectives found in the sample is c or fewer, the lot is accepted; if there are more than c defectives, the lot is rejected. Some use Ac (short for acceptance number) instead of c and Re (short for rejection number) for c + 1. The main question is how does one choose the values of nand c'? How large a sample size do we need? We will be taking samples that are only a small proportion of the lot, so that we can assume that the number of defectives has a binomial distribution. A small handful of items will not give enough power to discriminate between good quality and bad. Samples of about IOO or 200 will be the norm. It will not matter whether the lot size is 2000 or 5000, what will matter is the sample size. A common mistake is always to test a fixed percentage, say, 10%, of a lot. This can lead to ineffective samples from small lots and unnecessarily large samples from big lots. Cynics have suggested that an unscrupulous vendor. who realized that the fraction defective in his manufacturing process 180 ACCEPTANCE SAMPLING I had increased, might make it a point to present the substandard material in smaller lots. Then, since small samples have less power to reject null hypotheses than large samples, he could increase the chance that the poor material would pass inspection! The quality level of a process is the percent defective when it is under control-the center line on the control chart. We denote it by Po' The vendor wants a situation in which only a small portion of the product will be rejected when the process is under control. That is the vendor's risk, alpha. How small alpha should be depends upon the consequences of the rejection of a lot. If having too many lots rejected means loss of business to competitors who have better quality, or if it means large amounts of money spent on scrap or rework, thc vendor is under pressure to work at a low value of alpha. The buyer wants good product, with a low percentage of defectives. Zero defectives is a nice slogan and an important target, but it is not always practical. In some situations, one can obtain zero defectives by very slow and cautious workmanship at greatly increased labor cost. Perhaps it is in the buyer's intcrcst to accept a small fraction of defectives to obtain a reduction in unit cost. This will depend on the consequences to him of recciving a defective item. If your office manager buys pens by the gross and the pen that you pick does not work, you can throw it away and take another without too much loss. On the othcr hand, if the lot consists of vital parts whose failure is a matter of life and death, that is a different situation. The buyer has learned that he can manage comfortably with a certain level of defectives. This is the target level. It is convenient if this target level, which we denote as the acceptable quality level (AQL), coincides with the vendor's production level. On the other hand, the buyer also knows that the fraction defective in the lots received from even the most competent of vendors will vary; there is some upper bound on the fraction defective in a batch that can be accepted without extreme inconvenience or annoyance. This level is called the lot tolerance percent defective (LTPD). The buyer will want a sampling plan that gives a low risk, beta, of accepting so bad a lot. 11.5. SOME ELEMENTARY SINGLE SAMPLING PLANS A single sampling plan is fixed by two numbers, 11 and c. Why should one choose one plan over another? The decision should be made on the basis of the relative alpha and beta risks for certain values of p. In this section, we look at four plans with small values of n. These plans are quite unrealistic. They arc included only to give the reader some simple illustrative examples. In the next section, we look at some more realistic plans. For purposes of illustration in this section, we make the following basic assumptions to make the arithmetic easy. The vendor has a process that, SOME El.EMENTARY SINGLE SAMPLING PLANS 181 when it is under control, produces items that have p = 10% defectives. The buyer has learned to live with this rate of defectives and is prepared to accept lots that bave a slightly higher percentage, but does not want to accept lots that have as much as 20% defectives. Thus, alpha, the vendor's risk, is the probability that a lot with p = 0.10 is rejected; beta, the buyer's risk, is the probability that a lot with p == 0.20 is accepted, an unfortunate situation for the buyer. We use the symbol Pa ( 7T) to denote the probability that a lot with percentage defectives equal to 7T will be accepted. We can write and We assume that the number of defectives, d, has a binomial distribution with the appropriate values of nand p; a is evaluated at p = 0.1; {3 is evaluated at p == 0.2. Plan A. n =3, c = 0 Draw a sample of three items and accept the lot only if all three are good. If = O. to, q == 0.90, the probability that a sample will pass inspection is (0.9) == 0.729. r = 1.0 - (0.9)-' == 1 - 0.729 = 0.27 {3 = (0.80)3 = 0.51 . a Approximately a quarter of the vendor's "good" lots will be rejected. On the other hand, if the vendor presents lots that are 20% defective, about half of them will pass inspection and the buyer will have to accept them. Plan B. n =5, c =0 a == 1.0 - (0.9)5 == 0.41, {3 = (0.80)5 == 0.33. This plan is tougher on the vendor, but it gives the buyer better protection against bad lots. Plan C. n =5, c =1 a == 1.0 - P(d == 0) - P(d == 1) == 1.0 - (0.9f - 5(0.9)4(0.1) = 0.082. {3 = (0.8)5 + 5(0.8)4(0.2) == 0.738. This plan gives much more protection to the vendor than the previous plans, but it is completely unacceptable to the buyer. 182 Plan D. n ACCEPTANCE SAMPLING 1 =10, c =1 a == 0.26, /3 = 0.38. The two risks are more nearly equal, but they are clearly too big for practical purposes. We need to take larger samples. 11.6. SOME PRACTICAL PLANS In this section, we consider three plans that arc of more practical use. Two of them have II = 100; the other calls for 11 = 200. In comparing the plans, we assume that the vendor and the buyer have agreed on a target rate of p == 2% defectives. Plan 1. n =100, c =4 Draw a sample of 100 items and accept the lot if there are four, or fewer, defectives. To evaluate alpha, we need to compute, for p = 0.02 and q = 0.98, a = 1.0 - qlllU _ 100q~9p _ (100)( -' (lOO)('J'J)(9:)qY7p' - 91) q9Sp 2 (100)(99)(98)G~)q~Op4. This quantity is not easy to compute because of the number of decimal places required. During World War II, it was very difficult because electronic computers were not yet available. Statisticians turned to the Poisson approximation to the binomial, which is close enough for practical purposes and is, nowadays, easy to compute on a hand calculator. There arc now good tables of cumulative probabilities available, from which we can read a=5.1%. To obtain the Poisson approximation, we equate the parameter, /-L, of the distribution to np. In this example, /-L = 2.0, and P( d ~ 4) = ' e - p. ( 1 + /-L + = 7e- 2 = 2 ,\ 4) ~ + ~ + ~4 0.947 , a=S.3% . Table J J .6.1 gives both the binomial probabilities of accepting a batch for several values of p and the Poisson approximations. Table 11.6.1. Values of Pa for Different Percentages l>efective (n P Binomial Poisson 0.02 0.949 0.947 0.03 0.818 O.SIS (1.04 0.629 0.629 0.05 0.436 0.440 O,Cl6 0.277 0.2X5 =]00, c =4) 0.07 0.163 0.173 0.08 0.090 0.100 183 CHOOSING THE SAMPLE SIZE Table 11.6.2. ,,= 200, c =7 0.03 0.746 P" 0.02 0.951 Plan 2. n =200, c =7 p 0.04 0.450 0.05 0.213 (l.O6 0.083 0.07 0.027 (U)R 0.008 This plan also has a at approximately 5%, but it is more favorable to the buyer. The values of beta are less than those for plan l. The sample size has been doubled. The values of P" are given in table 11.6.2. Plan 3. n =100, c =7 This plan is the most favorable of the three to the vendor. It gives him enormous protection with an alpha risk of only 0.1 %; the buyer will accept 87% of those lots that arc 5% dcfc<.:tive. The sample size is the same as plan 1. but the acceptance number is the same as in plan 2. The prooabilitics are given in table 11.6.3. 11.7. WHAT QUALITY IS ACCEPTED? AQL AND l:flJD The buyer defines two particular levels of quality. The ac<.:eptable quality level (AQL) is the worst average level of quality that he is willing to a<.:<.:ept. Often this is a level of percent defective that the buyer has learned to live with over the years, and both the buyer and the vendor may agree that such a percentage is reasonable for both sides. There is also a lower level of quality (higher value of p) that the buyer finds particularly offensive (unacceptable). even in single lots. This is the lot tolerance percent defective (LTPD). The buyer wants a plan with a low probability. beta, of accepting a lot at the LTPD. The vendor wants a plan that has a low probability, alpha. of rejecting a lot that is at the AQL. Example 11.8.1 shows how one might design a sampling plan with the AQL at 2% and the LTPD at 7%. 11.8. CHOOSING THE SAMPI.E SIZE In chapter seven, we showed how to calculate the sample size and the cutoff point for a one-sided test of the mean of a normal population. The same principle is used to construct a single sampling plan. Suppose that we wish to Table 11.6.3. n = 100, c p p. 0.02 0.999 =7 0.03 0.989 0.04 0.953 0.n5 0.872 0.06 0.748 0.07 0.599 0.08 0.447 This means that it is subjected to 100% inspection in which every defective is (we hope) detected and replaced by a good item. When a lot is rejected.02. THE AVERAGE OUTGOING QUALITY.64Sv'o. When a lot is accepted.9. Example 11. on the average. and.n items in the lot may include some undetected defectives.07: V1i = l.. using the normal approximations to the binomial distributions involved. If P = P" the number of defectives is normally distributed with J. this assumption may be too optimistic. Thus. if P = Po.n)p defective items. and {3 == 10% when P = 0.05 n = 124 and 138 11. if P = PI' whence V1i= z"VPoqO+Zfl~ PI . AOQ AND AOQL Both the vendor and the buyer may also be interested in the average quality of the lots that pass inspection with a given plan. Then. The average outgoing quality is the quality that one gets by applying a sampling plan to a process in which the average percent defective is P and rectifying all the lots that fail inspection. We can calculate nand c for the test (sampling plan) in the following way. . o 11. Thus.q.28v'O:063T __ 0.• c=5. the defective items in the sample are replaced by good items and the rectified sample is returned to the lot.Po . The remaining N .L = np. and (T2 = np. In calculating the AOQ.8. the rejected lots are replaced by lots that have zero defectives. Two measures are used: the average outgoing quality (AOQ). it is rectified. Suppose that the engineer wishes to design a sampling plan with risks a = 5% when P = 0. each lot accepted by the buyer contains (N .184 ACCEPTANCE SAMPLING I test the null hypothesis Ho: P = Po against the alternative Ha: P = PI at given levels of a and {3. .Oi% + 1. we make the following assumptions about the inspection procedure. Given the problems that exist with 100% inspection. and the average outgoing quality limit (AOQL).1. bad lots come out better than good lots. The AOQL for plans 1 and 2 are underlined in table 11." We are saying that for plan 1. Note well "on the average"- Table ]1. the worst fate that can befall the buyer.06 0.50 0.07 0. 2. the AOQL is 2. Then.08 0.163 1. In plan 1.ISS THE AVERAGE OUTGOING QUALITY.9. Note that under this scheme.090 0.746 2.818 2. this quantity is well approximated by the formula (11.1) and so the AOQ is If nl N is small.05 0. 1. A fraction.52 0. and for the second. Note that we speak of "long-term average" and "expect to endure.07 0. will be rejected and rectified.66 0.008 0. Average outgoing quality values for plans 1 and 2 of section 11.05 0.02 0. few lots are rejected.19 0. When these are combined.14 0.90 0.08 0. 1·_· Pa . on the average.04 0.03.24% defectives. virtually all the lots will be subjected to 100% inspection and the AOQ falls to less than 1%. AOQ for Plans I and 2 Plan 1 Po 0.18 0.6 are shown in table 11. 9. of the lots will pass inspection. and so few are rectified. with this sampling plan and rectification. if the quality drops to 8% defectives. the A OQ.027 0. 90%. The AOQL represents the worst long-run average percent defective that the buyer should expect to endure when that plan is used.450 1.277 1. Pa .02 0.04.72 0. it is 2. the average number of defectives per lot is seen to be (11.1.45 0.03 0.24 0.03 0. on the average.24% at p = 0.9.1. On the other hand.9.98 .083 0.213 1. The average outgoing quality limit (AOQL) is the maximum of the AOQ for a given plan.629 2. will occur in the case when the vendor produces 4% defectives.07 0.2) AOQ= Pap.436 2.9. the buyer will receive. is thus only fractionally lower than p. For the first plan. AOQ AND AOQL A fraction.52% at P = 0.951 1.06 0.06 Plan 2 p Po AOQ(%) 0.04 0. with p = 2%.949 AOQ (%) 1.1.90 p 0. 91 % of the lots will be rectified. we make a few remarks about four other sampling schemes.ll. Dodge and Romig (1959) published tables for choosing values of nand c based on . and the overall quality of the product that the buyer actually receives will improve. We have only discussed one way of defining AOQ. In 1977 the Journal of Quality Technology devoted an entire issue to his memory. who pioneered the field of acceptance sampling during his career at Bell Laboratories. Our scenario implies that there is a supply of good items that can be substituted for the defectives. OTHER SAMPLING SCHEMES We have confined ollr discussions to single sampling plans for sampling by attributes. The average total inspection (ATI) is then given by (lJ. F. we will detect (and rectify) more of the bad lots. 1977). For further details. this procedure gives values of the AOQ and AOQL that arc very close to the values for the scenario that we considered. All four of these schemes-rectification sampling plans.186 ACCEPTANCE SAMPLING' some days the buyer will fare better and some days the buyer will fare worse! The interesting fact to note is that if the production line deteriorates and the vendor produces marc than 4% defectives. the reader is referred to the bookf> by Duncan (1986) and Schilling (1982) that were cited earlier. skip-lot plans. we chose a sampling plan based on desired values of AQL and LTPD.11. which in turn almost implies that the inspection is taking place at the vendor's plant before shipment. In the next chapter. There are other possihle scenarios. We can estimate the cost of inspection using this procedure. A different procedure for testing at the buyer's site would he for the buyer to carry out 100% inspection of the rejected batches and cull the had items (which would not be paid for). we consider multiple sample plans and sequential sampling. in eaeh rejected lot. N items will be inspected. Each accepted lot will have n items inspected. When the percent defectives is small. 11. chain sampling. the lot size N did not appear in the calculations.l) In section 11. RECTIFYING INSPECTION PLANS We assumed in the previous section that every rejected lot would be subject to 100% inspection.8.10. reprinting in that issue nine of his most important papers (Dodge. 1I. and continuous sampling-are associated with H. In the remaining sections of this chapter. Dodge. perhaps.13. The vendor must start again from scratch. The idea here is that after the vendor has established a past record for good quality. F. That is appropriate if the vendor is erratic. that claim of previous virtue is lost. 11. But if there is a defect.14. Which lots are chosen to be tested is decided by a process of randomization. provided there have been no defective items in the previous i samples. A pressure that is. 11. CHAIN SAMPLING PLANS We have built up a theory for handling individual lots. CONTINUOUS SAMPLING So far. we will still accept the lot. which appearcd in 1981. which also was introduced by Dodge (1955b). and by doing so save money on inspection? This is the rationale for chain sampling. should. The buyer might agree to test roughly half the lots and choose them by tossing a coin eaeh time a lot arrives. As long as the vendor's past record is free of blemishes. and sct c = O. 1I. or can. but nand i. we make substantial reductions in n. It is skip lot sampling. That is very different from telling the vendor that the buyer will. Dodge (1943) proposed a scheme for sampling by attributes.CONTINUOUS SAMPLING 187 other considerations. we have envisioned the vendor presenting the items to the buyer in (large) lots. Rather. we take that into consideration. we should skip some lots and test only a fraction of the lots that are presented. test only the odd-numhered lots that are presented. and N. . A listing of continuous sampling plans is givcn in MIL-STD-1235B. But if he is fairly steady.l2. H. If there is exactly one defective item in the sample. These plans were subsequently refined by Dodge and Torrey (1951). If the number of defectives exceeds one. not so subtle is put upon the vendor. building up a record of pcrfect samples until there is again a backlog of at least i without defect. They assumed that the objective would be to minimize the A Tl for given values of the AOQL. which was introduced by Dodge (1955a). we should not insist on inspecting every lot. p. The constants that define the plan are not nand c. a single defect will be excused. the lot is rejected. In chain sampling. SKIP LOT SAMPLING There is another way of rewarding past good quality. which could be applied to a situation in which the inspector saw a continuous flow of items passing by on the production line. in future. 03.3. What is the probability of accepting a lot with p = 0.03'1 What is the probability of failing to reject a lot with p = O. When it is appropriate.03? What is the AOQL for this plan? You may use the Poisson approximation. .1.188 ACCEPTANCE SAMPLING I 11. SAMPLING BY VARIABLES The procedure for sampling by variables is similar to the procedure for sampling by attributes. for example.: : O. Find the AOQL for plan 3 of Section 11. which is.2. A single sampling plan has n:. A single sampling plan has n = 200 and c = 5.08'1 What is the AOQ for p :.03? What protection does the buyer have against accepting lots with p :. What is the probability of accepting a lot with p = O. Design a single sampling plan that has a f3 = 15% when p = 0. The engineer accepts the lot if there are four or fewer defectives. which implies a multivariate testing situation. in turn. EXERCISES 11. compared to a critical value.:::: 150.06. = 5% when p = 0. say. There can be difficulties when thc conformity of the item calls for its being satisfactory in more than one characteristic.: : 0. sampling by variables has the advantage of requiring smaller samples.06? 11. The set of n resistances is reduced to a single statistic. for each item. One draws the sample and measures a characteristic.4.15. the resistance. and 1l. 1l. such as the mean or standard deviation.6. 12. n 2 = 2n\. If the number of defectives is neither so large nor so small as to make a decision obvious. The first has n] and the second n 2 • ]n many double sampling plans. and (. the buyer gets 100 fewer items per batch than otherwise. There are acceptance numbers: CI' c 2 . We take an initial sample from a lot.Statistical Methods in Engineering and Quality Assurance Peter W. If it is a very good sample. we take another sample to settle the issue. we reject the lot. It seems a foolish waste to keep on inspecting the whole of a large sample if it is obvious that the vendor is producing at excellent quality with a percentage defective well below the target value. either as a direct charge for the inspection at the buyer's site or as an overhead item figured into the vendor's bid for inspection at the vendor's factory. If it is a very bad sample. One way of getting around this problem is to adopt a double sampling procedure. John Copyright © 1990 by John Wiley & Sons. not a condition of the modern MIL-SID-105 tables. we accept the lot. This is necessary condition. In theory. On the other hand. THE COST OF INSPECTION Except for defining A Tl and mentioning the Dodge-Romig rectifying inspection plans. The extra inspection cost will be borne by the buyer.3' items not a but is three 189 .2. It used to be customary in military procurement. DOUBLE SAMPLING We will investigate sampling plans with two samples. M. it can be argued that it also called for inspecting n = 200 items rather than n = 100. one could continue this procedure to a third sample or even a fourth. If the testing is destructive.1. the buyer could argue in favor of plan 2 because that was the plan that gave him the best protection. This is more expensive. The ultimate refinement is the method of sequential analysis developed by Abraham Wald (1947). thus far we have ignored the cost of inspection in the discussion of sampling plans. In our examples. Inc CHAPTER TWELVE Acceptance Sampling II 12. The vendor who is producing good quality items is rewarded by a marked decrease in inspection costs..6288 0. We can have £I I = 4 and £1 2 :s 5.0989 and the probability of acceptance overall is 0. we note that P(d.2. PI = 0.0000 .0049 (I. If P = 0.028. 3 dz-. When p = 0. Such a plan requires us to take n = 190 every time. c 1 = 3. The expected number of items to be tested is E(n) == (0. a = 4.714)(300) = 243. If d l > c 2 • we reject the batch outright. we can show in the same way that the probability of acceptance is 0. and so + 0. the probability of acceptance on the second round is 0.0001 0.O 2 dz s 1 d2~'.1 %.0894 0.4315 0.958. or £II == 5.1. Acceptance after the second sample can take place in several ways. 0 0. For the first sample. "2 Example 12. The double sampling plan may require either 100 or 300 items to be inspected. we usc the binomial tables.0710 0. inasmuch as only two-thirds as many items have Table 12.2579 + 0.05.0175 = 0.141)(300) == 128.190 ACCEPTANCE SAMPLING" We draw the first random sample and note the number. (J007 0.02. Summing.0902 0.05. we give the lot a second chance. We illustrate this method of double sampling by an example. If c 1 < d 1 :s c z . d I ' of defectives. with n = 100 and P = 0. Let"l = 100.258 = 0.S59)(loo) + (0. The possibilities are listed in table 12.0114 O. there is probability 0. We then sum the products of the probabilities of li l and d 2 to obtain the desired probability. 0 A single sample plan with Po = 0.5% relJuires a sample size about II}O.(Xl31 0.2%.2. and find that the probability of acceptance is 0.7867 0. and c 2 = c 1 = 9.2351 0. and f3 = 27.859 that we will only need the first sample.0176 0.02.1. and so on.275..0002 d2 5 5 d z 54 d~:<. so that a = 4. > 9) E(n) == (0. d 2 :s 4.0007 0. Suppose that p ~ 0. Let d 1 denote the total number of defectives in the second sample.2.028)(100) + (0.02. and let £I = £II + d 2 • We accept the lot if £I :s ('3' If d> c J ' the lot is rejected. n l = 200.141. If dl:s c i • we accept the batch with no further sampling.0353 0. When p == 0.859 and the probability of nonacceptance is 0.1 First Sample dt lit dt dl dt dt =4 =5 =6 =7 =8 =9 Prohahility Second Sample Prohability Product 0.0222 0. We draw another random sample of items.05.02. and no decision. 12.. If (12. reject it. if L > k . It would seem that with bad luck. H Il . • . . Suppose that we wish to test a simple null hypothesis. THE SEQUENTIAL PROBABILITY RATIO TEST Wald's sequential probability ratio test was developed during World War II and published in 1947. 12.. this is theoretically possible. his procedure called for testing many fewer items than the traditional fixed-sample-size procedures described earlier. just as it is theoretically possible that we can toss a coin for months before getting the first head. The plotted points are joined by a line as in a control chart.THE SEQUENTIAL PROBABILITY RATIO TEST 1. The line starts in the no-decision region. These lines divide the paper into three zones: acceptance. The items arc sampled one by one.3. rejection. After each observation is made. We keep on sampling until the line crosses into either the acceptance region or the rejection region.4. identically distributed random variables: XI' x 2 ' . against a simple alternative. if L < k o.91 to be sampled. but Wald showed that in practice. ko and kl' to correspond to the preassigned values of alpha and beta. and We observe a sequence of independent. After each item. • Suppose that the density function for X is j~l(x) under the null hypothesis and II (x) under the alternative. we could keep sampling forever.4. The vendor with poor quality has to pay for inspecting more items. . The basic ideas have heen given in the previous section. we accept . SEQUENTIAL SAMPLING The procedure of sequential sampling was developed by Abraham Wald for the United States during World War II. The total number of defectives at each stage is plotted against the sample number on a chart that has two lines drawn on it. we calculate the likelihood ratio: We calculate two numbers. . we accept 110 . HI' about a parameter 4>: . we count the total number of defectives so far and make one of three decisions: accept the lot.x n .1 ) we take another observation. or keep on sampling. S.4. The testing stops when the path crosses one of the two boundary lines.5. . we assume that {j = ILl .~ ex' k =-I which are very good for small values of alpha and beta. In the first.5.4.3) As long as the path of Y remains between the two lines. After some algebra. corresponding to sampling for variables.5.lLo > o. so that in practice we use the approximations k o =--L t .5. We have and For convenience. The calculations of the correct values for ko and kl are very difficult. X will have a Bernoulli (binomial) distribution.lLo). It is also useful in practice to replace the expressions in equation (12.1) by their natural logarithms and to write (12.4.2) We now consider two examples. the inequality of equation (12. X will be normally distributed. the bounds on Y given in equation (12.2) is reduced to (721n(ko) l> < y n{) n-T< (T 2 In(k) {) (12. We write Yn for the sum of the first n deviations from lLo: Yn = E (Xi . we continue to take more observations. 1 sis n. 12.2) and (12.192 ACCEPTANCE SAMPLING II HI. in the second.1) If we plot the data with n along the horizontal axis and Y vertically.ex and 1.1) become a pair of parallel lines: Y= (72 In(k o) nl> +{) 2 (12. THE NORMAL CASE Suppose that X is normally distributed with known variance and unknown mean. Alternatively. Again.6. 0. .=x.105. 0. therefore.x.)-'2 the two boundaries become the two horizontal lines (12. accept the alternate hypothesis that JA. = 50. and (3 = 10%. In(k. JA-o = 40.90 = 0.= JA-l = 50.-JLr.89. l> = 10.::. we can write X Z Z 47 2 2 50 5 7 53 8 15 46 1 16 50 5 2] 47 2 23 54 9 32 48 3 49 4 35 39 The upper and lower limits are Z = 36 and Z = -28.0.95 = 0. Then the boundary limits are at the same distance above and below the center line Z = o. = --2. k. AN EXAMPLE Suppose that Vex) Then = 125.- + JA-u JA-. the graph enters the critical region at the ninth observation. 12. = 0. This is particularly true if a = (3. 2 lj and :.4) and and the plot looks like a Shewhart control chart.28 Y= 5n +36. The boundary lines of the critical region are Y=5n . and we.5.) = 2.25. we write z. JA-.193 AN EXAMPLE If instead of considering deviations from JA-o. and Consider the following sequence of observations: n X Y 5n -28 5n + 36 4 46 17 3 53 30 -18 46 -13 51 1 47 7 50 -23 41 2 5 6 47 36 50 46 -8 56 ··3 61 53 2 66 7 54 67 8 48 49 75 84 7 71 12 76 17 81 9 The path of Y goes across the upper boundary at the ninth observation.05 In(k o) a = 18.10 ko = 5%. 7.10.. the boundaries for Z are horizontal lines.2) where we have written h = In(PI) .6 )3 5 h . .06 f3 == 0. + g.·l'< I ' Puqo (12.) = ±1.7. HI: P =0. If we write ZI = X. the boundaries are two parallel lines. In(q1) Po lJo and (12. = 1 if the ith item is defective. = In(94/98) = -0 . has a binomial distribution.14 • -In 0.194 ACCEPTANCE SAMPLING II 12. Y n = L: XI' 1 S i :s n. In(k o ) h = ··2.197 • (l... k k 0<" Y.S. AN EXAMPLE OF SEQllENTIAL SAMPLING BY ATTRlBUTES Suppose that we wish to test P = 0. and XI = 0 if the item is nondefectivc.3) Again. SEQUENTIAL SAMPLING BY ATTRIBUTES We now tum to the case of sampling by attributes and let x.1) which reduces to In(k ) In(k1) h h o --<Y+ng<-- (12.t. Plq. The two hypotheses are and We continue to take more observations as long as Y n Y .7. 12.94) = 1.98.02 g (0.Ob)' = In ( 0.7.Y27.02 Ho: with a = vs.. In~. the limit is 2. The inequalities for the likelihood ratio define two pairs of parallel lines. . 1. we note that for n = 2. . 0. Similarly.927. It is unity for n = 80.0365n:5 Y:5 ] .19. The lower limit is negative if 11<52. If the first two items are defective. we continue sampling. there arc five defectives in the first 84 items. (ii) if there arc no defectives in the first 53 items tested. each defective causes a jump of one unit. have to be considered in terms of integers.(3)/(ex/2).93 + 0. 12.0365n = 3 when n = 29. and we reject Ho. The lines mark off a V-shaped region on the graph paper. arc broken into two single-sided tcsts. and so on.927 + 0. For the upper limit.79. We accept Ho if there are no morc defectives until n = 81. Thus. The path of Y is a horizontal line when there are no defectives.39. We again plot Y = E (x. we accept lIu .000. and so on. The bounds k() and k. Y = O. and the path of Y will cross the upper line if there are three defectives before n = 29. we will have Y = 2 when n :=. (iii) if there is exactly onc defective in the nrst 53 items.927 + 0. the path will not cut the lower-limit line until n = 53.9.#-to). (i) we cannot decide to accept Ho if n < 53. there are four defectives in the first 56 items. Corresponding lines in the pairs intersect on the center line.195 TWO-SIDED TESTS We continue to take more observations as long as -1.0365n ± 1. respectively. The limits.0365n . TWO-SIllED TESTS Tests of a simple null hypothesis against a two-sided alternative vs.39. are set at (31(1.(/2) and (1.: 2. there arc three defectives in the first 29 items. This means that if we have an initial string of nondefectives corresponding to the line Y == 0. HI bccomes or with the risk of type I error set at a 12 on each side. We therefore accept III if (i) (ii) (iii) (iv) the first two items are defective. .196 ACCEPTANCE SAMPLING n The lines above the center are and y == ( ~Z) In( k J) + (~) n ... The lines on the bottom half of the paper are their reflections in the center line: and 12.•. +I .. Then In(k o ) == -2...6.•....•..---------+---------~-----I I I I I I I . 11-0 = 40.+I •••••.25 and In(k J ) = 2. The pair of parallcl lines abovc the center line are I I I I I 60 I -------~---------~-----I I I I I I I 20 I I I I I -------.•• I -20 I I I I I t I I I -------+---------~---------T---------~-----I t I I I I I I I I I I I I I I I I I r I I I I I I I • -----~---------~---------~-----I I I -60 I I I I I I 2 4 6 Sample number- 8 Figure 12..••.. 10 12 .10. but with a = 10%.•••.... 0 = 10. AN EXAMPLE OF A TWO-SIDED TEST Suppose that we had elected to carry out a two-sided test in the example of section 12...10...•..1... Twu-sided sequential test for section 12.... We should have V(x) = 125.89 as before...... and (3 == 10%...10. a == 10%....... corresponding to 5% in each tail... 1 and y = -5n + 2R. crosses the line between n = 8 and n = 9. which appeared in 1963. MIL-STD-123SB is a collection of continuous sampling plans for attributes. In 1971. no tables arc included.4.12. 12. Y".197 MIL-STD-105D Y n = 5n + 36. Band C.10. It closely paralleled MIL-STD-105A. A complete account of these systems can be found in the latest edition of the book by Duncan (1986). 12.10. . the graph would be the upper half of figure 12.1. The tables consist of single sampling plans. in 1957.L = 40 should be rejected in favor of J.6. It went through two revisions. It has not been changed to keep pace with the changes in MIL-STD-lOS that came with the current version. This sampling system contains a collection of sampling plans. double sampling plans.1. This system. The U.! . MIL-STD-414.1 and y = 5n . called ANSI/ASQC Z1. before major changes were made in the current standard 105D.11. The plot of this two-sided test is shown in figure 12.1.S. parallels MIL-STD-IOSD.36. the American National Standards Institute and the American Society for Quality Control produced a civilian standard. D. ANSIIASQC Z1. we see that the cumulative sum of the deviations. MILITARY AND CIVILIAN SAMPLING SYSTEMS The standard for attribute sampling used in defense contracts in the United States is the official system called MIL-STD-I05D. A modern civilian version of MIL-STD414 was brought out in 1980. MIL-STD-I0SD The MIL-STD-lOSD system contains numerous tables of plans for sampling by attributes. The two parallel lines below the center line are Yn = -5n .9.28. It includes the CSP-l plans of Dodge (1943) and the CSP-2 and CSP-3 plans of Dodge and Torrey (1951). The 1981 revision of this standard contained the change in terminology for defective to nonconforming and similar rewordings. For the one-sided test of section 12. Although sequential sampling is mentioned in the accompanying handbook. In it. indicating that the null hypothesis J. and some multiple sampling plans involving more than two stages.L > 40. Department of Defense issued a system for sampling by variables. The 105 system first appeared in version A in 1950. can be considered satisfactory as a process average. reasonable costs for inspection arc included in the price agreed upon beforehand. 12.13. good records must be kept to which the buyer must have access. Code letter K calls for a sample of 125 items. The decision whether to use single or double sampling may be left to the producer. The buyer contracts for a quantity of items that meets a set of standards. The procedure calls for three levels of inspection: normal. indifference quality. then the buyer profits too by receiving the better quality product. for the purpose of sampling inspection." The handbook adds that "AOL plans arc designed to protect the supplier from having good lots rejected. " The term lot tolerance percent defective (LTPD) is replaced by limiting quality (LO). which is defined therein as "the maximum percent defective that. A new term. the producer hegins by looking in the master table for normal inspection.198 ACCEPTANCE SAMPLING II The tables are based upon the AQL. for the given AQL. This is the quality for which Pn (the probability of acceptance) is SO%. We discuss the situation in which the producer is making batches of the same size. The consumers' risks of accepting product of inferior quality are only indirectly considered by studying the OC curve for the AQL sampling plans. and tightened. Reduced inspection means increased profits. If the inspection is to be carried out by the producer in his plant. The implication is that the producer is paying for the inspection. for example. The buyer's inspectors must also be allowed access for occasional supervision of the actual inspection carried out by the producer. The producer of poor quality has higher costs. is introduced. a producer who is producing lots of ISOO items under general inspection level II. This can produce hard feelings. Some people design plans based on the indifference quality. The producer then finds the sample size code from a table and looks in another table to find the critical value. But if the producer is thus nudged to improve quality. The buyer and seller agree upon the AOL and the general inspection level (which is a function of the importance of the item being inspected). Consider. That lot size corresponds to sample code K. or values. but they are not in common use. day after day or week after week. . If single sampling inspection is to be done. reduced. The producer of good quality product is rewarded by reduced inspection costs. THE STANDARD PROCEDlJR"~ Once the buyer and the producer have agreed on the AOL to be used. the MIL-STD-IOSD procedure follows several steps. Continuing with the example of an AQL of 2.5%.199 TIGHTENED VERSUS NORMAL INSPECTION Table 12. 1 2 32 32 32 64 3 4 5 6 7 32 32 32 32 0 1 3 5 7 10 13 32 96 t28 160 192 224 (Ae) =2. If the producer chooses double sampling. the lot will be accepted if there arc three or fewer defectives and rejected if there are seven or more. Inspection usually begins at the normal level. This is as close as MIL-STD-105D comes to sequential sampling. the lot will be accepted if there are eight or fewer defectives in the 160 items and reject if there are nine or more. but the requirements for acceptance are more rigorous.5%) Rejection No. If the AQL is 2.5 and sample size K.1. If it becomes apparent that the quality level has deteriorated. the lot is accepted. If there are three or fewer defectives. (Re) 4 6 8 10 II 12 14 If the AQL is 1%. After the second sample.:es of severity of inspection-normal.15. Suppose that the AQL is 2.5%. III == 1/ 2 = 80. If it is clear that the quality level is very good. tightened inspection is required. TIGHTENED VERSUS NORMAL INSPECTION Tightened inspection usually calls for the same sample sizes. Criteria are given for switching from normal to tightened inspection and hack. the criteria for tightened inspection . if there are four or more defectives.13. SEVERITY OF INSPECTION There arc three degn. c (or Ac). which is used when there is no reason to believe that the quality level is either better or worse than has been specified by the AOL.14. and tightened. The acceptance and rejection numbers after each sample arc given in table 12.1. the supplier may he allowed to go to reduced inspection. After the first sample is drawn.13. This is a change from the older procedure in which 1/ 2 = 2n I' The multiple-sampling choice allows up to seven samples. the critical number. and from normal to reduced inspection and back. each of 32 items. the lot is rejected. reduced. 12. Multiple Sampling for Sample Size Code K (AQL Sample Number Sample Size Cumulative Sample Size Acceptance No. 12. is three. c = 7. Sample Code K. the sample size is reduced to approximately 60% of what is required for normal inspection.1.1.1.1 the acceptance number for the first sample under multiple sampling is replaced by an asterisk.5% Type Cumulative Sample Size Acceptance No. one can only reject it or take a second sample. Rejection No.16. REDUCED INSPECTION In reduced inspection. AQL =2. This also occurs in table 12.1. Single 50 3 6 Double 32 1 64 4 :. The use of this table and the apparent Table 12.200 ACCEPTANCE SAMPLING II Table 12.16.16. This denotes that one cannot accept a lot on the basis of the first sample. with a corresponding change in the cost of inspection. Tightened Inspection.5% Type Cumulative Sample Size Acceptance No.15. Single 125 5 6 80 2 5 160 6 7 32 '" 4 5 6 Double Multiple 64 1 96 128 160 2 3 5 8 192 7 9 10 224 7 9 are given in table 12. Reduced Sampling for Sample Size Code K.15. AQL =2. 12.16. We continue with our example and give the criteria for reduced sampling in table 12.15. Rejection No. 7 '" 5 Multiple 13 26 39 52 65 78 91 0 I 2 3 4 6 4 6 7 8 9 \II .1. In table 12. C.18. If five consecutive lots are accepted under tightened inspection. 12. and D.17.9. The producer is allowed to go from normal to reduced inspection if (i) the preceding ten lots have been accepted. If. three series of plans covering the situations in B. SWITCHING RULES We begin with normal inspection. has switching . the producer may return to normal inspection. the acceptance and rejection numbers are not consccutive integers. but inspection goes back to normal anyway. If the number of defectives faJls between the acceptance and rejection numbers. If the number of defectives is greater than or equal to the rejection number. the lot is accepted. ACCEPTANCE SAMPLING BY VARIABLES The MIL-STD-414 document has four parts. Plans arc given for one-sided and two-sided specification limits. therefore. Tightened inspection then continues until either of the two following events occurs. Part A is the general description. Notice that in reduced sampling. the inspection is stopped. in any lot. Part B has plans based on the use of the range to estimate the (unknown) standard deviation.ACCEPTANCE SAMPLING BY VARIABLES 201 discrepancies between the acceptance and rejection numbers are explained in the next section. There are. and (ii) if the total number of defectives in those ten lots is less than or equal to the acceptance number for the gross sample that is given in a second tablc. the process has not won its way back to normal inspection. The more modern civilian system. Part C has plans for use when (F is estimated from the sample standard deviations. and a search is made for an assignable cause. 12. If two lots in any sequence of five are rejected. Part 0 handles the situation when u is known. Inspection and production will resume when it is thought that proper quality has been restored. the producer must switch to tightened inspection. The producer may continue on reduced sampling as long as the number of defectives in each lot is less than or equal to the acceptance number in the table. the number of defectives exceeds the acceptance number. If after tcn lots have been inspected. the producer must return to normal sampling. We are speaking here of rejection on the original inspection. A lot that is rejected on the original inspection and then accepted after rectification is regarded as a rejected lot. ANSI/ASQC Z1. the lot is rejected. 6 103.= 103. the reader is referred to Duncan (1986).7 107.8 108.7 100.6 93.1 to the following data sets.3 87. Design a sequential sampling plan for sampling by variables with the following specifications.3 97.15 when J1. We wish to have a = 0.9 92. and rejects it if d>7.4 102.9 104.8 101.2 103.5 102. (ii) Items 1.05 when J1.1 111. 174.8 96.3.7 98.= 100.3 102. 135.202 ACCEPTANCE SAMPLING II rules between normal inspection and reduced or tightened inspection that arc similar to those of MIL-STD-105D.7 101.4 107.8 12.3 99.8 92.8 104. Apply your sequential sampling plan of Exercise 12. The engineer accepts the lot on the first sample if ds3.6 104.3 98.3 103.9 109.3 1O~U 95. Design a sequential sampling plan for attributes that has ex::: 0.2 113.0 95.4 103.9 98.8 106. The lot is accepted after the second sample if the total number of defects is eight or less.7 87.63. and f3 = 0.4 85.0 111.06.0.2 101. (iii) Items 75.4 96. 12. 12.1 103.7 (ii) 102.3 105.03? .20. When would you make a decision and what would it be? (i) The first three items are defective.9 92. 101.0 109. A double sampling plan has n I = n 2 = LOO. and 67 are defective.3 to the following three data sets.4 99.6 100.05 when p = 0. For further details.4 96.2.4.8 97.1. 175. 12.0 107.4 \03.42. What is the probability of accepting a lot with p = O.9 95.5 108. When would you make a decision and what would it be? (i) 98.2 109.3 99.0.3 109. 190.2 102.0 98.3 98.8 L05.2 105.2 96.6 102.0 109.1 103. 91.0 108.03.6 105.8 99.9 99.8 109. 17.5 101. The standard deviation of the characteristic is 17 = 6.37.5.9 106. EXERCISES 12. The target average for the characteristic being measured is 100. Apply your plan from exercise 12.5 108.2 93. and f3 = 0. 12.1.4 102.7 103. 101.15 when p = 0. and 202 arc defective. on combining rules 2 and 3 with rule I. Inc CHAPTER THIRTEEN Further Topics on Control Charts 13. o Rule 2. we now repeat the three rules that were proposed. The alpha risk is kept small. which is the risk that we will not stop the process when we should? In addilion to the standard Shewhart charts. M. a shift in the mean of the process occurred after the thirtieth run. We discuss three of these charts: the arithmetic movingaverage chart. John Copyright © 1990 by John Wiley & Sons. we can use several other charts whose purpose is to enable us to detect more quickly changes in the process mean. but at sequences of several points.1. focusing. we will rarely (one time in 400) stop the process when we should not. One point outside the three sigma limits. How can we speed up the detection of changes in mean? It is done by looking not just at the individual points on the chart. Rule J. It became clear that the traditional rule-take action if there is a point outside the outer control lines-did not do a very good job of detecting that shift. The Shewhart control charts arc conservative in the sense that as long as the process is rolling along with the process mean equal to the target mean.Statistical Methods in Engineering and Quality Assurance Peter W. We are not advocating replacing the Shewhart chart with these new chartsfar from it. the cumulative sum control chart (Cusum). and the geometric moving-average or exponentially weighted moving-average (EWMA) chart. INTRODUCTION In the first example of an x-bar chart in chapter nine. 0 203 . How about the beta risk. For convenience. Two consecutive points either above the upper warnmg line (two-sigma line) or below the lower warning line. We begin the chapter with another look at the stop rules that were considered in chapter nine. That problem was even more conspicuous in the example of the c-chart in chapter ten. since we needed 15 more runs to trigger it. We are suggesting that the engineer should keep several charts simultaneously. in particular. More generally.204 U. FURTHER TOPICS ON CONTROL CHARTS Rule 3. . Suppose that it takes t points until a run of seven occurs. Using the results that are derived in the next section. if the probability of a point above the line is p.8413f (0. rule 2 calls for a run of two points.) What can be said about the probability distribution of t? The derivation is beyond the scope of this book. Rule 3 calls for action if there is a run of seven points.(0.30. which we round to 15.8413)7 = 14.2.8413. The normal procedure is to combine them into a single rule. The probability that the next seven points will constitute a run that triggers rule 3 is P =: (0. we follow the argument of Page (1954). DECISION RULES BASED ON RUNS Rules 2 and 3 are both based on runs of points above or below a line other than the outer control lines. t is a random variable. (Beware of the double use of the word run in this sentence: the word run in ARL stands for t.3. whose expectation is the ARL when this rule is used. rule 2a: RuLe 2a. the probability that the next seven in a row will be above the line is p 7. 13. COMBINED RULES Rules 1 and 2 are rarely applied separately. But we need to ask a different question.1587)(0. but we can derive a formula for its expected value. the ARL for the rule of seven is given by the formula £(1) = 1 . so that the mcan changes from JL to JL + if/Vii.8 . Take action if either there is one point outside the outer control lines or there are two successive points outside the warning lines and on the same side. A run of seven points either all above the center line or all below 0 13.8413)7 = 0. In developing the ARL for the combined rule 2a. [J The table given in chapter nine considered only one rule at a time. The probability that a point falls above the ccnter line on the chart is now increased to 0. we can show that when there is a shift of aVn in the mean. Suppose that there has been a shift of + a/Vii in the mean. and we stop with L' = 1. in which case we start again. Let the probability that a point falls in each of these three regions be p (good).3. there are three possibilities: (i) the next point is good.1) If we start from a point in the warning region. (iii) the next point is in the action region. q (warning). The process will stop if there are two successive points in the warning region or one in the action region. and the action region consists of values outside the action lines. (13. For rule 2a. the warning region consists of values between the warning lines and the outer (action) lines. We combine these possibilities into the equation L = p( 1 + L) + q( 1 + L') + r . ( 13. and action.3.3. warning. and L' =1 + L. let L' be the ARL when we start from a point in the warning region. (ii) the next point is also in the warning region. and r (action).2) Eliminating L' between the two equations gives L = _l_+---=q_ 1 . there are again three possibilities: (i) thc next point will be in the good region. and we have the expectation of L more points: (ii) the next point is in the warning region.P -. which we can call good. The corresponding formula for the ARL is 1 -q L = 1 _ (p m + q) + pqm m = 1 -q In r+ pq • (13.205 COMBINED RULES The set of possible values of i can be divided into three regions.pq (13. and we expect L' more points. (iii) the next point is in the action region.3) A more general form of the rule would require either a run of m points outside the warning lines or one point outside the action lines. We combine these possibilities into a second equation: L' = p( 1 + L) + q + r . and we stop. and we stop with L' = J. with probability p. the good region consists of values of i between the two warning lines.3. Let L denotc the ARL when we start the sequence of points from a point in the good region. where p + q + r::: 1.4) . If we start with a point in the good region. 1'( 1. The hasic Shewhart chart with rule 1 uses only the single latest .0606.2.5) .. Then we apply equation (13.6. A TABLE OF VALUES 01<' ARL Table 13. and p 1. q L q=l-p-r. p = 1 .1 gives values of the ARL for various values of the shift in the mean. AN EXAMPLE Let the shift be dlTlv'Tt = O. We discuss three such charts.F(d). F(3 . 13. but rule 3a has larger values of the alpha risk.0062 . r =1- In this example.9332.493 . We calculate the ARL for the two rules in the following way.5) = 0. 13.5.4) with m = 7.d) .3085.206 FURTHER TOPICS ON CONTROL CHARTS We can also combine rules 1 and 3 into rule 3a: Rule 3a.50'Ivn. 0 To calculate the average run length with rule 3a for a positive shift in the mean.5) r = 1 ~ F(2. MOVING AVERAGES The charts that are presented in this chapter for detecting shifts in the mean eome under the general heading of moving averages.0062.c/ 2 = 103. q = 1'(2. = 0. 13. we let the "good" region consist of the negative values of x and let the warning region run from the center line to the upper three-sigma line.q7 = r -I pq 7 = = 0.06. For rule 2a. r ~ 0. Rule 3a is triggered hy a run of seven or a point outside the outer limits. whence 33. If we let r = 0 and m = 7.4.5. r+ pq L = For rule 3a. 1 .5) = 0. we derive the formula that was used for rule 3 in section 13. The values of ARL for rule 3a are shorter than those for rule 2a.6R53. = 0. p = F( 1. Take action if there is eithef a fun of seven succcssvie points on the same side of the center line Of a point outside the action lines.3. the variance of y is denoted by (1"2.I values of Y" The value of k used in the chart is the choice of the engineer. but with weights that decrease as the points grow older. 1/ k. Y . Average Run Lengths to Detect a Shift d(u/v'li) in the Mean..0 1. For this reason.. k = 4 and k = 5 are commonly used. Arithmetic and EWMA charts have applications beyond traditional quality assurance. we change our notation to make it more general in its application..7. The ith sample is the sample drawn at . .6 0.4 1._--------- 85 44 44 25 17 26 17 J3 10 11 R 7 H 6 5 6 5 observation to make the action decision and pays no heed to any of the previous data. The arithmetic moving average uses the average of the k latest observations with equal weights.2 1. the observed response is denoted by Y. I. Sample number. The geometric moving average (or exponentially weighted moving average. the ordinate for t = 5 is z. along the horizontal axis. The cumulative sum control chart uses all the observations that have been made with the same weight. = _t/ and V(y. 13.) = (T. ARITHMETIC-AVERAGE CHARTS The first of our charts for detecting shifts in the mean is the (arithmetic) moving-average chart. In the special case of Shewhart charts.6 1. If k = 4. Against this.4 0. notably in econometrics and in engineering process control.5.8 2. but Z" the mean of that and the previous k .1. and neglects ohservations made before then.8 1. The moving-average charts use the earlier data to different extents.2 0. = Yz + )'3 + Y4 + Ys 4 . through the rest of the chapter. is measured along the horizontal axis. or time. We plot the sample number as discrete time.207 ARITlIMETIC-AVERAGE CHARTS Table 13. for the ith sample. One-Sided Tests Rule 3a Rule d 2a 194 571 273 142 77 0.0 0. we plot not the value of Y. Rule 2 employs only the latest point and its immediate predecessor.0 _. EWMA) chart uses all the previous observations. .... On the other hand..13 t: « .208 FURTHER TOPICS ON CONTROL CHARTS Figure 13. For low values of k.I~--'--'-.-'-.J.L.L.I--I.2 10..7. Because we are now plotting averages of four samples.. 9.. beginning with Xl + x2 + X3 + x4 = 10.. I I I I I ..8.13 4 4 = to.<> ~ ::- ~ . 10 9..1 shows a moving-average chart with k = 4 for the data of figure 9. the smaller changes are easier to detect with higher values of k.4 ...3 E -5 1+-+~--T-4---+--+--+--i-----. these charts detect large changes in the mean quickly. I I ----+---------~-----I .L-.1.-.I.1. but at a 11.'---~ 10..<L-. this is the first average that contains a point with the "new" mean..... 1525.J.467 50 .. I I I I I I I I I I ---------~---------~---------~-----I I I I I I 9.-.. 10. but they are slower to detect smaller changes..6 .. I I I I .793 <> Q/.7.-j""". .-'--'--'-.l-.} + x30 ) /4.7 I 1 I . The twenty-seventh average is (X27 + X2l! + x2 .70 + 9..I-.) 10. the outer control lines are pulled in to The shift in the mean that took place after the thirtieth sample is now more apparent than it was with the traditional x-bar chart.50 + 10....28 + 10. There are 47 averages on the chart.j""".I~ o 10 20 30 40 Sample number Figure 13. ..~ '.9 10..J... Arithmetic moving average. The procedure for using these charts amounts essentially to performing a sequential probability ratio test backwards. using previous data on the performance of the plant. it is usually the value of the process mean under steady state. f. Often.). Two parameters are involved in the test procedure: the lead time. In Cusum charts. Lucas has written several papers on the use of Cusum charts including parabolic masks (1973. The similarity to the two-sided sequential probability ratio test with its V-shaped critical region will be apparent. Cusum charts. which is a piece of cardboard or plastic with a large V-shaped notch cut in it. Page (1961).L() ± 8. between the edges of the V. S. Hence. The lead time is a function of the values of the alpha and beta risks that are acceptable. . The horizontal line joining those two points bisects the angle of the V. The choice of the angle depends on how large a deviation from the target value the engineer is prepared to tolerate. The damping effect of using high k values slows the detection of the larger shifts. CUMULATIVE SUM CONTROL CHARTS Cumulative sum control charts were introduced by E. which will happen if i has been rising or falling sharply. . i = 1. Thus. the apex of the V-mask is placed at (d + c/' z. The process is declared to be out of control if the cumulative sum disappears beneath the mask (crosses the edge of the V). the two critical values.. implies that the engineer is carrying out a two-sided test. 13. are symmetric about the target value. i. d.Lo is sometimes called the target value. !-Ln..209 CUMULATIVE SUM CONTROL CHARTS price. 28.c. 1976. these charts may accidentally introduce some specious trends.. they are established by cut-and-try methods. The use of the V-mask. The latest point to be plotted on the chart is (I" Z'.8. The values of these parameters can be calculated as one would with a sequential test procedure. they are commonly called by the shorter name. thc two values that are used to decide that there has been a shift in the mean. we plot along the vertical axis the sum of all the deviations of the sample average from some predetermined value since an arbitrary starting time. The apex of the V is at the same level (z value) as the latest point all the chart with a lead distance. which is symmetric about the horizontal line. (13. d.1) where f. corresponding to the center line on the Shewhart chart. The engineer uses a V-mask. It has also been shown by Nelson (1983b) that in their effort to smooth out actual time trends in the data.8. say. and the angle.). I . 1982). The mask is placed on the chart with the V opening backwards (». and are at. The engineer should then choose the angle H so that the . 88 JO 10. where kJ = (1 . is made with the following parameters: 1-1-0 = 10.23 2.10.13) -0.25 -.23 The lead distance.96 II 11.30 -0.65 -·1. Example 13. a ::= 0. (0. which is shown as figure 13. Suppose an engineer makes a cumulative sum chart of the data in table 9.01 4 9. is calculated as d = 2(0.05.79 13 11.60 3. we can follow the argument of section 12.()6 0.6.30 .1.1. and so we take For a cumulative sum chart that is equivalent to a one-sided test.76 15 10.40 9 10. d.10.1.210 FURTHER TOPICS ON CONTROL CHARTS edges of the V have slopes corresponding to ±8/2.88 --1.50)' The engineer takes a mask with half angle (I = arctan(0.1 and wishes to check for a shift in the mean after 15 points.8.43 -0.25) and places it on the chart with the apex at a distance d to the right of the fifteenth point. we recall from section 12.1. In calculating the lead distance./3)/(0/2).5.35 -1.51 12 5 10.60 -0.13 and CT. otherwise. = 0.9 that the outer lines of the acceptance region in the two-sided sequential test arc z = ( CT~ 2) In( k 1) + ( ~) n . tan (I = 8/2 will be taken. If the scales on the two coordinate axes are the same.18 2 3 10. and 5 = 0.8.13 10.58 -0.08 HU() -0. The data points and the Cusums are as follows: Sample number Sample average 9.5. The lines intersect at The V in the Cusum chart is facing backwards. tan (I will be modified accordingly.15 8 9.442.8.88 7 10.28 1.27 14 11.442Y In(36) = 5.60. The . The chart.25 Cusum E (i . which were the process values calculated earlier in section 9.66 6 9. and /3 = 0. 33.251. it is Y = 0.. (13.3 Sample number Figure 13.25t - 1.8..: G) "0 G) .66> 1. = O..23) with slope 0.92 = 1.00 + 5.::. n .7 "3 § u -1.S.251 + y" .1. Since z 12 = 0.1 = 1. Cumulative sum chart.60. lower arm of the V intersects the plot of the Cusums between samples 12 and 13. below 0. i.1. When the mask is used at the nth point. 0..33..92. y. > . 3.251 . The fifteenth point has z = 3. when t = 13.92 = 1.d) . When the V-mask is positioned with its apex a distance d ahead of the fifteenth point.92. its arm will intersect the Cusum graph if there is a value of z.08 and z l.25.1.:.40 - n 4' . and so the equation of the lower arm of the mask is the line that passes through the point (15.08.1. For this example.. The engineer concludes that there has indeed been a shift in the mean.23. the arm of the mask is the line y = Yn + ( ~) (t . 0.§ . the ann of the mask intersects the graph between t = 12 and t = U.2) .25t . When (= 12.1.211 CUMULATIVE SUM CONTROL CHARTS 4. we have O.7 .51 < 1.e .7 r-r-r-T"" 2. 95 -0.45 1. = z" - 5d T' (13.92. the arm of the V-mask becomes the horizontal line z=z.40.25 12 44.55 3.25 10 44.00 45.10.H5 6 42.1) The mask cuts the graph if at some previous time t. The plant has been running at an average rate J.~) = 2: (x.5 with a = 0.55 7 Observation Deviation Cusum Observation Deviation Cuslim 17 IH 43.2 0.20 II 42.-1. The general form of equation (13.40 9 43.4 0.30 5 44. . (13.70 J3 42.05 2.0 and standard deviation (T = 1.45 8.9.9 2. A HORIZONTAL V-MASK If in the example of section 13.3 \. Z 15 = --0.9.7 1.60 .3 -0.65 6.9.9 0.9.Lo .2) Example 13. The plant manager wishes to be able to detect an increase in production ratc of 0.7 -0.10 8 44.9.55 -0.60 15 43.f. we let Z. z.L = 43. = 2: (x. . The data in table 13.7 -0. .1.45 3.7 0. (n this example.9.4 2.1 arc daily production rates (conversion percentages) of product in a chemical plant.10 3 42. Data for Example 13.7 0.15 0.95 -0.49< -1.38) = y.4 -0.15 14 43.25 -1. Table 13.7 0.35 16 45.15 -0.85 -0.25t .1.1 2 43.75 4 43.0 --1.45 3.65 0.8.212 FURTHER TOPICS ON CONTROL CHARTS 13..1) is z.7 1. and zl2 = -2.35 44.O.15 IUS 1 Observation Deviation Cusum 42.0 -0.05 and f3 = (UO.25 3.52.45 -0.1. < z" .4 \.1.40.9.9. moving-average charts were introduced by S.13 I I ' .~=----:lOt-I 10 _ _-+-_ _--I10.4 j 1--=\-++-+-.1.25).2) becomes 1. -r---------~-----t Sample number Figure 13.99.r)z'_1 (13. W..JIrl.30 (13.10. = zn .25).10. 10. we plot the ordinate Z..2) are usual. GEOMETRIC MOVING AVERAGES Geometric.10. = zn - Equation 6.21 ) z.. Geometric average (r .9. The choice of r lies with the engineer. Values in the range 0. 0. Roberts (1959).8 10.213 GEOMETRIC MOVING AVERAGES The plant manager calculates deviations z.1) against time.5 In(18) = E (x.631 ~~ '5 10.43.( 0. = ry.. or exponentially weighted.10. + (1.10:5 r ::.~ 0.. .-. In these charts. (13. o 13.r.. The plant manager will report the shift in the rate after the sixteenth run. 11. we note that the variance of z.. Substituting for c. .3. r measures the depth of memory that the process itself has. is given by The sum of this series for large I is (13.YI' of the predicted response at time t from the ohserved value. Figure 13.1) where e. EWMA AND PREDICTION J. in terms of the responses: Z( = t ry + r( 1 . 0.3. These charts are usually preferred to the arithmetic moving-average charts.2 + .1.r) v .10. Hunter (19X6) approaches the EWMA from the point of view of prediction in a stochastic process.1. . is the deviation. e... . The lowest value of ): occurs at r = 0.10..2.1 shows these calculations for a set of 20 points for r = 0. The engineer who wishes for a value with more decimal places can use quadratic interpolation with the values 0. instead of z" and y.2 is good enough for most purposes.1). hence the name. The value 0. we get Y'+l = ry. Again. He lets y.r)ly I . This is the same as equation (13.1. The first observation is 23.214 FURTHER TOPICS ON CONTROL CHARTS The weights given to earlier observations drop as t increases. (13. + (1 (13. or time series.2. in equation (13. S. and he estimates the response at time t + 1 by the prediction formula (13. the predicted value of the first observation e. and may be estimated from the data by choosing the value of r that minimizes ~ Table 13.25 for the data of figure 9. one notices that this chart detects the shift in the mean faster than the Shewhart chart. like the terms in a geometric series.2) . 0..r)y..11. Equation (13.11.1) may be transformed to give z.10. and 0. y.2. The sum of squares increases as r becomes larger.10.1 shows a geometric chart with r = 0.3) To computc the control lines for the EWMA chart.10..1) with y. denote the response that is observed from the tth sample.6.4) 13. instead of Z'_(' In this formulation. and 0.11. I I + r( 1 .8.11. 40 0.69 23.64 0.5 24.32 0.24 3.18 24. on the same piece of paper.67 23.33 0.43 -0.63 23.3 24.HO 23. .76 23.3 23.16 y.7 23.1 23. Hunter (1986) suggests that we plot both )" and y. We arc seeing a drift upwards in both the observed and predicted values.20 -0.54 (l.03 24.45 -0..74 23.11.12 -0.56 23.215 EXERCISES Table 13.80 23.18 24.~4 23.28 0.35 24. - y.04 24.62 y.8 23.49 0. --0_20 0.56 0.45 0.Y.3 23.08 0.94 24.26 0.25.06 24.1 make the following charts: (a) an arithmetic average using three samples in each average.79 23. The reader is referred to his article for his insights into the usc of EWMA for control purposes.00 -·0. e~ " 23.74 0.22 24.()4 0.35 -0.70 r_.RO 23. Notc that in the arithmetic mean chart with span 3 and the EWMA chart the last point on the chart is close to the uppcr control line and the graph is heading upwards.5 24.92 24. -0.40 O.4 23.16 24. (c) a geometric (EWMA) chart with r = 0.79 23.76 23.04 0.70 0.)4 0.0.76 23.11 0.07 0.65 0.75 23.4R 0.85 23.1 23.81 0.40 -0.03 24.05 0.14 0.4 23. The change of emphasis is important.18 24.14 0. This is not true for the arithmetic .00 24.75 23.1. It turns our attention to the usc of the quality control chart as a tool for process control.71 23.29 (l.24 0.7H 0.06 y.39 -0.34 0. (b) an arithmetic moving average using four samples.45 1.27 0.66 23.!H 23. 23.1 24. EWMA I<'orecast Computations for Estimating r Data r == 0.1 24.1.31 0.81 23.76 23.~0 23.79 23.23 24.86 23.27 0.12 0.07 -0.71 23. For the data set in exercise 9.12 0.9 24.1H 3.14 r = 0.76 23.78 23.39 -0.14 -0.1.75 23.05 24.7H 13.39 -0.8.71\ 23.41 1. EXERCISES 13.14 0.1 24.fJ6 0.9.20 -0.2 23.21 24.3H 1.1 24.91 23.0 23.7 24.68 is the target value.15 -0. . -.16 24.02 24.3 24.73 23.31 0.6 23.80 23.15 24.32 24.6H 23.22 -0.3 24.84 r =0.OR --0.9 24.90 24.12 3. 13.8 from a target value of 39. Repeat exercise 13.216 FURTHER TOPICS ON CONTROL CHARTS averages with span 4. and continue with the 16th.0 in the data of exercise 9.831 0.827 0.1. etc.806 0.850 0. 13. 13. assuming that you wish to detect a shift of 0.892 0.846 0.792 0.825.786 0.863 0. Make a Cusum chart to test for a shift with a = 5% and {3 = 10%. Treat the data of exercise 13.878 Make the same three charts as in exercise 13. the standard deviation of the sample averages was 0.4. B. We wish to detect a shift in the mean of as much as 1. The process mean was 0.841 0.4. How many samples will you need before you detect a shift? 13.L =: 39. In this example assume that you wish to detect a shift of 0.801 0.3.85 and (J' =: 0. 13.862 0.868 0. Note also that the one sample that falls above the Uel in the x-bar chart does not cause an outside point on the average charts.78.828 0.910 0.871 0.855 0. assuming that J. Make the same three charts for the data in exercise 9. 13. Make the same three charts for the data in exercise 9. The following data set consists of 40 sample averages.819 0.6.4.2 as jf they were single observations and make an I chart. 0.909 0.5.0 from the target value of 64.. 13.7. and also without a mask.808 0.855 0.85 with (J' = 0. Make the same three charts for the data in exercise 9.5.S.3.803 0.905 0.850 0. until you spot a significant shift.827 0. Start testing at the 15th sample.841 0.754 0.6 for the data in exercise 9.837 0.3.L = 17.9.6 for the data in exercise 9.03.797 0. using a V-mask.871 0. Repeat exercise 13.865 0.834 0.838 0.3.2.797 0.842 0.78.8 from an original mean of J.882 0.811 0.837 0.800 0. .5. That does not mean that they necessarily rank below the average 217 . we have dealt primarily with univariate data. his scores in the Scholastic Aptitude Test (SAT) or the Graduate Record Examination (GRE) for English (verbal) and mathematics (quantitative). there are exceptionsthe short. By and large. perhaps. We can only give an introduction to the topic here. Smith (1981) and Fitting Equations 10 Data: Computer Analysis of Multifactor Data by C. but this is not true over a narrow range.. We do have a considerable number of arts students who are weak quantitatively when measured against the average for graduate students. the connection is not so clear. Wood (1980). In this chapter. In the next two chapters. When the range of students is truncated and consists only of those who are accepted to study for master's or doctoral degrees. fat man and the long. and his weight. measured his height. Inc CHAPTERFOURTEEN Bivariate Data: Fitting Straight Lines 14. R. our data will be a set of " points. John Copyright © 1990 by John Wiley & Sons. but not by some algebraic function. x. the better students tend to have high marks on both parts of the test.Statistical Methods in Engineering and Quality Assurance Peter W. INTRODUCTION In the previous chapters. the breaking strength of a strand of fiber. M. Prominent among them arc Applied Regression Analysis by N. y" or else. each one of which consists of observations on a pair of variables. Similarly. tall men are heavier than short men and heavy men are taller than light men. The reader who wishes to go further will find numerous books on the topic. the variables are obviously connected. Of course.1. We may have taken n male undergraduate students and for the ith student. In both these examples. and science students whose verbal skills are low compared to the graduate student average. Draper and H. Daniel and F. Each observation was a single measurement-the length of a rod. we turn to multivariate data: bivariate data in this chapter. when we look at the whole range of students. and more than two variables in the next. thin beanpole. The general topic of fitting lines and planes to data by the method of least squares is called regression. it looks as if there is a straight line around which the points fall. to predict his weight. Our basic problem in this chapter is to fit a straight line to bivariate data. - n-l n x for E(X) (14. Y will tend to decrease as X increases. and so the covariance is positive.2. We would like to be able.E( Y)]} . is a strong tendency for Y to increase as X increases.2.218 BIVARIATE DATA: FlTnNG STRAIGHT LINES for the man in the street. The reader should check with the local computer center to find out what is available.15. cloud of points running from the southwest to the northeast.2) .2. . They vary only slightly in dctail and in the availability of optional features.. When the covariance is estimated from a data set.1) Supposc that we draw coordinate axes through the centroid (x . we introduced the concept of the covariance of two random variables: Cov(X. In this chapter and the next.E(X)][ Y -. as with height and weight. if we are told that a student is six feet tall. the covariance is multiplied by (3. and from kilograms to pounds. Thc covariance is then estimated by L (x. Often. and the covariance will be negative.23. the cross product is positive.iust that they rank below the average for the subgroup that we have chosen. if Y is the yield of a product and X is the percentage of impurity. or correlation coefficient. the data will tcnd to fall in the southwest and northeast quadrants. If the scale of X is changed from meters to feet.205) =:: 7.y for E( Y). Every standard software package has subroutines for fitting lines and planes. In each of these quadrants. we use in the illustrations edited printouts from the Minitab software package.v) of the data. Y) = E {[ X . On the other hand. It is defined by Covi~. we will get an elliptical. The value of the covariance depends upon the scale of the variables. The data will tcnd to fall in the northwest and southeast quadrants where the cross products are negative. If we take our sample of undergraduates and make a scatter plot of height against weight. we substitute and . If there. THE CORRELATION COEFFICIENT In section 4. The correlation.28)(2. or cigar-shaped. of X and Y is a scale-free version of the covariance. 14. Y) tT)(tT y (14. A more practical usc would be to predict the yield of a chemical process given the percentage of impurity in the feedstock. - x)(y. with the aim of being able to use it to predict future values of Y given X. 3..3) . or vice versa. and 5.205. but the standard deviation of X was multiplied by 3. where 2 r = lL: (x. if X and Yare independent. If the line is horizontal. the sign of r being the same as the sign of the slope of the line. CORRELATION AND CAUSALITY One should avoid the trap of confusing correlation with causality.2.0.219 CORRELATION ANI) CAUSALITY It is estimated from a sample by r. then when Y> 0.28 and the standard deviation of Y by 2. -- 2: (x.2 = 0. More generally. with one . U) = corr(X. The population of Russia and the output of cups and saucers in the United States have both increased. and that Y = X2. It can be shown algebraically that r must lie in the interval .0. then corr(W. Y). It does not follow that if r is close to unity. On the other hand. over the past half century.3.9626 and r=0. 14. If the points lie on a straight line. and their correlation is zero also. - (14. the correlation is zero. Then we have x y 1 1 2 4 3 9 4 16 5 25 Applying equation (14. then X must cause Y.i)(y. the relation between X and Y is close to linear. It does not follow that if X and Yare highly correlated. and.98. Nor does it follow that if r = 0.205).. .1. by and large. we arc as likely to have X> 0 as to have X < O. - i)2 2: (y. The covariance of X and Y is zero. hence.1. it is a simple algebraic exercise to show that if W = a + bX and U = c + dY. . Suppose that X takes the values I. The correlation coefficient is invariant under changes of scale and changes of origin. the correlation is :!:: 1. . We have already seen this in example 4. In the case where we switched from meters and kilograms to feet and pounds. One must be careful of converse arguments here. 2. They arc both positively correlated with time. the variables are independent.2. The multipliers cancel out in the correlation coefficient.3).28)(2. the covariance was multiplied by (3.16. 4.Y)2 and r has the same sign as the covariance.0 s r S +1.y)f --~-- . and if X is much cheaper to obtain than Y. if X and Yare two chemical measurements that are highly correlated. represents the random noise in the system. . = a + {3x. There are numerous examples of these nonsense.4. perhaps. (14. or vice versa. the observed values of Y will vary. the values of x. correlations.). Our model implies that the expectations E(YIX) lie on a straight line. The weights of students who arc 70 inches tall vary about a value E( YIX = 70). is a random variable with zero expectation. University enrollment in the United States has increased steadily since the end of World War II." The name regressiun has stuck to the technique and is generally used by statisticians to describe fitting lines and planes to data.220 BIVARIATE DATA: FlITING STRAIGHT LINES another.4. 14. = a + (3x. + e.2) and then Yi =: p-. but we could use the enrollment figures for men to estimate the female enrollment. or spurious. FITTING A STRAIGHT LINE Suppose that we have a set of n bivariate data points. . This line is called the regression line of Yon X.4. in particular of the "law of universal regression. + e. that e. we may. The numbers of both men and women have increased. (14. but we cannot rationally argue that the one caused the other.3) We now make our first assumption about the behavior of e" namely. feel comfortable about using the cheaper X as a proxy. (14. e.4. . Then p-. The model is y. or surrogate.) . The noted English biometrician Sir Francis Galton fitted a straight line to the plot of the heights of sons versus the heights of their fathers in his studies of inheritance. are taken as given. = E(YIX = x. We propose to fit a straight line to the data by the method of least squares.1) The coefficients a and {3 arc unknown constants to be estimated from the data. On a more practical note. for Y. If we take several observations at the same value of X. with the primary idea of predicting the value of Y given the value of X. Let the ith data point be (x" y. We can write p-. . It would not be appropriate here to speculate on whether the increase of women caused an increase in male enrollment. On the other hand. An edited version of the printout is given in table 14. per 12-ounce can.2. Y.52 3. 5. The derivation of the square of the correlation coefficient and the table of sums of squares that follows it are in section 14. 6. The first eight beers were classified by the brewers as light beers and the last eight as regular. The details of the printout are discussed as this chapter proceeds. The data are the number of calories. To fit the line on Minitab. The printout begins with a statement of the fitted equation. The standard deviation of Y refers to the standard deviation of the error terms. 70 R.36 3. for the regular and light types of beer brewed by eight different producers.14. The t value for the slope of the line. Beer Data y l.89 Y 2. AN EXAMPLE The data of table 14. 4. 6R 5. and we reject the hypothesis that the slope is zero. 13. For the moment.27 3.11. A plot of the data is shown in figure 14.43 2. 10. in C2.H9 3.221 AN EXAMPLE It is an unfortunate accident of history that we usually refer to e j as the error. 14. If our straight-line model is correct. as it would be if we were fitting a straight line to points that lie on or about a parabola. 12.22 3. and the percentage of alcohol. e. 110 70 124 130 lhO X 3.17 3.5. 140 Y X 2.5. Then the coefficients in the equation are repeated and their t values are given.0.1 appeared in the Austin American Statesman newspaper in the fall of 1978. we entered the data in columns 1 and 2 of their work sheet and gave the command REGRESS Cl 1 C2." We are not going to discuss that situation yet. 134 14.90 .13.12. and we do that later. 97 11.61 2. is comfortably larger than 2. 15.5. if our model is wrong. This suggests that it might be appropriate to fit a line through the origin.79 3. X. the t value for the constant term is small. Table 14.5.32 3. will be a combination of both noise and "badness of fit. 7.40 3.5. telling the program to regress the Y variable that was in Cl against one predictor variable.08 3. It is discussed in section 14. we assume that our model is correct. 96 96 134 140 140 129 X 3.76 3. X.Hl 3. 16.1. It is not error in the sense that someone has made a mistake. 9. The calculation of those t values is discussed in section 14.1. e j represents random variation. 27 3.18 '2. PRED. numbers 5 and 6 (number 2 comes very close to earning an X rating).83 2. they stand a long way from the main group. One might argue that the three weak beers should be excluded. -0.8087 ANALYSIS OF VARIANCE DF 1 14 15 DUE TO REGRESSION RESIDUAL TOTAL SS 8138. DEY.66 4. beer 13.2 12453.86 5.81.65 104.17 ST. and the t value for the slope falls to a nonsignificant 1.32 2. One argumcnt would be that we wish to confine our investigation to beers of a certain minimum strength. with some details.5 308.8 MS = SS/DF 8138.04 9.D.RES.00 70. the correlation coefficient between calories and alcohol content is r = 0. say. R-SQUARED = 0. Printout of Minitab Regression for the Beer nata (Edited) THE REGRESSION EQUATION IS Y=-23.654 R = 0.f. has a had fit to the data. They are very weak beers. DEV.3+41.5.2.384 ST.12 The printout ends with a listing.25 8.4 X COLUMN X C2 COEFFICIENT -23. number 13 had 140 calories.15).5 4315. For the whole group of 16 beers. The point with the R rating. 3%. the correlation coefficient falls to 0.00 140. those three beers are below specification.47.14 and 14. Y RESIDUAL ST. There are two of them. 27.222 BIVARIATE DATA: FITTING STRAIGHT LINES Table 14.054 T-RATIO = COEF/S.30 41. The fitted line predicts only 104 calories for beer with 3.30 --0. of three "special" points (see sections 14.08% alcohol.00 PRED.2 (BACK SOLUTION FOR SPECIAL POINTS) ROW 5X fiX 13R X C2 2. 14 d. OF COEF'. Y VALUE 72. When the three lightest beers are excluded.08 Y Cl 70. It seems to have too many calories for that amount of alcohol. . 9.41. they pull the line down.71 70.14 THE ST. OF Y ABOUT REGRESSION LINE IS S = 17. The X ratings denote beers that have especially large influence on the fit.86 35.56.65 -0. DEY.71 -0. 5 3. The method of least squares is to take for Ii and /3 the values that minimize the sum of the squares of the residuals: ( 14.' I I I I I • 1 I I I I I .2) This difference is called the ith residual. we have an estimate of P-" y.223 THE METHOD OF LEAST SQUARES • I I 148 ______ I I I 1__________ I ~ ~ I I I I I I I I I 1 I I I I I I 128 ~ _ I I I . ]4. . I I I I I -------~---------+---------+---------~------I I I I I I I I I I I I I I I I 1 1 1 I • I I I I I I I I • • -------+---------~---------~I I 88 ~ 1 I I I I I I 108 _________ I _________ I _________ L I ______ _ I I I I I I I I I I I I • • I -------ot----I I 2. THE METHOD OF LEAST SQUARES Suppose that we estimate a and f3 hy Ii and ~.2).. (14.3) .4 3.8 3..1 2.6. = Ii + ~x. Regression of calories on alcohol.6.y.6.4. Substituting those estimates in equation (14. (14.7 4 Alcohol Figure 14.1) The difference between this estimated value and the observed value is the deviation e. = y.1.6.5. 12) .. or the residual sum of squares. . we have 2: x.) == 0 2 L x.a .6.6) and dividing equation (14. := (14.5) and (14.) ./3x .6. (14.a . the fitted line passes through the centroid of the data.6. 2 which simplify to 2: Y.5) by n.11) The prediction line can now be written as (14.6./3x.9) so that (14.( Y. i.6.6).(y.6.6. Equations (14.6./3xJ = 0. If we multiply equation (14.x L x. (14. = L xja + /3 2: xi .10) We then substitute this value in equation (14. we have (14.6.6.6.224 BIVARIATE DATA: FITl'ING STRAIGHT LINES The quantity Se is called the sum of squares for error. gives two equations a and to /3. .6.6. and equating the L (Yi . .6.2).6.6.7) by E Xi and subtract it from equation (14.6) are called the normal equations. -. ( 14.e.8) We rewrite this equation as (14. 2: xiY. Substituting the estimates a and /3 in equation (14.y) /3 2: (X. y= a + /3X.5) (14.7) and obtain a = y .4) Differentiating Se' in turn. = na + /3 2: x. with respect to derivatives to zero. y) lies on the fitted line.7) This says that (x. 0 -1. .niy + nx.6 +9.0.2 27.1 19.0 3.9 25.y.0x . in the algebraic identity. = -23. Sxy = 196. + niy = 2: x.297. - i)y.8.niy . .i 2: y. The prediction line is y = 10.0 -3.1.8.y 2: x.658.3R9.65RI4.9) is SXY = 2: (x.ji).0 6. . 1.0 0.v = 2: x.5 x-x y-y -3. = 2: x.7. = L (y.2 +3.0 .0 7.3 16.0..7 31. and so the term on the left of equation (14.752 = 41. so that 22.0 +3.1 +5. SXY = 84.nxy . (14.y.1 x 1. .0 +1.1-12. = y. i=3.0 +2.0 4.0 21.0. y= 114.7.225 A SIMPLE NUMERICAL EXAMPLE For th~ beer data. and Sx = 28. = We can show in the same way that L (x.752.2) 14.y) 2: x.0.8 --6.0 5. If we set x. and fl = .0= to. /3 = 3. . and Sx =4.1 + 3. Example 14. AN ALGEBRAIC IDENTITY 2: (x.4 i = 4.niy .875.0 2.y.7.i)(y.1 -0.v 13.0 -8.y.0 -2. .3844. Y = 22. - y)x. Then f3 = 196. we also see that the multiplier on the right of the equation is (14. a 14. - i)(y.6. .1) . A SIMPLE NUMERICAL EXAMPLE This example has been constructed to illustrate the ideas of the previous section. I 31.2 27.1 2.1 25.1) Also.9 25. The observations arc uncorrelated: for h =.'" L.0 21.f i .1 16. and the sum of squares of the residuals has n .1 22.1 -0. One can as well say that 2 d. have been lost because two parameters.e. = 2: (y.2 +0. That follows from equation (14. (14.0 7.6. (14. '\.0 3. (i) This assumption is enough to make {3 an unbiased estimate of the slope of the line. {3 2: (x. The residual sum of squares is Sr = 0.2 degrees of freedom. A := 0.1 x y . from equation (14..0 4.1 -0.9.3 16. y) .8.f.1 The predicted values and the residuals are shown in table 14. Three more assumptions arc usually added.1 19. Note that the sum of the residuals is zero.2 0.0 5.5 13.6.0 6.1 19.226 BIVARIATE DATA: FJ1TING STRAIGHT LINES Table 14.8.Y 1. THE ERROR TERMS So far we have only made one assumption ahout the nature of the error terms: E(c. - x) = 0 . (ii) .1. 14.J x.0 e +0.12) since and so 2: e..0 -0.4 2R.42.2) So there arc two restrictions on the residuals.8.)=O.0 13.4 +0. a and {3. have heen estimated from 0 the data.7 31.8).8. The variance of ~ is computed with assumptions (i). that and y.N(/J-" (/). it is a linear combination of the observations. y...X w'=-S.E(y.1) This is intuitively reasonable.10.y" S is an where X. we have only used the first assumption about the error terms.. (ii). E(~) = 1:'(2: w. Even if the observations .x.ex + f3 L w.) 2: w. THE ESTIMATE 0:1" THE SLOPE In this section.) = 2: w. ~. and (iii). For convenience. The last assumption is that the errors are normally distributed. x unbiased estimate of the slope. Under the normality assumption.10. is a linear estimate of the slope. (iv) and that the observations arc independent. ~ We now show that = 2: w. This implies.e. Note that in showing that ~ is an unbiased estimate of the slope. 14. (iii) Assumption (iii) is called the assumption of homoscedasticity. The estimate. {J is a linear combination of normal random variables and is itself normally distributed.. of the slope of the line. .). the more precise is the estimate of the slope. The wider the spread in the values of x.227 THE ESTIMATE OF THE SLOPE The observations have the same variance: for all i . we develop the properties of the estimate. in conjunction with the other three assumptions. i. we write S. = = f3 . (14.. 6. the quotient Se/(T2 has a chi-square distribution with 11 . It can be shown. /3.2 d. and the locus of E( YI X) is a straight line. In the printout for the beer data. the central limit theorem justifies us in acting as if /3 were approximately normally distributed.. after some algebra. 17.1) has a t distribution with n . f.8..2)0'2 • so that S .12. The reader should note that this is the first time that we have made use of the normality assumption. We have already seen that the estimate. 2 Sr =-It - 2 is an unbiased estimator of (T".2) that there are two linear constraints on the residuals.2 d. the sum of squares of the residuals appears on the second line of the section headed by "Analysis of Variance. which leads to the estimate 4315114 = 308 for the variance.12.8.f.12.2) ." It is 4315 with 14 d. 14. The square root of that quantity. The statistic /3 t=--- VS2/Sx (14. that £(S. 14. With homoscedasticity (homogeneous variance). appears earlier in the table as the estimate of the standard deviation.11. I·TESTS If the errors are normally distributed. 2 (14.1) and (14. of the slope is a normal random variable. We have seen in equations (14. It leads to the tests that are derived in the next few sections. ESTIMATING THE VARIANCE If our original model is correct. the residuals provide an estimate of the error variance. The residuals reflect only the noise in the system. It follows that /3-{3 t=--- Vs /S.228 BIVARIATE DATA: FI1TING STRAIGHT LINES are not normal. £( Y) = E( Y) at each point.t. This t statistic can be used to compute confidence intervals for the "true" slope.) "" (n . {3. therefore. X.752 .2 d. or sum of squares of residuals.f. The reduction is obtained at the cost of 1 d. (14. or. can be measured by comparing SR to the estimate of the variance.13. about the horizontal line is Sv == ~ (y. . for Se' If the "true" line is indeed horizontal.. We recall from equation (14.f. Smaller values of t would lead to accepting the hypothesis that the true coefficient is zero. to the error variance. The scatter. a is 14. which confirms that the line of best fit is not horizontal. It is called the (-value for ~. This is the F-test. The effectiveness of the reduction.f. corresponding to the slope..13.3) We now look at the sum of squares for regression more closely. Fitting the predictor.12) that . to arguing that the evidence is not strong enough to reject that null hypothesis.0 in absolute value as significant. at least. i. Sy. the model becomes E( Y) = constant. Similar calculations lead to the (-value for the other coefficient. The I-value for only -0.6/V 4. which means that X is of no use as a predictor of Y.f. the I-value for the slope is t == 41.2) The total sum of squares.1) and the constant is estimated by y. In the beer data. a. has n .e. The analysis of variance describes this division of Sv into the two components: (14.13. We argue that if (3 = 0. the apparent reduction should just be equivalent to 1 d.y)2. {3. worth of noise.86. SUMS OF SQUARES Another approach to testing the significance of the slope of an estimated line is through the analysis of variance.1 d. A common rule of thumb is to regard values of t that exceed 2.3844 = 5 14 17.229 SUMS OF SQUARES can be used to test the hypothesis that the "true" slope is zero. That leaves n . .6. which suggests refitting a line through the origin. reduces the scatter to Se' The amount of the reduction is called the sum of squares for regression and is denoted by (14.13. 230 BIVARIATE DATA: FlrnNG STRAIGHT LINES so that and Se = 5 y . f=--=t 2 S2 (14.2. Sn s ( 14.13.5.6) ' and so the F-test is equivalent to the t-test that was described earlier.8138.5) F=2 .t. described earlier.5 = 4315.. However.55/12.654. Under the null hypothesis that f3 d. which is explained by adding X as a predictor.658) = ~138.2) . the easiest way to compute S.453.2 and S2 = 4315. .453.13. But we can also write = 0.8 .2114 = 308. the fractional reduction is scale-free: (14. = 12.8 = 0. {3s.).7) This gives a physical interpretation of the correlation coefficient in terms of the fit of the linear model.4) Computationally.{3s-"y . The F statistic.13. This quantity depends on the scale of Y. The analysis of variance also leads to the correlation coefficient. = 8138. is to compute S.13. F has the F distribution with (1. whence ( 14. Sn whence r2 Sc = (41. For the beer data. n . is ~bmct. and SR and .3844)(196. We have derived the sum of squares for regression as the reduction in scatter.2~S"y + ~2Sx = Sy . The last column contains the standardized residuals. it is not always easy to spot the influential points with only a visual examination of the data. In the particular example of the beer data. INFLUENTIAL POINTS In the heer data.1) with s substituted for the unknown (T.xi)~ J.1) that Y(jl. = . Such action can sometimes be justified. A similar rating can be used in multiple regression.). STANDARDIZED RESIDUALS It follows from equation (14. each residual is divided by the estimate of its standard deviation.i)zY(f3) (T2[~ + ~. a case could he made for setting those points aside if. The quotient is essentially a {-value for that residual. The actual criterion of how far is a long way is a matter of choice. They earned these ratings because they were a long way from the centroid of the data hase. they were considered to be outside the limits within which the scientists wished to work in future. the authors have changed the cutoff value and those two beers no longer earn that rating! One should be careful not to fall into the trap of automatically dropping from the data any points that have X ratings.14. obtained from equation (14. in retrospect. when Y is predicted by several X variables. Its choice of cutoff point gave X ratings to beers 5 and 6.231 INFLUENTIAL POINTS 14. Beer 2 missed an X rating hy a whisker. (14. This kind of action has to be taken on a case-t~-case basis. Then they might be justified in using a prediction line based only on the stronger beers.15. the two weakest beers received X ratings to indicate that they had a large influence on the slope of the prediction line.14.1) A similar formula can be derived for y(e. In the current version of the Minitab package. and it is flagged by the letter R. . In the beer data. This computer run was made with an earlier version of Minitab. . In that situation.10.0 for the standardized residual indicates a point whose fit is questionable. and only with due consideration. point 13 carns an R rating.14. A value greater than 2. on the argument that they must surely be extreme. In the last part of the printout-the back solution for the "special" points-there is a column for the standard deviation of y. 14. In this column.) = Y(y) + (x. 3) This can be appreciably larger than V( Yo)' and the confidence interval is correspondingly wider. It takes its minimum value when Xu = x.2) where t* is the tabulated value of t with n . The sum of squares for Y is now the raw sum of squares. ILf). (14. it has not been adjusted by taking deviations about the mean value.7.f. The variance of the V(Yo). If we now take an observation. so that the estimate of the variance has actually fallen from 30R to 302. CONFIDENCE INTERVALS FOR A PREDICTED VALUE Suppose that we predict the value of Y when X predicted value is = Xo. TIlis is done in Minitab by two commands. X. Yo' at X = xu' we may wish to compare the observed value with our estimate. A confidence interval for the "true" value. of the data base.16.16.16. first REGRESS c1 1 c2. and so we fit a new line. DEY. constrained to pass through the origin.17.1). followed by NOCONSTANT. Notice the change in the analysis of variance table. It is important to remember that there will be noise in our observation too. Y" in the printout.16.1.X)2 ] co n • (14. and S* is the standard deviation of Yo' obtained from equation (14.86.232 BIVARIATE DATA: FITTING STRAIGHT LINES 14. V(Yo) increases as Xu moves away from the center. LINES THROUGH THE ORIGIN In the printout for the beer data. The variance of the difference is (14. The printout appears in table 14.2 to 4540. The sum of squares of residuals has increased from 4315.17.1) ""x- This is the formula that is used to calculate the entries in the column headed "ST. Clearly.16. from 14 to 15. 14. PRED. but there has been an increase in the number of d..:::: V(y) + ~(xo . when X = xo' can be obtained in the usual way by Yo - Y t*s*:5 ILo:5 u + /*s* .. the I-value for the constant term is -0.6. E l. No reference is made to the correlation .x) i)2V(~) = V(y) + (xo =0' 2[ -+ 1 (xo . We cannot reject the hypothesis that the constant term is zero. y.2 dJ. 0 26.) leads to (14. THE REGRESSION EQUATION IS Y = +34.6 Xl COLUMN NOCONSTANT Xl C2 ST. DEV. and that . It is the loss of that constraint that adds one more degree of freedom to the error term.1) It is easily shown that j3 is an unbiased estimate of the slope. V(f3) fT =L 2 X2I . We are no longer asking whether including X as a predictor is an improvement over the line y :.2) The residuals are no longer constrained to add up to zero. - '2 f3x. COEFFICIENT 1. Now we ask if it is an improvement over the line y = O. Minimizing the residual sum of squares "\. and so we .1.90 MS = SS/DF 219053.17. (14. That is the only constraint. The least squares estimate of the slope is obtained from the single normal equation. RESIDUAL PLOTS We have already mentioned that the residuals (when there is a constant term) are constrained to sum to zero.J (y.7 coefficient.17. T-RATlO = OF COEF. NOCONST ANT.588 ANALYSIS OF VARIANCE DUE TO DF SS REGRESSION 1 219053.286 34.~ L. 14.18.233 RESIDUAL PLOTS Table 14. COEF'/S.D.4 302. Beer Data (Line through the Origin) REGRESS cl 1 c2. because there is no normal equation that corresponds to a constant term.:: y.17.4 RESIDUAL 15 4540.6 TOTAL 16 223594. .1. perhaps. Suppose that the data are entered in ascending order of X.5 3. A glance at the list of residuals should show a random ordering of plus and minus signs. Roughly half the residuals should be positive and half negative. or close to. : I : t I I : I 1 I I I I _____ l _______ • _______ + _______ + _______ L ___ _ I I I . I I I I 33 _. all the points but one.:. we were wrong to think that the variance was constant'! Maybe the variance is a function of Y.1 Alcohol I'igure 14. : -----1-------~-------~-------~-------t--~• I I I I I .2 2. Some like to make a normal plot of the residuals and see whether they lie. Before plotting on the computer was as convenient as it is now. If you have one large positive residual and the others are small and negative.7 4 . Nowadays. we may have fitted a straight line to a parabola or a hyperbola (see the example in the next section).: : : I : I :. :' I -7 I _____ J _______ ~-------~-------~-------~----I I I I I : 13 I I I I : : • I. these precautions of checking the randomness of the plus and minus signs were standard practice. In that case. the ohservation with the smallest value of X being number I and so on.234 BIVARIATE DATA: FlTIlNG STRAIGHT LINES should expect them to be more or less randomly and normally distributed. If the residuals are alternately plus and minus. it is easy to plot the residuals against X. I I I I I • I I I I I I I I -27~~~~~~~~~~~~~~~~~~~~~ 2. . on a straight line. (A plot of the residuals for the beer data is shown as figure 14. and perhaps something else. ReSIduals plot for beer data. we may have fitted a straight line to a sine curve! If there is a string of positive residuals followed by a string of negatives and again a string of positives. Do the residuals seem to get bigger as X increases? Do they get bigger as Y increases? Does this mean that. more or less.18. that one point is an outlier somewhere above the line.) One can notice the phenomena that have just been mentioned. 3.:. I I I I I I I I I I I I I : : : I I • : I I I I I I 1 :. perhaps there really is a straight line passing through.18.1. shows that the end points lie above the fitted line and the points in the middle lie below the line-a distinct pattern. J. In the 1970s. Week DeL 50 5 41 30 14 l) 6 4 <) 21 13 17 15 6 10 14 33 18 11 6 3 IS 22 26 to 21 25 8 4 3 . we fit a straight line where a hyperbola is more appropriate.1 Week DeL 3 7 65 11 15 16 10 12 Il) 7 23 5 20 24 27 Week 4 X 16 Def. With so large a value of r as 0.875.19. OTHER MODELS So far.1. one might be tempted to think that the straight line is the "correct" model.19. 14.2 shows a summary version of the printout when a straightline model is titted to the data. Week Def. In the foIlowing example. we have assumed that our model is correct.19. such as Draper and Smith (1981). The consideration of that complication is beyond the scope of this book. The points appear to lie on a fairly smooth curve that looks like a rectangular hyperbola. Example 14.19. A company sets up a new process for producing parts. some economists plotted Table 14. The data set is given in table 14. The straight-line model gives a reasonahJc approximation to reality as long as one stays within limited ranges of the X variable.2. A plot of the residuals against X. Does this mean that fitting a straight line to this data is useless? Not necessarily. The reader may wish to refer to a book that specializes in regression. figurc 14.19. Thcrc is enormous potential for trouble if one tries to extrapolate-to use the equation to predict values of Y outside the range of the data. figure 14. reveals this to be incorrect.19. they start to take a sample of WOO parts each week and to record the number of defectives.1.235 OTHER MODELS one should be using weighted regression or a transformation. After two weeks.19. A plot of the data. A large value of the correlation coefficient tcnds to give credence to that idea. If one extends the line far enough. it will cross the X axis and will predict a negative number of defectives. Table 14. There is no question that Y decreases as X increases. The management is delighted to see that the number of defectives decreases. which is bcttcr than even the most enthusiastic quality engineer would predict. and that there is some underlying linear connection between the two variables-not necessarily a functional relationship. 993.~ ~ I 8 20 I f ----~-------~-------. Extrapolation.-------+-------~----I I I I I . very dangerous. .236 BIVARIATE DATA: fTITING STRAIGHT LINES Table 14." Regression of defectives on week.3586 -2.948 and r = 0.48 R:::= -0.2. The new "line" is y:::: -3.-------+-------T-------~-------L---I I I I I --~-------~-------t---I I I I I I I I I . some of us are finding that hard to believe now. Predictor Constant Coef 46. I I I I I I I I I I I I I I I I I I I : I 40 I I I I I I I I I --. is very easy-you just keep drawing the line.00174 x Stdev 3. etc. in particular. linear extrapolation.81 (~) with s = 1.875 the price per barrel of crude oil against time.2 The regression equation is Y = 46.19.792 0. I I I I I I I I I I I I I : : : : 1 I I I I I I I I I I I I I I I .94 + 216. reciprocals. and fitting a straight line with 1/ X as the predictor instead of X. I I I 15 Week Figure 14. Engineers are used to converting curves to straight lines by using special paper in which X and/or Yare replaced by their logarithms. and.22 -8.004 t-ratio 12. square roots. I • 60 .766 s:::= 8.19. We can fit a hyperbola to the data by introducing a new variable.00 X.236 R2:::= 0. It is also very.36 . That straight line extrapolation forecasted that oil would cost $80 a barrel by 1990. 1/ X. I . this model is not a linear model.20. a and {3. by whieh we mean that the right-hand side is not a linear function of the two unknown parameters. Residuals plot for defectives data.2. We take natural logarithms of both sides of the equation and get Z=c+dX (14. make it linear by a transformation. This procedure does some violence to our assumptions about the error term.: ..2) where Z = In(Y). however. c = In(a). • I I ~ • • I • I I I 25 30 Week Figure 14.81 .94x = 216. This is obviously a good fit to the data. 21 11 I _____ I _______ +-_____ I _______ I I I I I I I I I I I f I I I I I I I I I I I I I I I I I I I I I . but it does enable us to fit the model.237 TRANSFORMAnONS . None of the usual assumptions has been violated by using 11 X instead of X as the predictor of Y.20. and d = {3. JL -------------T-------T-------T ·---. I ~ I • . We can. TRANSFORM ATIONS Another model that might be appropriate for a data set that looks like this is the exponential model: Y= ae llX • (14. . Multiplying both sides of the equation by x gives the rectangular hyperbola: xy + 3. We fit the straight line and then .19.-------L-___ _ I I I ~ ----1-------t-------t-------r-------~---: I I ' I ' 1 : I I I I or I : • I I .1) As it stands.20. I I . o 14. y) = 87.28 - 0. to the following data: X= Y= 13 2 3 4 17 19 12 5 6 7 8 14 29 34 27 9 LO 36 33 Plot Y against X and plot the residuals against the predicted values. 1: y = 810. The problem of finding the best fit in these nonlinear problems without using linearizing transformations is full of mathematical complexities. b. 124. we have E x = 200. Fit a straight line. Test the hypothesis {3 = 2 against the alternative f3 ~ 2.0. c = In(a). and z = In(X). The new regression equation is In(Y) = 4. y)2 = 104.20. Y" y. y = a + f3x. • The fit that we have made is equivalent to plotting the data on semilog paper and drawing the best line through the points. For a set of 20 observations. Note that the correlation between the predicted values and the residuals is zero.238 BIVARIATE DATA: FllllNG STRAIGHT LINES take exponents to return to equation (14. . E (y -.v" Compute the correlations bctwecn . Fit the line y = (r + f3x to the data and test the hypothesis f3 == 1. a = exp(c)..1).1. EXERCISES 14.£)( y .124x .O.2. 14.. and the reader is again referred to Draper and Smith (1981). that minimizes the residual sum of squares in the log-log world is the same as the pair that minimizes the sum in the untransformed world. which corresponds to y = 72. The results that we get with these transformations are only approximations to the best fit.0 against the alternative f3 /.24e. 1: (x £)2 = 92. It does not follow that the pair of coefficients. and ~ (x . We can also get the equivalent of log-log paper and fit the model y= ax h by taking logarithms of both sides and fitting the line w = c + bz where IV = In( V).1. and e. as reported in the Austin American Statesman. the line is not an implausible model at first glance. city 53 49 46 46 46 43 40 38 38 37 m. You will see that for x> 4. m.g. The residual plot shows a clear parabolic pattern. 14. Test that hypothesis. highway 58 52 50 50 50 49 44 42 39 43 Obtain a least squares prediction line to predict the highway mileage from the city mileage: highway = f30 + f3 1 (city) It could be argued that the line should pass through the origin. This can be used to check your model. Note that R2 ~ 0. .. X= 1 2 3 4 5 6 = 3 3 5 9 15 23 Y 7 8 33 45 9 10 11 12 59 75 93 113 14. This illustrates how extrapolation from a model that seems to be plausible can be a trap.4. The table shows the miles per gallon for the ten most economical cars in the 1990 EPA fuel economy survey.3. Test that hypothesis.239 EXERCISES 14.5.p.g. Make plots of Y against X and of the residuals against the predicted values. Fit a straight line to the following data.3x + 5. and 15.4 has repeated observations at city = 46 and 38. Verify that for each point y = x 2 . and compare them with the actual values from the quadratic. which is very impressive.p. Since the data in exercise 14. 14. Readers who are using Minitab should usc the commands REGRESS Cl I C2. there is an estimate of pure error between those duplicates. You have fitted a straight line to a parabola! Estimate from your fitted line the values of Y when X = 13.e. PURE. to test whether the straight-line model is adequate. i.915. 50 19. X (77K) Y (300K) 3270 3960 3920 3860 3330 2330 3760 10460 9970 7.70 16 15.8.20 12 10.10 14 13.50 26 25.40 5. This a reason for some statisticians to prcfer to plot against y. i. which is not trivial.50 5. Fit a straight line to predict the mobility at 300K from the mobility at 77K. 4.10 7 3. rather than )'" e 14. Fit a line Y = ll' + {3x to the data and comment about your results.10 23.70 9 3.50 28 27.20 3. therefore. 3.50 11.78 5 2.00 16.531..50 22.240 BIVARIATE DATA: FIniNG STRAIGHT LINES 14. reasonably drop it. Fit that model and compare its fit to that of the first model.40 23. you showed that corr(e. 9.20 2 1.90 2.40 17 15.60 6 2.10 16.00 20 19. X (77K) Y (300K) 3990 3940 3980 3930 3980 3770 9020 8620 9670 9830 11200 7880 l.80 22.80 23. A plot of the data will explain why the twelfth sample has large influence and also a large residual.20 4 l. 12.50 23.50 19.) = v'l=? In exercise 14. Do that and fit a new line.tO 24 22. 2.35 21 21.10 22 22.60 18 19 9.80 15.50 13 13.9.2.80 27 25. It can also be shown that corr(e. X Y = = 1 0.1.40 5.40 17. 10.45 3. 6.00 4.00 22. Suppose that you felt comfortable in arguing that for the data in exercise 14. 5. )'.718 and so corr(l\ )'.60 IX.80 3 1.50 23 22. Prove that this is always true when there is a constant term. 11.XO 3.35 10 7. when the line is not forced to pass through the origin.40 25 23. y. 8110 4890 2200 14.6. In a study of the growth of InP.) == 0.70 15.7.20 2. in the model.) = O.. engineers recorded the van der Pauw measurements of mobility at both 77K and 300K.10 9. In exercise 14.40 17.6. the twelfth sample was far below the range of interest and that you might. The measurements in meters for 32 trees follow.35 8 3.30 15. Interpret this. ll'.2 =0. Is the fit markedly better? 14. Two scientists wish to develop a formula for predicting the hcight of an aspen tree (Y) standing in a forest from its diameter at breast height. 8.20 22.30 9.90 15 15.50 Il . It is suggested that they should use Vx rather than x as the predictor.e. 6 15 84.8 16 83.10 23. Express the regression line in terms of the new units and prove that the correlation between price and length is unchanged.52 + O.9 97.5 25 91.9 27 93.1 105.12.7 95.80 31 32. i 14.10. 1 2 X= 105. Suppose that for a set of 12 observations.9 19 92.40 22.62) per meter.5 106. Another estimate of the slope of a line through the origin is ffi = i/y.4 Y= 106.l) 98.7 98.2 93. when i ¥ O.2 93.1 29 85.8 20 85.2 13 86.5 101.7 5 88.1 89.4 31 85. The data follow with Fl as Y and F2 as X.9 95.8 30 87.3 12 83.10.1 7 98.3 91. Fit the model y = {3x.20 23.1 99. E xy = 200. 14.7 32 87. In an octane blending study that will be mentioned again in the next chapter.6 83. Caleulate its variance and show that it is greater than the variance of the least squares estimate.0 99. 14. Fit a regression line to enable an engineer to predict the F 1 number from the F2 number. Snee (1981) determined the octane ratings of 32 blends of gasoline by two standard methods-motor (FI) and research (F2).0 90.0 against the alternative {3 ¥ 2.6 8 92.0 94.8 89.83x . Suppose that the priee had been computed in Swiss francs (1 S.7 6 91.4 97.3 9 94. and x is the length in feet.5 90. E x 2 = 91.8 96.50 14.3 97.0 21 22 23 84.7 93.13. E y == 60.4 99.4 99.7 103. R.S. and ~ == 801.2 JOl.0 81.5 11 86.2 17 86.7 88.8 85.0. = $0.4 94.7 28 90.3 24 87.0 92. Do the calculations for the data of exercise 14. E x = 21.2 96. Fr.5 . Show that {3 is an unbiased estimate of the slope.3 14 87.3 100.7 97.4 26 90.40 23. dollars.11. A purchasing agent compares the prices of a certain kind of wire and fits a regression line y = 0. D.4 10 18 3 91.241 EXERCISES 29 30.4 4 84.2 99.50 32 35. Test the hypothesis {3 = 2.50 30 32. where y is the cost in U. 22 different asphalts were compared. X is the percentage of resin. Is this plausible? Is there a marked change in tit when number 11. X is the softening point in degrees F. too. Drop them from the data set and rerun the regression. Y is the percentage of asphaltene in each asphalt. Asphalt 4 is seen to be very influential. the old asphalt. number 5. It is suggested that number 22 is somewhat different from the others. Fit a straight line to predict Y from X. U) = . of those asphalts in practice? Y= X= 77 158 149 106 119 150 142 14.242 BIVARIATE DATA: FI1TJNG STRAIGHT LINES Blends 1 and 10 stand out from the others. Fit a straight line to predict the percent asphaltene from the percent resin. Y is the penetration at U5°F.16. They have by far the highest octanes because of a high aromatic content. These are measurements on II samples of asphalt. docs that improve the fit? What justification would you have for dropping one. Do you get a better fit to the data? 14. Prove that if W= a corr(X.15. and it is reasonable to set them aside. stands out as influential. Suppose that you drop it. V). then corr(W. Is this suggestion plausible? Is there is a marked change in the prediction line if number 22 is deleted? When that is done. or both. 136 III 153 49 194 93 156 153 112 100 145 147 145 + bX and U ~ 57 172 76 154 c + dY. Docs the fit improve when point 4 is dropped? When you fit the data without number 4. too. In a survey of roofing asphalts. number 11 has a large residual. Make a plot of the data and explain this. is deleted? X = 22 Y = 35 2 21 35 12 26 33 13 23 31 3 25 4 23 29 32 14 24 15 22 31 36 5 29 27 16 27 26 6 7 26 25 8 27 9 25 29 28 31 17 29 32 18 24 31 10 11 30 21 36 24 39 19 20 21 22 24 29 27 27 24 34 27 23 14.14. We use two methods. we express the equation of the last line in terms of the original variables. we make a two-stage fit using straight lines. We fit the model to the set of seven points shown in table 15.Statistical Methods in Engineering and Quality Assurance Peter W. John Copyright © 1990 by John Wiley & Sons. Later. we progress from the straight line. M.3)/2. when the equation was transformed to x = (y . This goes back to the terminology of elementary algehra. There is no requirement that the X variables be independent of each other in any broad 243 . x was called independent and y dependent. the sheer weight of the calculations involved prevented this technique from being used very much. Thirty years ago. finally. INTRODUCTION In this chapter. one has to invert a (p + 1) x (p + 1) matrix. and then fit a line to the adjusted values. In that procedure. At first. As in chapter fourteen. In order to handle a problem with p predictor variables. but nowadays. or fitting Y to a single predictor variable. we begin by adjusting XI for Xl and Y for X 2 . Inc CHAPTER FIFTEEN Multiple Regression 15. the term e denotes the random error. mUltiple regression is one of the most commonly used statistical tools. the labels changed. We start with a manufactured example to illustrate the underlying theory. Sometimes Y is called the dependent variable and the X variables are called the independent variables. it is subject to the same conditions. When one took the equation y = 2x + 3 and substituted a value for x to obtain y.2. Before the advent of electronic computers. The second approach is the method of least squares. we use matrices to develop the least squares method for the general case with more than two predictor variables. even if no errors crept into the calculation.1. inverting a 5 x 5 matrix on a desk calculator took several hours. x became the dependent variable and y the independent variable.1. to the more complicated problem of handling several predictors-multiple regression. y for u and X [ .Xl. Our procedure accommodates the polynomial equation by setting X =::: XI' and X2 = X 2 • It also handles the trigonometric model Y = {3Q + (3l sin(wt) + (32 cos(wt) by setting sin(wt) == XI and cos(wt) == X 2 • We do require that XI and X 2 be linearly independent. More is said about this later. we chose to "adjust" Y and Xl for X 2 .104x 2 4.244 MULTIPLE REGRESSION sense. The three lines. which means that we cannot have XI being a linear function of x 2 .2. involves fitting three lines: 1. u == (3v. It is left to the reader to show that we would have arrived at the same equation if we had started by "adjusting" Y and X z for XI instead. If there were three predictor variables.XI y + O.2. calculated in this example.724 + O. Xl = 2X2 + 6.30x 2 ) • In this procedure.35x 2 . we could start hy adjusting for X 3 • and then adjust for Xz' This is a process that soon becomes tedious.819v. we cannot have.0 + O. The data for an example are shown in table 15.0 .819(x l l - == 1.30x z .0 + O.819x[ + O.819(x u == O. 3. and then to fit a line to the adjusted values. for v in the last line gives the final fitted . . arc y == 5.0 + 0. Xl in terms of Xl with residual v = Xl .3Sx 2 ) + O. 2. Substituting y model: y= = X[ == 4. The procedure for fitting Y to two predictor variables. Xl and X 2 .>" where>.1. for example.0. THE METHOD OF ADJUSTING VARIABLES This is an old-fashioned method that has only pedagogical value in this age of computers. is the fitted value. 15. Y in terms of X 2 with residual u = y .i\) (5. 30 12.I + ffiz L X.4 10. The summation is made over all n observations.70 -3. We differentiate Se with respect to each ffi. Adjusting Variables y Xl X2 Y U Xl U 15 12 14 13 13 9 8 14 12 11 23 IS 19 23 18 22 17 13.!31' and !32' respectively. For the data in our example.3 -0.9 9.95 0.2 X tl X.70 -2. This gives us a set of three equations that are called the normal equations: L L X.3.2.2 is e The corresponding residual is = Yi .30 11.y.05 11. and set the derivatives equal to zero.ffilx tl - ffi1x'l)2 .65 13.95 1. in turn. and obtain (4) 34 = 40ffiJ + 12/32 (5) 14 = 12/31 . == L (y. We subtract 10 X(l) from (2) and 20 x(l) from (3) to eliminate ffio.2 ' X.4 9.35 --0.6 J.6 9. The estimated value of Y when Xl = x tl and X 2 = X.05 11. THE METHOD 01<' LEAST SQUARES We denote the estimates of PlI' PI' and /3z by !3o.lY L X.245 THE METHOD OF LEAST SQUARES Table 15. = L X. .2 X. + 40/32 • ..ffiu .1 J() 9 7 7 15. and we choose the estimates so as to minimize the sum of squares of the residuals: S.9 -0. the equations are (1) 84= 7ffio+ 7offil+ 140ffi2' (2) 874 = 70ffiu + 740!31 + 1412ffiz ' (3) 1694 = 14offio + 1412ffil + 2840Pz .1 2.70 10.l + ffi2 L = ffiu L + ffil 2: + ffiz L y.05 1.9 9.95 10.70 2. as before.2 • == ffiu + ffi l L X.1 3.4 -3.6 -2.2 .2Y nffio + ffil X tl X.1.7 10. 2715 = 3.59 -1.3. The estimated values.3.64.23 -0. 3. 0. In this case. }: = 10.71 12. Table 15.02. /32 = 0. 0. and for /32' 12 = 0.43 1. The sum of squares for the residuals about this line is 11. We will see later that in this example. changed when x 2 was dropped. /30 = 1. which is estimated by (0.01 (it should be identically zero but for roundoff error).72 .23 I) R . 0.31 LO.03 -0. : :. the new least squares equation is the straight line: y :::.870x t • The coefficients . (T2.97 9. We note that L e.02750"-.29 0.8t) = V(.819. for /3" t. The addition of x 2 to the model with x I alone docs not improve the fit significantly. and to drop x 2 from the prediction equation for y. 2.59 13.40(T2 A V(.50 + 0.82 could be zero. it would be reasonable to conclude that . which is only a 3. :::. is estimated by SO e.1 Y .43 12. Also. i = n L~ 3 = !0~72 :::.10.75 ':1.8. and that the fitted equation is indeed the same as the equation that was obtained earlier by the first method.80 and .75 -1.5% increase over the sum of squares for the model with two predictor variables.819/0.1. :::.2715)2. y" and the residuals are shown in table 15.39.68 with s = 1. When this is done.104.246 MULTIPLE REGRESSION whence /3. The variance. ~ .v d 15 12 14 13 13 15.72.69 2.82) = 1456 =0. Thus. The matrix X is called the design matrix. numbered 0 through p.2Y'Xp + P'X'Xp . In matrix notation. eaeh of whieh contains p + I coordinates (variables). that they are independent normal variables with zero expectations and the same variance.4. corresponding to the coefficients. otherwise. we write this model as: Y = X{J + e. (1'2. XI"~ y. leads to the normal equations X'Y = from which we obtain the estimates (X'X)P . (15. the important point is that the model is a linear expression in the unknown betas. The zero (leftmost) column of X is a column of ones.3) The estimation procedure calls for minimizing the residual sum of squares. = Y'Y . for the ith data point. As we said before. Xu' which is always unity.4.4.•• . It has N rows. MORE THAN TWO PREDICTORS We now turn to the general case. e" namely. sines or cosines. {3" by the be the vector of estimated values of the method of least squares.4.Y= Y - XP .1) with the usual assumptions about the error terms.4) . the element X'I in the ith row and the jth column is the value. xI' x 2 ' . this corresponds to a dummy variable. The vector of residuals is Xp e= Y . We wish to obtain a vector. (15. and p + 1 columns. Let Y= dependent variable. and is associated with the constant term. it matters little to us whether the x variables are squares or cubes. 13 is a vector that consists of the p + 1 coefficients. Suppose that there are N data points. of estimates of the coefficients.4. 13. in which there are p predictors.2) where Y is the N-dimensional vector of the observations. We wish to fit the linear model (J5. and e is the corresponding vector of errors. one for each data point. of X. (15. This model is called a linear model because it is a linear expression in the unknown coefficients.247 MORE THAN TWO PREDICTORS 15. It can be shown that differentiating Se with respect to each of the ~" and equating the derivatives to zero. 1) P. (15..[ 7~ 140 140] 70 740 1412 .1 degrees of freedom. For f3" we have t. It is used to compute I-values for the coefficients. is the jt~ diagonal element of V.. if Vj. /3.2 are as follows: x'x.. The mathematical justification for the use of the method of least squares is found in the Gauss-Markov theorem. N-p-l . 15.5. in order for its inverse to exist. is estimated by the unbiased estimator S 2 = Sr. which means that they must not have a linear relationship among themselves. Ys (15. f3" is a linear combination of the observations.5.3. E( P) = covariance matrix is p.5) The reader should confirm equation (15. the least squares estimates are the (unique) estimates with the smallest variances. This states that of all possible linear estimates of the unknown coefficients. = . (1'2.5.5. An example of this problem is given in section 15. . It is obviously necessary that the matrix X'X should not be singular.P .2.6. MORE ABOUT THE EXAMPLE The X'X matrix and its inverse for the example in section 15. They are sometimes called the BLUE estimates (best linear unbiased estimates). Y 1 • YZ' . f3" that are unbiased. the variance of is V ii u 2• Each of the estimates. This implies that the X variables must be linearly independent.13.4.248 MULTIPLE REGRESSION (15.. 1412 2840 . PROPERTIES OF THE ESTIMATES The least squares estimates of the coefficients are unbiased. YN' Such estimates are called linear estimates.4. Their (15.4) for the example in section 15.2) This estimate has N .3) VII 15. In particular. The error variance. Much of the analysis consists of noting the percentages of various geologic components that are present in the sample.58 X2 . which is in c7. THE PRINTOUT REGRESS c7 5 c2-c6 This command means regress the Y variable.72. 15. We also see from the matrix X'Xthat VII = vn = 280/10. which is (84. The regression equation is Y == 102 -.1.192 = 0.03 X5 Predictor Constant Xl Coef. 1694)" we get the I vector of coefficients (1.7.462 O. as mentioned earlier.8.26 .82. A GEOLOGICAL EXAMPLE A geologist sinks several deep holes into the ground.7. 0.0275. We wish to ohtain a linear prediction equation to predict the amount of spherulites from the other variables. 280 Multiplying X'X hy the vector X'Y.249 THE PRINTOUT and 10192X'X I == 107856 -1120 [ -4760 -1120 280 -84 -4760] -84 . so that. A sample of the composition of the earth is taken at various depths and analyzed.0.012137 St.7.dev.tlO5373 t-ratio 41.1. Tahle 15.31 -2.1.0. 874. 2. 0. as in table 15.0121 Xl . against five predictor variables that are in columns 2 through 6.0.816 X3 .] contains such a set of data. The remaining columns contain the percentages of the components. 101.720 -0. Xl is the depth in meters at which the sample was taken. The data was read into the Minitab worksheet in columns 1 through 7.10).993 X4 . 15. The next section consists of an edited version of the printout. There are 37 samples (rows) and six measurements on each (columns). The first column is the observation number and plays no further part in the discussion. Y = % spherulites.4 1.3 0.5 285.0 6.0 7.4 85.3 0.6 83.3 1.1 0.0 86.3 S3.6 94.5 0.0 2.8 1.9 6.3 1.7 254.1 405.1 1.3 11.3 1.3 350.1 91.6 266.6 3.TJPLE REGRESSION Table 15.7 0.0 0.-~~~~~~~~~-~~-~---- 0.7 86.8 1.1 1.2 86. X5 = <}~) amygdulcs.1 92.6 289.0 310.1 1.7 6.3 1.5 1.9 83.8 90.8 0.9 0.0 0.1 2.3 1.4 0.5 0.0 83.1 S3.0 0.1 3.6 185.0 2.9 1.6 S9.0 0.9 0. X4 = % granophyre.2 1.4 6.0 80.2 9.9 277.0 1.5 0.7 88.7 30S.5 3.0 0.6 91.2 84.8 1.4 1.O 88.4 4.3 9.2 5.0 93.1 0.6 4.2 7.0 90.0 0.0 0.6 4.9 2.6 0.1.0 0.3 153.0 0.3 474.8 1.5 0.S 254.4 323.6 2. X3 = % lithic fragments.2 0.2 2.7.6 1.9 10.5 9.2 3.8 22o.S 3.6 7.9 89.6 6.7 0.9 6.3 0.1 0.4 81.0 0.2 13.1 230.0 2.7 1.3 0.1 193.0 2.2 4.8 13.8 291.6 283.7 81.8 258.9 1.1 212.0 193.6 1.2 0.0 0.3 1.4 0.8 - 1.0 6.7 1.6 94.7 2.5 1.1 lH.7 432.4 0.8 4. .3 1.5 .8 1.7 2.2 7.7 220.0 3.8 445.1 5.4 5.1 376.8 86.7 1.4 1.2 90.0 291.6 93.6 XI = depth in meters.5 0.0 1.9 0.7 359.8 1.0 0.5 272.9 314.250 MUJ.9 3.8 339.4 7.6 4.3 483.0 1.1 89.3 0.S 2.3 77.9 1.1 1.8 4.6 90.2 0.1 0.6 10.8 1.1 1.7 1.5 10.1 1.2 1.4 0. X2 = % phenocrysts.S 0.8 4.5 0.6 326.2 0.4 2.4 319.8 82.7 0.1 0.0 0.0 5.9 84.8 303.6 89.6 188.0 0.2 0.8 0.2 91. Analysis of Geological Samples 1 2 3 4 5 6 7 S 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 2S 29 30 31 32 33 34 35 36 37 Xl X2 X3 X4 X5 Y 155.5 6.2 5.1 84.7 0.1 10.0 0.0 \. 2781 = 0. the value of F is clearly significant. Later. The ratio 472. All except the coefficient of X2 have I-values that exceed 2.498 5. at least for X2 . The sum of squares for regression is obtained by suntraction.86 We note that the coefficients of all the predictors are negative. when the percentages of the other components increase.21 -7.f.y)2. This is compared to the tabulated value of F with P d. dropping X2 from the model.735 = R2.1 == 36 d.735 -1. that tells us nothing new necause we have already established by the I-values that several of the coefficients arc not zero.508 The total sum of squares.L The error sum of squares has 36 .f. The mean square for error is /.0 in absolute value.71 R =0.491/643.240 MS 94. equal to the number of predictors.8161 -0.749 643. The analysis of variance table shows how the total scatter of the observations has neen divided.5 = 31 d. in the denominator.f.X6. The column headed MS contains the mean squares. we make a second run. .347 -1. 1: (Y. ..!Q -3. It has 5 d. They arc the sums of squares divided by their degrees of freedom.f. has 37 . the amount of spherulite decreases.491 170.9935 -1. In this example. R = 0.0314 R2 1. This is reasonable.498 F = 5.86 is called the coefficient of multiple correlation.1270 0.2544 0.15 -3.p -. in the numerator and n . Analysis of Variance SOURCE Regression Error TOTAL DF 5 31 36 SS 472.240 = 0. 94. However.373 0.-508 = 17. The F test for the hypothesis that uses for the test statistic the ratio of the mean square for regression to the mean square for error.16.I d.251 THE PRINTOUT X2 X3 X4 X5 s = 2.578 -0. we should not be surprised if about 5% of the residuals earn R ratings. REG RESS c7 4 c2 c4-c6 The regression equation is y = toO .94R The residual for observation 30 is about 10% of the observed value.120 Residual -4.(H1673 -0.dev.657 1.0117 Xl . The presence of a large residual should prompt an engineer to question an observation and to check it.959 23 327 83.498 30 483 80.40R -2. there may have been an error in the analyis.0. we can conclude that X 2 is not needed as a predictor.30 5.01 X4 .17 -3.29R -3.137 0.198 -8.Fit 1.000 85. .169 St. Three points have large standardized residuals: Unusual Observations Obs.700 88.56 R =0.0.120 -O.dev.0069 -0.041 OJJ05385 0.38 7.0. There is a trivial increase in s and a trivial decrease in R2. -2.869 St.56 All the i-values are now significant.2533 0.8556 -1.359 Coef.02 643. More formally.987 Predictor Constant Xl X3 X4 X5 s = 2. If we regard the standardized residual as a I-value with 2.9.856 X3 . THE SECOND RUN Since the t-value for the coefficient of X 2 is not signHicant. There are small changes in the remaining coefficients (but not all in thc same direction) and in their I-values.0.Resid.92 -3.723 Analysis of Variance SOURCE DF Regression 4 Error 32 TOTAL 36 St.04 -2.9866 R2 =. we test and accept the hypothesis that {32 = O.959 -5. The geologist should check this observation if possible.85 SS 465.24 MS 116.1271 0. One should be wary of discarding an observation just because it has a large residual.2768 X5 t-ratio 49.1.252 MULTfPLE REGRESSION 15. or even to repeat it if possible. XI Y Fit 10 154 81. 100.22 178.300 88. 2.0 corresponding roughly to one chance in 20. 862.1.1O.93 + 0. which means that we have to carry several places of decimals in our estimate of the coefficient f32. and carry out the usual multiple-regression procedure with two predictor variables. hut there arc prohlems. The X variable is the reactor temperature. The first is the size of the numbers involved.dev.99 R = 0. This hecomes even more of Table 15.1 come from a chemical engineering process for polymerization.116 -0. X 2 = X2.995 • t-ratio -6.0.82 -6. Suppose that we wish to predict the yield at a temperature of 255 degrees.10.4 1. at higher temperatures. there is a tendency to overdo the polymerization and to produce pentamer instead of the desired product. Y is the yield of tetramer.5094X with s = 4. A plot of the data is shown in figure 15.3 7. At lower temperatures. increasing temperature increases the yield.10. but that could be a mistake.6 260 86.0 270 85.10. When the model Y = f30 + f3I X is fitted to the data. the regression line is Y = -47.9 230 71. However.l. 127. yield (Y) 220 58.013482X 2 Predictor Constant X X2 Coef. we let XI = X.116X .300 R2 = 0.70 6.3 + 7. What we have just done is theoretically correct.2 .0 250 81.253 FInING POLYNOMIALS 15. It is clear that fitting a straight line explains a large fraction of the scatter in the data and one might be tempted to stop there.270 and R2 = 0.013482 St.025.34 The significant (-value for the quadratic term shows that it was a good idea to fit the quadratic and not to stop with the straight line. X2 = 65. -853. Polymer Yields Temperature (X) Ave.6 240 78.002128 s = 1. The regression equation is Y = --853.043 0. FITTING POLYNOMIALS The data shown in table 15. To fit a quadratic model. We substitute 255 for X in the fitted equation. and we are interested in changes in the yield of a few percent. tetra mer. Then Z\ and will have zero correlation. . . .. -\.e. a problem when the X variable is time and is recorded as X = 1982. X and X2. The problem can be mitigated by changing the coordinates. In this transformation. 245 is the average of the X values.1. -245)1 5 = -5. 1984.254 MULTIPLE REGRESSION ~1I-'111-'111-'1~1-'1~1-'1~1-'1~I-rl~I-rl~I-r~~~~~ 83 I I I I I I : : I I I I I I I I 78 I -------~---------~---------~----I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I -~---------+---------~-----I I I I 68 63 I -~----------~-----I I I I I I I I I I I ------""-----+ I --~-----I I ------i---------1r----I + I I -4----____ I _________ I ~ 1 +---------~------ I I I I I I I I I I I I I I I I I I I I I 1 I I -------+---------~---------~---------~-----I t I I I I I I I I I I I I I I I I I 1 230 240 250 260 270 Temperature FiKUre IS. The more serious problem lies in the fact that when the values of X are positive integers. 1983. The average value of z~ is 35/3. +3. we use Z\ = (temp. of dubious value.. Even though nowadays we can do the inversion on pes. Instead of using as the predictor variable X = temp. they will be orthogonal. -3. and so we elect to work with the more convenient numbers. and +5. are highly correlated. i. Subtracting this from Z~ gives 40/3. Thirty years ago this would have been a problem because of the inability of the small "mainframe" computers to actually invert the matrix and solve the normal equations. -8/3.. Regression of yield on temperature. the estimates are highly correlated and.. the denominator 5 was chosen to make the values of ZI small integers. while "independent" in the sense that neither is a linear function of the other.. therefore. zi .. + \.. It is more convenient computationally to modify Z~.IO. which is merely a matter of convenience. This causes the matrix X'X to be badly conditioned. 255 A WARNING ABOUT EXTRAPOLATION , Z2 = 3zi - 35 =5, --g-- -1, -4, -4, --1,5; 2J and Zz are a pair of first- and second-degree orthogonal polynomials. Using the orthogonal polynomials has two consequences; the calculations are easier, and we can assess the contributions of the linear and quadratic terms independently. The 111 to this quadratic model is Y = 76.8833 + 2.5471Z1 - 0.898822 with s = 1.300 and R2 = 0.99 as before. We can substitute back for X in terms of Z, and Z2 to obtain the carlier equation. However, if we only want to predict the yields for particular temperatures, we can more casily work with Zt and Z2' The predicted yield for 235 degrees is obtained by substituting Zt = -2 and Z2 = 5 to obtain y = 74.4. 15.11. A WARNING ABOUT EXTRAPOLATION Table 15.11.1 shows the predicted values of the yields at three temperatures by the two models. The two estimates differ by 2.6 at 235 degrees, but we note that the observed yield for 230 degrees was 71.6 and for 240 degrees 78.0; both estimates fall between those two observations. A similar remark can be made about the estimates at 245 degrees. However, at 280 degrees, the estimates are far apart; they differ by 12.6. As the temperature increases, the linear model keeps climbing at a steady rate of 0.51 per degree, and will continue to do so. The quadratic model, on the other hand. is beginning to lose steam; the estimates are starting to drop. Here lies the trap of extrapolation. For some sets of data with relatively small variance, the linear and quadratic models give predictions of Y, within the ranges of the independent variables for which data were taken, that are rather close to one another; the engineer might not get into serious difficulty by using either model. However, the engineer who extrapolates is asking for trouble. It is then that the linear and curved models diverge faster and faster. Unfortunately, the simple first-degree model is a very appealing model. Sooner or later, somebody, at a safe distance from the plant, is going Table 15.11.1. Predicted Yields Temperature Linear Quadratic 235 71.8 74.4 245 76.9 R{U~ 280 94.7 82.1 256 MULTIPLE REGRESSION to notice that for the reported data, yield increases at a rate of five points per ten degrees, and will surely draw a straight line and expect you to match that line outside the range! 15.12. TESTING FOR GOODNESS OF FIT In the previous sections, we fitted our polynomials to the means of observations at six temperatures. There were actually four observations at each temperature, and we can use this data to illustrate a method of testing the adequacy of the fitted model. The residual sum of squares is the sum of the squares of the deviations, L (y, - .9,)2. If the model is correct, this deviation is due entirely to noise, both the noise in the observation Yi itself and the combination of noises in the observations that went into computing y,. If the model is incorrect, the deviation also contains a component because E( y,) is not the same as E( y,). As we square and sum the deviations, these discrepancies from the inadequate model accumulate. The residual sum of squares thus consists of two components: noise and a lack of fit. In the usual situation, we cannot separate the two. However, there are two cases in which we can examine the residual sum of squares and obtain a test for "goodness of fit," or, more appropriately, lack of fit. The first case appears in the next section in an example on octane blending. With octane measurements, experience with the testing procedure has given the engineer a good estimate of u 2, both from previous work and from the work of others. The engineer can compare the mean square for residuals, S2, to the prior value, (T2. If i is considerably larger than u 2, that is evidence of a lack of fit in the model. In the case at hand, we can use the variation between the duplicate observations to obtain what is called an estimate of "pure error," the true value of u 2• The actual data arc shown in table 15.12.1; some minor adjustments have been made to eliminate roundoff problems. The differences between the four observations at 220 degrees have nothing to do with whether we chose the right regression model. They are due entirely to noise. Their sum of squares, L (y, - y-,)2, is an estimate of Table 15.12.1. Polymer Yields Temperature 220 230 240 250 260 270 Yields 60.2 59.2 58.9 57.3 72.6 70.3 72.3 71.2 77.1 77.5 77.5 79.9 80.7 80.7 81.9 83.1 86.4 85.5 84.4 86.6 84.1 85.7 H5.6 86.5 257 SINGULAR MATRICES with 3 (U. Pooling this with the similar estimates from the other temperatures provides a sum of squares for "pure error" with 18 d.f. When a straight line is fitted for yield against temperature using all 24 observations, we obtain the same fitted line as before when we used the means. The sum of squares for error is 313.2 with 22 d.f.; the sum of squares for "pure error" is 21.44 with 18 d.f. The difference, 313.2-21.4=291.8, measures error and a lack of fit with 22 - 18 = 4 d.f. We divide each sum of s'l,uares by its d.f. If th~re is no lack (!f fit, we should ha~e.two estimates of (7-. We compare the raho of these estllnates to the F statistic. In the present example, we have 3(72 291.8/4 F= 21.44118 = 61.2 to bc compared to F"'(4, 18) = 2.93. In the general case, jf there are q d.f. for pure error, we denote the sums of squares for error and for pure error by Sc and 51'<' respectively; the test statistic is F = (S, - Spe)/(n - 1 - p - ..q) Sp,lq with n - 1 -- P q and q d.f. 15.13. SINGULAR MATRICES We have already mentioned that the matrix X'X must be nonsingular (a singular matrix is one that does not have an inverse). This means that the columns of X must be linearly independent. The following example illustrates how such a singularity can occur in practice. Suppose we are mixing three component gasolines to make blends, and that the volume fractions of the components in any blend arc denoted by x" x 2 , and x]; the sum x I + x 2 + x] is identically equal to unity. If we try to predict some response, such as the octane number. by the eyuation the X matrix will have four columns, but its rank will be three. The sum of columns 1 through 3 is a column of unit elements, which is identical with column O. In this case, we will not be able to find a unique solution to the normal equations. If (3 = «(30' (31' (32' (33)' is a solution, so is the vector «(31) + 3/, (31 - I, (32 - t, (3 .. - t)' for any value of t. There is clearly one variable too many. The solution to this difficulty is to drop one of the columns from the model. 258 MULTIPLE REGRESSION 15.14. OCTANE BLENDING The following data are taken from an octane blending study reported by R. D. Snee (1981). The complete study involved six components. We confine ourselves to the blends involving only three components: alkylate (ALK), light straight run (LSR), and light cat-cracked (Lee). The response recorded is the research octane number of the blend in the presence of an additive. The data are given in table 15.14.1. The usual approach, recommended by Scheffc (1958) is to drop the constant term from the model and fit the new model: Minitab achieves this by adding the subcommand NOCONST ANT. The data Y, XI' X 2 , and X3 are read into columns I, 2, 3, and 4, respectively. The variables arc denoted by Y = OCT, Xl = ALK, X 2 = LSR, and X) = Lee. REGRESS c1 3 c2-c4; NOCONSTANT. The regression equation is OCT = 106 ALK + 84.1 LSR + 101 LCC Predictor Noconstant ALK LSR LCC Coef. St.dev. t-ratio 106.054 84.0741 100.876 0.779 0.6873 0.686 136.15 122.33 146.95 s == 0.9608 Table 15.14.1. Octane Blends y X, Xl X, 106.6 t\3.3 99.7 94.1 l.00 0.00 0.00 0.50 0.50 0.00 0.00 0.00 0.33 0.00 1.00 0.00 0.50 0.00 0.50 0.25 0.75 0.33 0.00 W:U 93.6 97.4 8ltS 97.4 D.()O 1.00 0.00 0.50 0.50 0.75 (US 0.34 259 THF. QUADRATIC BLENDING MODEL This formulation is the usual linear blending model. The individual coefficients are estimates of the octanes of the pure components. For the prediction of the octane of future blends, we can act as if the octane number of the light straight run is 84.1. We could, instead, have dropped one of the gasoline columns and fitted, for example. Y = f3() + f3 I + f3~ . With this model, the fitted equation is OCT = 100.9 + 5.2 ALK - 16.8 LSR. This model amounts to using the light cat-cracked as a base fuel; its octane is estimated by the constant term. The other coefficients represent the differences hetween the component and LCC. Thus, the estimated octane of pure alkylate is 100.9 + 5.2 = 106.1, and of light straight run 100.9 - 16.1; = 84.1. IS. 15. THE QUADRATIC BLENDING MODEL The fit of the linear blending model to the data in table 15.14.1 had s = 0.96. The standard deviation of an octane determination is about three-tenths of a number. Clearly. there is strong evidence that the linear model is not adequate, and that there is some curvature, or synergism. The engineer should try a quadratic model. The complete quadratic model. under the restriction that LX, = 1.00, is Y = f31X1 + f32X2 + f3.lX~ + f312 X 1X 2 + f3n X I X 3 + f3UX2Xl . (15.15.1 ) A term such as f3Xi is redundant, because X7 can be written as its inclusion would only lead to another singular matrix. Let us continue with the octane blending example. The fitted quadratic equation is Y == 107.0 ALK + 83.2 LSR + 99.8 LCe - 2.84X1X 2 + O.05()X1X 1 + 8.73X2 X 3 with s = 0.1955. This is clearly a more satisfactory 11t. 260 MULTIPLE REGRESSION EXERCISES IS. I. The data set used in exercise 14.4 contained a third column. This was a composite mileage, obtained by combining the city and highway mileages: compo = f30 + f3) (city) + 132(highway) M.P.G. CITY M.P.G. lII(iHWA Y M.P.C;. COMPo 53 58 49 55 45 46 52 50 50 50 43 49 40 44 42 46 46 47 47 47 45 42 38 38 39 39 38 37 43 40 Use least squares to calculate the coefficients f3u' 131' and f32' In retrospect, the composite value for the second car is clearly wrong, probably a misprint; why? Drop that car and recalculate the values. 15.2. The plot of the data in exercise 14.6 shows that the appropriate model is a parabola rather than a straight line. Fit a parabolic model to the data. 15.3. The values of x and x 2 in exercise 15.2 are highly correlated. What is their correlation? 3700 is a good approximation to i. Fit the parabolic model: Notice that the t-value for the quadratic term is the same as it was in exercise 15.2, and that the fitted curve and i are, of course, the same. Why do the t-values for the first-degree term differ in exercises 15.2 and 15.3'1 15.4. An engineer wishes to predict the mean annual runoff of rainwater. He has observations from thirteen regions for the following variahles: 261 EXERCISES Y: mean annual runoff in inches Xl: mean annual preciptiation in inches X2 : area of watershed (square miles) X3: average land slope (%) X 4 : stream frequency (square miles)-I No. Y XI X2 X3 X4 1 2 3 4 5 6 7 8 9 10 17.38 14.62 15.48 14.72 I!U7 17.0] 18.20 18.95 13.94 18.64 17.25 17.48 13.16 44.37 44.09 41.25 45.50 46.09 49.]2 44.03 48.71 44.43 47.72 48.38 49.00 47.03 2.21 2.53 5.63 1.55 5.15 2.]4 5.34 7.47 2.10 3.89 0.67 0.85 1.72 50 7 19 6 16 26 7 1.36 2.37 2.31 3.87 3.30 1.87 0.94 l.20 4.76 3.08 2.99 3.53 2.33 11 12 13 11 5 18 21 23 5 Fit the linear model y=~=+~~+~~+~~+~~. Is it reasonable to regard point 7 as a special case? Assuming that it is, refit the model, omitting that point. 15.5. Coking coals were investigated by Perch and Bridgewater (Iron and Steel Engineering, vol. 57 [1980], pp. 47-50). They observed Xl = percent fixed carbon, X 2 percent ash, X J percent sulfur, and Y = coking heat in BTU per pound. Fit the prediction model The first coal seems to differ from the others, and to earn an X rating from Minitab. How does it differ? Omit the first coal from the data and rerun the regression. Do you get a better fit? In what way? 1. 2. 3. XI X2 X3 Y 83.8 78.9 76.1 11.2 5.1 5.3 0.61 0.60 1.65 625 680 680 14. 15. ]6. Xl Xl X.l Y 55.6 54.7 53.4 6.6 6.5 4.5 0.90 1.54 1.10 715 705 730 02 0.9 61. 25.8 70.2 7.8 26. II.0 25.0 59. Fit Y to all five predictors.9 26.8 61. You will notice that XI is the predictor with the highest correlation. Yet the best subset of four predictors does not .2 11.4 60.8 55. Continue this process.7.0 73.1 5. to. 6.I : percentage of insoluble n-pentane Xi: flash point (OF) y 71 149 III 49 57 76 106 119 93 153 112 100 XI X2 X3 X4 X5 158 136 153 194 172 154 150 142 156 145 147 145 2] 18 20 16 14 12 17 10 18 35 38 34 26 27 26 28.6.2 30.03 0.07 725 710 710 700 715 710 740 730 730 15.6 68. (A continuation of exercise 15.3 l.41 1.6 9. 8. 12.0 7.3 8.6 7.06 1.1 59.6 27. 72. called backwards elimination.3 27. These are measurements of properties of twelve samples of asphalt.4 9.6.3 33. 18.262 MULTIPLE REGRESSION 4.3 56.: penetration at 71°F X.2 8. until all the predictors left in the model have significant [-values. 9.90 0.0 70.8 6. What is your final prediction equation? 15.85 0.36 1. 20.91 1.8 61.7 63.1 73. 21.70 0.8 635 590 550 625 595 625 625 590 550 550 545 595 18 38 26 32 41 14 29 12 32 We want to predict the penetration at 115°F from the other variables.09 0.4 56.74 1.31 1. The measurements are Y: penetration at ] 15°F Xl: softening point CF) X z: penetration at 32°F X.4 7.8 10.76 1.1 33. 19. 5.71 1.08 1. 23. 7.9 61. 24.0 27.75 0.9 lU 6.4 31. Then eliminate the predictor that has the smallest I-value and fit again.8 57.8 6.99 0.6 6.8 6.93 710 710 685 705 685 680 700 720 705 730 17. 60. 13.6 1. 22.5 6.9 30.6 8.) Find the correlations of each of the predictors with Y. 9.8.10. This peculiar behavior docs sometimes happen. Xl' X 4 . Prove that they give the same equation for any linear model with two predictor variables. Prove that the least squares estimates are unbiased (section 15. In the previous exercise.5).EXERCISES 263 contain XI' Find R2 and S2 when Y is fitted to Xl alone and compare them to the corresponding values for the best subset of four. 15. .2 and the method of least squares give the same fitted equation for the data in table 15. Drop that point from the data set and repeat the backwards elimination procedure used in exercise 15.7.6.0 and y = 124. The method used in section 15.76. and X s . (A continuation of exercise 15. 15. you might have decided that Xl was of no use as a predictor.2.1. 15.) When you fitted Y to Xz. point 3 had a large residual: y = 111. 16.1. which correspond to different sources of variation. John Copyright © 1990 by John Wiley & Sons. At harvest time. In those plots. Inc CHAPTER SIXTEEN The Analysis of Variance 16. A. these arc the data. the same variety might be planted in each plot. one from each population. THE ANALYSIS OF VARIANCE The analysis of variance was introduced by R. such as different fertilizers or different methods of cultivation.Statistical Methods in Engineering and Quality Assurance Peter W.2. M. This is done by the analysis of variance. this is called the total sum of squares. We then subdivide the total sum of squares into components. and then different treatments applied. The basic idea of the chapter was that the engineer took two samples. and compared the sample means by the two-sample I-test. The two different catalysts compared in chapter eight can be 264 . It is convenient to continue to use the word treatments in this chapter as a general expression. Alternatively. INTRODUCTION In chapter eight. In this chapter. The reader will notice the very strong connections between the analysis of variance and multiple regression through the underlying principle of least squares. or grand. The analysis of variance was originally used in agronomic experiments. and so the jargon of agronomy continues in the field. Fisher about 60 years ago. mean. The essence of the procedure is that we take the sum of squares of the deviations of all the observations in an experiment about the overall. usually one variety to a plot. and test certain hypotheses using F-tests and I-tests. we extend those ideas to the situation in which we want to compare more than two populations in the same experiment. different varieties of some crop are sown. A fuller treatment of this subject can be found in John (1971). we considered experiments to compare two populations. An agronomist plants crops in a field divided into small plots. the yields of each plot arc recorded. That procedure included obtaining an estimate of the variance from the experimental data. 2) . The jth observation on the ith treatment is denoted by y".97 41. observations on the ith treatment provide a sum of squares of deviations. observations with a total of N = ~ 11. usually denoting a "run" made in a chemical or other plant under a certain set of operating conditions.8 45.9 43.t.0 43.yy. Suppose that the engineer has samples of observations from each of { treatments and that the sample from the lth treatment contains tI.3.72 43. We can call y" the response observed on the jth run with the lth treatment. the six temperatures in the example in chapter fifteen can be regarded as six treatments. = L L (y" .0 42.6 42.8 44.9 42.7 45.5 45.1 44.1 42.1 44.1 45. observations.97 .4 41. E (y" .3 44.C.1 A Averages 45.2 45. . the e" terms arc the random noise with the usual assumption that they are statistically independent normal variables with the same (also unknown) variance u 2• Let y. It is now common to replace the name analysis of variance by the acronym ANOVA.9 46. we have S.3. THE ONE-WA V LA VOUT One-way layout is jargon for the simplest experiment in which we compare { different treatments.y.1 41. The n. Table 16. . (16. Later in the chapter. Table 16. denote the average of the observations on the ith treatment and y denote the average of all N observations.1)u 2 and 11.3.6 42.5 42.1 shows a set of data for an experiment with six observations on each of five treatments.9 41.1 d.3 42. The mathematical model for the data is (16.265 THE ONE-WAY LAYOUT called two treatments. are the (unknown) treatment means that we wish to compare.H 46.25 42.1 42.3 43.1 ) The parameters J.3.60 B C D E 42.H 42. . The agricultural word "plot" is often replaced by the word rUI1.)2 .3 41.5 43. more complicated experimentstwo-way layouts and three-way layouts-arc introduced.1 43.3. with expected value (n.3 43. Summing these terms. 16. the expected value of the numerator of F is (T2). = E (T.f.4) and that its expectation is E(SJ = 2: n. error TOTAL DF SS 4 49.08 to the tabulated value of F for (4.05.05/25 = 0. The mean square for treatments is M.76 at (~ = 0. SS for error = Se = Sy~~ S. Table 16. with n . which is 2.3. observations on the ith treatment and that their total is T.I. .I) = 49.5) where ji is the average of the "".J(t . = S.33/4::=.2. M.C. s S.f. We compare the value 18. Then we compute the sums of squares: The total sum of squares = Sy = L. ( 16. These slims of squares are combined in the analysis of variance table 16. 1n. and reject the null hypothesis.L. Their ratio is the F statistic: F= 12. ANOVA Table SOURCE treat. with a little algebra.2. It is easier to proceed as follows.333. The F statistic tests the hypothesis that all the treatment means.("". """ are equal (in which case. L. - ji)2 + (t . which is called the correction for the mean.682 = 18.682 F 18.3. It can be shown. y~ .1)(T 2 ..t) d. Suppose that there are n.y)2.08.08 .t d.t) = 17.050 66. SS between treatments = S.3) The term S. The formulas that were given hcfore are not the formulas that are used in practice. The sum of squares.3.25) d. The F statistic. S" has t.3. Let G be the grand total of the N observations.266 THE ANALYSIS OF VARIANCE We now write Sy ::=. and (16.3.C.I(t-1) 2= has (t .682.) . N . = Ny.333/ 0..1 d. is called the sum of squares for treatments. G = 1: 1'.333 0. that (16. We compute C = G 2/ N. L (y. 12.380 25 29 MS 12.330 17. 52 = Sf/(N .f. i" in increasing order. WHICH TREATMENTS DIFFER'! We have just reached the conclusion that it is not true that all the population means. E>B. as they arc often called.4. . if. f.1 lists the treatment averages.2) The term on the right side of equation (16.60 16. In the example.4. With / = 0. we have for ex = 5% with 25 d.97 44. Applying this LSD to the averages in table 16. 16.06..1) reduces to ( 16. C.4. E. equation (16.4. Some software packages compute these difcrences automatically or on request. Table 16. and Duncan.) . we conclude that A> 8.1. This does not imply that they all differ from each other. D. the LSD is 0. (16.4. 1L2.5. we declare that Ilh "" Il. 1* = 2.4.682 and n = 6.4. C> B.t d.f.t. We would reject the hypothesis ILh == ILl if where (* is the tabulated value of ( with probability a 12 in each tail and N .97 42. or.25 45.98. we cannot conclude that IL4 "" ILs or that IL4 . Sample Means B D E c A 41. It is called Fisher's least signifkant difference (LSD).72 42. I.I > l*(sYl Inh + I In. and only if. THE ANALYSIS OF COVARIANCE We can sometimes sharpen our ability to distinguish between treatments by taking into account and adjusting for c()variatcs. Details are given in the manuals. IYh - Y.4. where the sample sizes arc all equal to n. D. Other formulas for significant differences have been derived by Tukcy.267 THE ANALYSIS OF COVARIANCE Table 16. Newman and Keuls. E. Some statisticians argue that Fisher's LSD is too liberal and that it declares too many differences to be significant. Equivalently. such as the example of the previous section. Il" arc equal.2) applies to all pairs of treatments.1) In the particular case. 38 3. The model implies two things.6 51.42 3.2.79 3.16 concomitant variables. The adjustment for the covariate is linear.1) and write (16. contained an impurity.48 3.9 52.2.1 53. and the weight gains are compared. This could be called a first-order adjustment. One-Way ANOVA Ignoring Covariate SOURCE Treatments Error TOTAL OF 2 21 23 SS \6.7 52.3 50.17 2.1 is a plot of Y versus X. The plot of Y versus X is a straight line (except for random noise) for each treatment and. Three reagents for a chemical process were compared in a pilot plant.2 49.48 3. Example 16.9 A X Y 2. Nevertheless.23 3. Figure 16. In animal nutrition experiments where some young animals are fed ration A for a few months and others are fed ration B.78 2.1) where x" is the value of the covariate for the ijth observation. furthermore.31 2. Yields and Impurities y 54.86 100. does not Table 16.99 F 2.45 3.73 2.5. We add a term to our earlier model in equation (16.1 51.1 51.6 54.5. A common situation occurs when each run is made on a separate batch of raw material and we may want to "adjust" Y for the amount of some impurity in the batch.16 2.41 2.5.1.81 MS 8.5.1.1J6 83. by and large. The raw material.268 THE ANALYSiS OF VARiANCE Table 16.4 50.5.1.1 47. unfortunately.4 51.43 2.19 2.48 3.12 . Y denotes the yield and X the percentage of the impurity in the raw material for each run. Eight runs were made with each reagent (treatment). one might take the initial weight of each animal as a covariate.7 5\.72 3.3 53.3.1 52.7 51. The data are shown in tahle 16.29 2.17 3. We illustrate the analysis of covariance with the following example.22 51. which ignores the levels of the impurity.5.9 50.40 2. It is clear that the yield decreases as the amount of impurity increases.9 C X 3.5.6 54. and that.76 54.99 2. the one-way ANOYA in table 16.3 B X Y 2. the slope of the line is the same ({3) for every treatment.57 3.0 55.5 53.6 56. the yields with treatment A are lower than the yields with C. which is much lower.83 14.6512 = 10. however. Plot of yield vs.6. namely. a value that is exceeded by three runs with A and by five runs with B.30 l'igure 16.81 MS 70.269 THE ANALYSIS OF COVARIANCE e 56. The treatment averages were A = 51.43 and 3. we show how the figures in table 16.6512 4.00 Percent impurity 3.32 13.400.99 . and we cannot conclude on the basis of that analysis that the treatments differ. = 14. which is somewhat lower than before.40 2. impurity. and S2 = 0. c = 53.02 100. Analysis of Covariance SOURCE Covariate Treatments Error TOTAL OF 1 2 20 23 AD] SS 70.5.60 3. which is significant. we adjust for the amount of impurity on each run by the analysis of covariance.5.3832 F 108.5. Table 16.32.77 10. In the next section.3 are obtained.0 e B Y A A e B B e 52.5 e B A A A e e Be A B A8 B 49. After adjusting for the covariate.99.437. Table 16.0 A 2. The two smallest yields for Care 51. and we are able to conclude that the three catalysts are not all equal.3 shows the printout from an analysis of covariance program. B = 52. The F ratio increased to (14.16. S. give a significant value of F.1589 0. but those values of Y occur with relatively high values of X.5. 3.8331 7.1.6512. If.3.70 3. we can establish that there are differences between treatments.32/2)/0.162. 3. one for each treatment.34 . Let Xl denote the percentage of the impurity present in a run.6023 0.6.s rather than just one line is 27. table 16.4035 0.5. X3 = 1 for C.3454 0. equation (16.13.2).32.10.92 .02 = 14.871 Analysis of Variance.789 13. This is the (adjusted) sum of squares for treatments in table 16. The following model fits three parallel lines to the data.S070 Coef.1.048 0.5. If. and b 3 = 1L3.60 Xl + 0.94 P 0.66 = 0.6.813 MS 29.ILl. The reduction in S" achieved by considering three separate parallel line. Sr drops to 13. THE ANALYSIS OF COVARIANCE AND RE(.88 X3.744 X2 + 1.5.000 .1. Regression The equation is Y = 61. see the accompanying table 16.2.6.6.919 -3.3 can be seen if we approach the problem from the point of view of regression.RESSION The rationale for the analysis of covariance table 16.270 THE ANALYSIS OF VARIANCE 16. We next introduce two indicator variables: X 2 = 1 if the observation is made with catalyst B and zero otherwise.1) The sum of squares for error about this line is 27. The reduction in Se that occurs when we include the covariate is Table 16. we refer back to the one-way analysis of variance. SOURCE OF Regression 3 Error 20 TOTAL 23 SS R7. X2 == 1 for B.ILl' With this model. We tlrst fit a single straight line through all 24 observations with the model (16.263 0. b o = ILl' the intercept for the line for the first treatment.7445 1.8799 R2 St.651 F 44. Xl = Impurity.6.3.07 .86.dev. 1. (16. 61.85 4.2) The common slope of the three lines is hi = {3.43 1.6. on the other hand.02. the sum of squares for error when we ignored the covariate was 83.024 100.342. b 2 = iL2 . X3 = I if the observation is made with catalyst C and zero otherwise. Predictor Constant Xl X2 X3 s = 0.4038 t-ratio 59. 12. A similar argument defines the quadratic contrast as L (Z\I.5471. The coefficient in the regression equation was E (z I. We notice that the I-values for the two indicator variables arc 1. We look at it again as an example of one-way analysis of variance with a factor.4 86.12.6 86. or that J.801.3 71.271 QUANTITATIVE I'ACTORS 83.4 85..9 80. with E = 70. 16.5 77.2 77.6 84. We conclude that catalysts A ami H do not differ significantly. := 4(linear contrast)2 = 1816. The corresponding slim of squares is z.1 77.2.1 86.6 70. We recall.84. and Chevrolet.1 for convenience.1. that has six levels. which is repeated as table 16.5 and the corresponding sum of squares as 5 Q == 4(quadratic c()ntrast)~ 22.00. e.= 271.3 72. we cannot conclude that b l "" 0. The data were shown in table 15.3 as the (adjusted) sum of squares for the covariate.YJ 170 == 2.00 " = -84.7.2 58.7.5 84. and 5. temperature.V) = -75.5.66 for C. from section 15.7 .2 59. The analysis of variance table for all 24 observations is shown in table 16.Lz . the linear polynomial. QUANTITATIVE FACTORS In section 15.3. Out of the sum of squares between temperatures. This quantity (allowing for some roundoff error) appears in table 16. were -5.85 for B and 4. Since 1. make of car with levels Ford. 1.9 83.3 72. We can also establish by similar calculations that C gives signifkantly higher yields than B. we introduced an example in which a quadratic polynomial was fitted to the yields of a polymer plant. The term quantitative is used for a factor whose levels are expressed as quantities.Il-\ "" O.g.7. 22 zi the factor 4 in the numerator is the numher of ohservations that were made at each level of temperature.9 57. 51. ? L zi Table 16.1 85. Plymouth.85 < 2. but that C gives significantly higher yields than A.86 . we can take components with single degrees of freedom for the linear and quadratic trends.02 == 70.7. -3.7 80. as opposed to a qualitative factor whose levels are not numerical.62.5 85.13.1.44 . -I.5 79. Polymer Yields Temperature 220 230 240 250 260 270 Yields 60.10. that the values taken by Z\.7 8\. and Toyota.76 1. Analysis of Variance SOURCE Temperature Residual TOTAL OF 5 18 23 SS 2108.67 1. with n ohservations on each make. Residual TOTAL OF SS I I 1816. Cubic etc. indicates that the model with only the quadratic trend is not entirely adequate and that there is still some unexplained curvature. 18)::: 6. A = E ad.3.272 THE ANALYSIS OF VARIANCE Table 16.7.3.7. Quad.19 5.62 F 271. Temp. and B = E b. Two contrasts. The F statistic.Y" are orthogonal if E a.29 21.01 2129.44 6.35 21. Suppose that we have as the levels of a qualitative factor six makes of car: Cadillac. Its sum of squares is nB2 nB2 SIl=12=~2' "-' b.7. Mercedes.68 . Single degrees of freedom for orthogonal contrasts can also be isolated with qualitative factors. Its sum of squares is nA2 nA2 SA=6=~ 2 ' L. The contrast compares the American cars as a group to the foreign cars.44 MS 421..44 3 18 23 20.79 MS 1816. The reader will notice that this is the same as the pure error test for goodness of fit that was used in the previous chapter to test the adequacy of the simpler first-degree model.68. Analysis of Variance SOURCE Lin. Table 16. The orthogonal contrast compares the luxury cars to the ordinary cars. Temp.76/1. Chevrolet.62 271.44 2129. Mazda. Plymouth.19 F 354.2.79 We can now amend the analysis of variance table in table 16.19 = 5.b. F(3. = O. a. 12) which is not significant.8. 16. The two new contrasts arc orthogonal.1.88 lUIS 11.06 with 12 d. a one-way analysis for Table 16.273 THE TWO-WAY LAYOUT Indeed.8. ordinary for American cars) and (luxury vs.54 12.13 .8. A.48 S. and S. ignoring the fact that the observations were made at different pressures. The data set for an experiment like that which has just been described is shown in table 16. The analysis is illustrated in the following example. B can be divided into two contrasts: (luxury vs. We could just as well have five temperatures and six pressures.47 41. temperature and pressure. = 3. ordinary for imported cars) .1. = 20.55 2 3 4 Sum 9.70.28 44. Example 16. If we were to make a one-way analysis for the temperatures.32 47. This is called a two-way layout.76 9.f. We can analyze the data as if there were two one-way layouts--one for rows and one for columns.27 11.W 40.99 40. Similarly.06 43.89 11. The corresponding Fstatistic is F(3.I7 13. A Two-Way Layout Temperature Pressure 1 2 3 4 Sum 8.67 = 1. at four levels each.95 to.03 11. 12) 2.59 9. TIle engineer can arrange the data in a rectangular array.52 with 3 d.8. P(3. we would get S( = 8.84 = 1. The fact that in this example the two factors each have the same number of levels is irrelevant. THE TWO-WAY LAYOUT Suppose that a chemical engineer wants to investigate two factors.97 174.96 11. and makes one run with each combination of temperature and pressure-16 runs in all.11 49. both to one another and to the first contrast. with the columns corresponding to the four levels of temperature and the rows to the levels of pressure.f.44 12.49.60 9.10 40.1. 180 9. or we can. Two-Way ANOVA SOURCE temp.8.2. In the general case. and we now conclude that there are differences both between temperatures and between pressures.2.126 4.8. press.042 0. we let T. contrasts for linear temperatures. table 16. Notice that the value of . for the sum of squares for error. denote the sum of the observations at the ith level of A and T. quadratic temperatures. However. ignoring temperatures. h .548 from its earlier value of 1.f.200 the pressures. The F statistics are compared to the tabulated value F*(3.518 15. J denote the sum of the ohservations at the jth level of B. if appropriate. and (0 .86. ORTHOGONALITY The reason that we were able to take out the sums of squares for each of the two factors in the prior example is that the experiment was balanced. error TOTAL DF 3 3 9 15 SS 8.. the two factors arc said to be orthogonal to one another. We can go ahead and apply the methods of section 16.67 when we looked at temperature alone.I d.l)(b . Orthogonality is a very important property in .7 and take out single d. the pressures "cancel out.L for the sum of squares for A.274 THE ANALYSIS OF VARJANCE Table 16. for B." and vice versa. A and B. there will be a . 9) = 3. To compute the sums of squares for A and for B. This balance has the technical name of orthogonality.4 to see which temperatures differ. with a and b levels.f.580 MS 2. and cubic temperature.839 5.f. and similar contrasts for the pressures. also does not give a significant F value.548 F 5.1) d. and so when two temperatures are compared. apply the methods of section 16.9. where there are two factors. we can take out of the total sum of squares both the sum of squares for temperatures and the sum of squares for pressures and obtain the following ANOVA table.. respectively. Each temperature appeared exactly once with each pressure./ has dropped to 0.I d.936 28. Then the sums of squares for A and Bare and The sum of squares for error is obtained hy subtraction as o 16. but the fourth is fixed because the column total must be zero.570 0.017. average. We can use the following model for the observation in the ith row and the jth column: (16.9.275 ORTHOGONALITY Table 16. . and the residual in that cell is 8. there is a total of nine cells that can be filled freely. m is estimated by the overall.615.625 -0.160 -0. Thus. 174. '2 = = -0.325 0.485 0.055 0.0.8.570. In each of the first three columns.701 1.040 -0.1. is the extra contribution for being in the jth column.484 an experimental design.520 = -0.061 0. -0. and c. The sums of the residuals in each row and in each column are identically zero.9. The complete set of residuals is shown in table 16.883 .277 -1.105. '4 = c.1. The sum of the " and the sum of the c. or grand. and we will see morc of it in the next three chapters.911 = 10.2. arc both identically zero.082 0.1) where m represents an overall mean.858. the first three entries are random.495 = 9. In our example.0.603 0.Ml . We can now see why the sum of squarcs for error has nine d. For we subtract the grand average from the avcHlge of the observations in the first row: 'I' 'I = ·-0. CI We can go further and calculate an estimated value for the observation in row i and column j as so that .048 -0.868.520. C2 = +0. given the first three entries in each row. The sum of the squared residuals is indeed 4. Residuals -0.868 .9.337.883.=+1.95 -.495 . similarly.13/16:: 10.5S7 -0. the fourth is fixed. C4 = -0. +1.936. " = -O.f. as was shown in table 16. " is the extra contribution for being in the ith row.257 (J.9. Similarly. the entries in the other seven cells are forced by the conditions that the rows and the columns sum to zero. 0 52.0 8H. and so on.0 65.11. we should. split each unit in half.0 52.10. An agronomist who is carrying out an experiment to compare v varieties of barley would divide a field into plots.1). then a column term. and the experiment is called a randomized complete block experiment.0 34. Instead of taking 2n units of raw material and assigning n to each treatment.0 55.0 51. RANDOMIZED COMPLETE BLOCKS In our discussion in chapter eight of the two-sample (-test.0 49. equation (16.0 65. Our engineer. INTERACTION The model. Filling Weights of Cans 68. we noted that the precision of the experiment can be improved by pairing.0 84. might take n batches of raw material. Uti.1. Table 16. Thc technical term used for the strips is blocks.0 50. 16. that we have used so far is a simple additive model. The portions of raw material play the role of the agronomist's plots. The name carries over to engineering applications. divide each batch into ( equal portions.O 52. we add to the mean first a row term. We can return to our example and replace the word pressure by the word block and go through exactly the same calculations. and so the field might be divided into strips of u plots each that arc fairly homogeneous and each variety planted in one plot chosen at random from each strip.0 52. and run one-half with one treatment and the second half with the other.0 87. if possible. A semiconductor manufacturer could take a boat that holds 32 wafers in an oven and load it with J6 wafers that have received one treatment and 16 that have received another.0 70. similarly for each of the other varieties.0 75.0 73. Again. In the analysis. and assign one of the portions to each treatment.0 61. he would call it a split lot.0 70. it is important that each treatment appear the same number of times in each block in order to achieve orthogonality-to make treatments orthogonal to blocks.0 57.0 50. One experimental design would be to choose the n plots for A at random from the total number.0 32. and the batches are the blocks.0 56. n for variety 8. take n units.0 45. of plots available. we treat the blocks as if they were the levels of a second factor.0 45.0 . Then the agronomist would take n plots and sow variety A in them. who wants to compare t treatments.0 4!U) 36.0 47. and. But often there is considerable variation in fertility over the whole field.9.11.0 40.276 THE ANALYSIS OF VARIANCE 16. however.90 -4. It can be argued. we see that head 5 exhibits some very erratic behavior.70 5.1.27 -0. Perhaps so.13 -4.47 -26.33 4.1 are the coded weights of the amount of drink that the head put into the can.90 -7. R. Example 16. with the benefit of hindsight.2. But when we look at the residuals. Residuals 4. table 16.l. E. This implies that the improvement in yield by increasing the temperature from level 1 to level 2 is the same.10 6.27 1. which is discussed in the next section.47 -2.2. It is also rather restrictive because it implies that the amount added to each observation because it is in the first column is the same in every row.13 8.07 130. we have a test for the existence of interaction.67 -5.20 23.93 7. ANOVA Table SOURCE times heads error TOTAL OF 4 5 20 29 SS 2601.40 -6. like the residuals in a regression.70 -4. pattern is called interaction.217 INTERAC'TION Table 16.3. should not show any pattern. The data shown in table 16.97 MS F 650. spot interaction by examining the table of residuals. We can sometimes.47 6428.00 ·-1. but rigid.13 15. Ott and R. The residuals in the table. When there is only one observation in each cell. They took samples of cans filled by six heads of the machine every hour for five hours. we have no such test. no matter what the pressure.20 -0.20 -8.10 0. The sum of squares for error measures both interaction and error.11.30 -0.47 -8.88 finally.70 0 . Snee (1973) reported an investigation of the heads of a multi head machine for filling cans of a soft drink. but it is easier to spot this kind of anomaly from the table of residuals.60 -6.11.3. When there are several observations in each cell of our array.00 l.11. The failure of the data to follow this simple. There is no significant difference between heads.28 245. O. an error term.12 5. The analysis of variance table is table 16.07 9. or nonadditivity. that we could have spotted this peculiar behavior by looking carefully at table 16.11. Table 16.50 "17. and is correspondingly inflated.11.13 1225.30 7. because its column contains the four largest residuals.50 2.11.11.30 7.37 2602. 278 THE ANALYSIS OF VARIANCE Table 16. where we divided the residual sum of squares into two portions: pure error and goodness of fit.00 MS 21. i. This is done in the same way as in regression. Example 16. totals 16.2.Sa .1.Sab .12.e. and two observations in each cell.12.00 70. The data are shown in table 16. A (rows) with three levels and B (columns) with four levels.. in the ijth cell.12. R3 RI R6 R4 85 82 90 88 679 2.79 3.1.2.29 . Data II III IV Ruw sum J. The sum of squares for error is then obtained by subtraction as Sc "" Sy .12. RO 7'15 81 78 8R 86 Rl 657 85 477 491 503 521 1992 Cu\.1.13 58. INTERACTION WITH SEVERAL OBSERVATIONS IN EACH CEU~ When there are several observations per cell. To compute the sums of squares for A and for B. one due to interaction and the other to error.Sb .00 330.00 11.58 F 3.12. we let T" denote the sum of the observations at the ilh level of A and the jth level of B. Table 16. Consider an experiment with two factors.75 43. Then the sums of squares arc the following: Sb = L ( :r: J ) - C. The analysis of variance table is table 16.12.25 174. we can split the residual sum of squares into two components. Analysis of Variance SOURCE A B AxB Error TOTAL df 2 3 6 12 23 SS 42. 77 78 80 R2 80 83 '157 89 656 3. 12. In the general case of two factors.1) d.1. and the kth level of C is denoted by Y Ilk' The sums of squares for A. the analysis proceeds in the same way as for two factors except for some extra terms in the table. for the slim of squares for error. b. and so on .3. 12) = 3.l)(b . the interaction sums of squares are given by .L for the total sum of squares.f. and there is dear evidence of interaction. /c' Then So = 2: T. b . B.5 (row 3.5 The value of the F ratio. and c . perhaps. and Care computed as before.L for the AB interaction and the remaining ah(r . B.1) d.5 79 81 81. we have a third subscript on the totals. is significant. the last cell is appreciably higher than the others. T.5 88 83. and C at a.5 86.b ' .3. One suspects that the observations R6.12. which arc now denoted by 1'. with a and b levels. .5 (row 3. the mean in the first row is at least three points higher than the mean in the last row. and T .1 d. In the Ilrst two rows. o 16. . and have a . A. The observation at the ith level of A.279 THREE FACroRS Table 16. table 16. for A. . This will be divided by the ANOVA calculations into a-I d.f. together with (a .1 d. A and B. 3.-.C. column 3) and..13. f. column 4) should be looked at carefully.5 89 77. THREE FACTORS With three factors. J.1. . F*(6. we look at the cell means. h . there will be ahr ..29.1 for B. We note that in every column save the third. To see this more clearly. Cell Means 82 85 83. and r observations in each cell. and c levels each.00. 83.5 79. the jth level of B. respectively. as before. c Similarly. .. five.280 THE ANALYSIS OF VARIANCE If there are m observations per cell. lbr become T.. the remainder is the sum of squares for error.1.and three-way layout can be extended to several factors. Furthermore. but they point out a fundamental difference in experimental strategies in the two fields. The engineer makes five determinations of the octane number of each gasoline. then the agronomist has to wait until the end of the season to harvest the data. of course. one after another.14.. Do the gasolines differ significantly in their octane numbers? Can you recognize any of the gasolines as being different from the others? . we can also take out a component for the three-factor interaction.1) d. Nowadays. The residual sum of squares is then with abc(m . terms such as Tj2. they are easily done for us on a Pc. The cost per plot in the agricultural experiment is usually much cheaper than the cost per run for an engineer. these calculations were tedious.f. These remarks are. EXERCISES 16.f. The engineer makes runs. The data are given in the following. In the next chapter. with m> 1. The large factorial experiment with several factors at four. Years ago. Anything omitted has to wait until next year. and is therefore more able to conduct a series of small experiments. In the formulas for the sums of squares. There arc two reasons for this. the agronomist has to plant all the plots together in the spring and the plants grow side by side.Ibrm. or six levels is more appropriate to agriculture than to engineering. pausing after each to decide where to go next. After all these six terms are subtracted from the total sum of squares. in sequence. . we turn to a group of factorial experiments that are commonly used in industry--experiments with several factors at only two or three levels-and develop some of the strategy of sequential experimentation. A research engineer for an oil company wants to compare four gasolines. SEVERAL FACTORS The methods presented in this chapter for the two.2. 16.1) d. Sabe' which will have (al)(b -l)(c . generalities. Carry out the analysis of variance calculations and find which acid differs from the others (if any).0 86. There were seven samples from the first source and four from each of the other two.5 71.5 85.2.7 80.7 80.8 80.0 71.9 72.0 70.281 EXERCISES B 80.8 87.4 80.4 80. the engineer was not able to make the same number of plant runs with each of the acids.3 A 81.9 86.2 80.9 86.9 79.4.6 87.0 72.4 16.7 II 86.6 C D 79.3. B A 73 75 78 73 C 78 74 79 71 76 78 74 75 79 78 80 81 77 71 16. the engineer wants to compare five gasolincs.1 80.1 85.7 89.6 87.0 ~n.6 80.4 73.6 86.5 72. Unfortunately. In another octane study.2 86.4 SO.3 86.5 71.5 88.1 IV 86.6 86.9 III 88.8 72.7 71.3 80. An engineer compared acid catalysts from thrcc suppliers in a chemical process at the pilot plant stage.0 V 87.0 79. The cnginccr takes live cars and makes one determination of the octane number (on the road) for each gasoline in each car.2 80. Is there a significant difference between cars? Which gasolines differ? Cars Gasolines A B C D E I 88. Do the sources differ in the amount of product? A = B= C= 72.0 78.2 .2 69.6 80.5 72.1 72.8 86.6 16. The pcrcentages of product from three sources of raw material follow. Thc figures shown are the percentages of raw matcrial converted to product.2 85.3 81.1 86.1 86.6 86.5 81.7 71. 5 23. Positions (columns) 1. 127.8. the rows correspond to four machines and the columns to nve levels of operating temperature. 129 125.129. 127.6. 2. In exercise 16.2 25.282 THE ANALYSIS OF VARIANCE 16. The measurements in column 5 are made at the center of the wafer.127. Three components are tested under each set of conditions.6 23. were compared for thickness.7.6 22. C. Analyze the data.7 16. A company wishes to consider changing the way in which one of the materials that it uses in its fabrication process is made. Temperature Machine I II III IV A B C D E 21. The response is a measure of the tensile strength of the material and a higher value is desirable. Is there cause from this data to think that there are significant differences between positions on the wafer? Docs the thickness at the center differ from the average thickness on the edge? 16. 12R 16. The following data are the lifetimes of components.5 22. 122 12R.7 26.5 21. 126 135.7 21. Is there a difference between the methods? Is there a significant difference between suppliers? Is there significant evidence of interaction? Method Old New Supplier .7 21. 3. B. Two subcontractors are invited to make five samples of the material prepared by the old method and five samples prepared by the new method. 134.9 21. Five wafers. 124.3 24.) 1736 1665 1623 1696 1711 1725 1657 1647 1725 1712 1726 1674 1685 1735 1742 1730 1693 1721 1735 1744 1659 1637 1738 1735 1701 16.9. A. and 4 arc 90 degrees apart on the edge of the wafer. Do the wafers differ significantly in thickness? (Wafers are rows.2 24. the columns represent different positions on the wafer. and two power ratings.1 24.5. 136. 133. In a two-way layout.3 23. The components come from two different manufacturers.4 Supplier B 130.8 22. They are tested at four operating temperatures. 128. fabricated under different conditions. Five measurements were made on each wafer. 125. We may assume that the . 137.2 21.0 22.6.9 23.6 24. 126. Which are the important factors? Make a two-way table of means to explain the B*C interaction. BI 82 B3 B4 1252 2952 2892 2184 C.283 EXERCISES lifetimes arc normally distributed. 2544 2288 3076 2828 C2 1848 2092 2328 2416 2664 1960 2896 3200 3024 3360 3108 2668 290R 2212 3368 3148 Manufacturer A l C2 CI BI 2088 B2 2452 2824 3172 B1 114 2316 2124 1836 2412 1796 2088 278() 252R 4068 2188 2520 4176 3448 2548 3200 3236 3472 2392 2764 3U64 . Manufacturer A. S. Taguchi. especially in manufacturing engineering. An experiment with 11 factors. W. The engineer carefully decides beforehand to make. M. The factorial experiments are designed experiments. P. That design gives a good experiment with the important property of orthogonality. factors with more than two levels. who was using them for experiments at the Rothamsted Agricultural Station near London. each of which has exactly two levels. and in chapter nineteen. P. N. The reader who wishes to go further will find fuller discussions in the books Statistics for Experimenters by G. and Statistical Design and Analysis of experiments by P. applications that have come to be associated with the Japanese engineer and statistician G. G. for example. His first example dealt with using fertilizers to help in growing potatoes. one at each of five levels of A <lnd at each of four levels of B. These experiments playa very important part in modern engineering experimental design.1. Inc CHAPTERSEVENTEEN Design of Experiments: Factorial Experiments at Two Levels 17. John Copyright © 1990 by John Wiley & Sons. in chapter eighteen. and W. 20 observations. applications to chemical engineering were developed by George Box and others at Imperial Chemical Industries (ICI). We only have space for an introduction to factorial experiments and their fractions. Now they are widely used in 284 . It ended with a discussion of factorial experiments. In this chapter. We turn now more specifically to applications of experimental design to engineering problems. Hunter. phosphorus. is n called a 2 factorial experiment. and potassium. we consider factors at two levels. K. Box. using factorial experiments or fractional factorials. They were first discussed in a systematic way hy F. During and after World War II. Hunter. Yates (1937). M. His three factors were nitrogen. INTRODUCTION Chapter sixteen was a general survey of the analysis of variance. J. which were applied at two levels: zero and four pounds per plot. John and in articles in stich journals as Technomelrics and the Journal of Quality Technology.Statistical Methods in Engineering and Quality Assurance Peter W. E. 2. which were introduced by David Finney (1945). .1 factorial would be an experiment in a pilot plant with three factors: temperature at 200 or 300 dcgrees. With four factors. There arc nonorthogonal fractions that are useful (sec. it is necessary to assume that some of the interactions are zero. We will see in scction 17.525 25 ' rate . This means that in the analysis. with five factors. so we turn to fractional factorials. This important property holds for the 2" factorial and will also hold for the fractions that we will use. but thcy are mathematically equivalent. 17. As we noted in the previous chapter. pressure at 500 or 550 pounds per square inch. which we denote by 2 7 . or balance. they have been adapted successfully for many years by G. we can run a well-planned half fraction. Each can be reduced to experimenting on the vertices of a cube with coordinates ±1. the latter standing for seven factors at two levels each in eight points. 64 runs. sometimes denoted by t' \ with 32 points. we can use averages.4 or 2 7 / / 8. factorial experiments have the important property of orthogonality. In Japan. or a quarter replicate with only 16 points. and with six factors. In the chemical experiment. the factors are independent of each other. EXPERIMENTING ON THE VERTICES OF A CUBE The two experiments that were mentioned in the previous section come from vastly different fields.12 that when we use fractions. That many runs may tax the budget of time or money too highly. 1971). thc book by John. 32 runs. we can let the three factors be A (tempcraturc): B (pressure): C (flow rate): XI = x2 = X3= temp. We discuss a one-sixteenth replicate of a 27 factorial. so that when we arc estimating their effects. .250 50 pres.10 --2---' . Instead of running all 64 points in the factorial. or half replicate. but they are beyond the scope of this book. and flow rate at 8 or 12 gallons per minute. for example. Taguchi and his associates. and it also makes the analysis much simpler. A chemical engineering example of a 2. Orthogonality makes it easier for us to understand what is going on in the experiment. This reduction in the number of points does not come without cost. Each level of factor A appears the same number of times with each level of factor B. which increases the precision of the estimates. the complete factorial calls for 2 x 2 x 2 x 2 = 16 runs. His ideas arc now being applied extensively in the West.EXPERIMENTING ON THE VERTICES OF A CUBE 285 chemical engineering and in the semiconductor industry throughout the world. bd. as we will see shortly. The point with A and B at high levels and C at its low level is denoted by abo The point with all three factors at their high levels is abc. sometimes called a treatment combination.ls by 1 and 2. c. bcd. Taguchi denotes the leve. a. The point with A. The letter b denotes both the point with high B and low A and low C. Many agronomists use levels 0 and 1 rather than ± 1. h. ah. such as a choice he tween sodium hydroxide and potassium hydroxide as a catalyst. cd. In the three-factor experiment. 17. Each vertex of the cube represents a set of operating conditions. With only two factors. h. and abc. and ah = 54. With more than three factors. = (amount . and abc. In any case. we can let A. Yates gave a useful notation for denoting the various treatment combinations. we are. P. we are experimenting at the vertices of an n-dimensional hypercube. respectively. The notation serves double duty. ac. aed. we can arbitrarily call one of them the low level of the factor "catalyst" and the other the high level. and C correspond to N. we add eight more points: d. and K. a = 48. at its high level and the other two factors at their low levels is denoted by a. then. a. This formulation of the experiment was introduced by the scientists at leI. b. and abed. he. abo To add C. The extension to more than three factors is obvious. we multiply (1). We can present the data at the vertices of a square as follows: lib = 54 (1) = 36 a = 48 The (main) effect of the factor A is the average of the runs at high A minus the average of the runs at low A: .2)/2. A~ EXAMPLE OF A 22 EXPERIMENT Suppose (1) = 36. with X. h = 52. and also the observation (or average of the observations) made under those conditions.3. the one point with all three factors at their low levels is denoted hy (1). abd. ae. getting c. and ab by c. ad. EXPERIMENTS AT TWO LEVELS In the potato experiment. In a 24 factorial. be. The experiment has eight points: (I). we can talk about the high and low levels of the factors. If a factor is qualitative. experimenting at the vertices of a unit square.286 DESIGN OF EXPERIMENTS: FACTORIAl. B. The points are usually written in standard order: first (1) and a. multiplying by b to include the second factor. the effect of A is II . at low 13. and abo We usually write + and . Is the A effect the same at low 8 as it is at high 8'! Do we get the same improvement in yield with a ten degree increase in temperature at high pressure as we do at low pressure? At high B.(1)l = 54 . We could just as well have compared the effect of changing the level of B at high A and low A.3. The A B interaction is a measure of the failure of the simple additive model. b.b = 54 . If there were no interaction. 17. the effect of A is ab .36 = 12. ll. w-----------------------------------~=:~ ab54 b 52 48 (\) 36 High b Lowb Figure 17.-10.1'. each of these effects has I d. The difference is 2 . the interaction is (2 .instead of + 1 and -1.1.[b . In this example.12)/2 = -5. Interaction plot.12"'. figure 17. . is an array that shows the levels of the factors at the points of the design. THE DESIGN MATRIX The design matrix.4.48 .52 + 36 = -10 . The main effect of B is similarly defined: 36 + 48 _ 1 0 _ 52 + 54 2 2 -1 . The interaction can be displayed graphically in an interaction plot.52 = 2. the two lines on the plot would he parallel. X.0-44. and ohtained (lib . .0=7.1.a) -.287 THE DESIGN MATRIX A= 48 + 54 36 + 52 2 2 =51.3. X has two columns corresponding to A and B and four rows corresponding to (1).0. B'- Since there are two levels for each factor.(1) = 48 . This is orthogonality. -1). be. c. (-1. -1). We note also that so that AB corresponds to the product X I X 2 . 17.2. abe . ac. (l ). In the general 2" design. There are seven effects .: (XI = -1) 2 md 2.: (XI == +1) . The numerators of the fractions that give the estimates are called the A. B. b. + 1) exactly one time. We can add another column to the design matrix to correspond to the interaction term. (+1.L (x 2 = -1) 2 -.: (X2 = +1) . ab.288 DESIGN OF EXPERIMENTS: FACTORIAL EXPERIMENTS AT TWO LEVELS A B (1) X= y= + + + + + a b ab The estimated main effects can be obtained by multiplying the vector Y /2 on the left by the transpose of X: A= B= 2. An important consequence of the orthogonality of the design is that we are able to increase the precision of the estimates by using averages in this way. which can be represented as the vertices of a cube. the estimates of the effects are obtained by dividing the appropriate contrast by N /2 or 2" -I. +1). THREE FACTORS With three factors there are eight points. a. Xl X2 XI X2 + X== + + + + + Notice that every pair of columns contains (-1.5. and ( + J. and AB contrasts. .5. ignoring B.1 A (XI) IJ (xJ y C (Xl) (1) (l b ah 40 48 54 56 + + + + + + + + + + (" lie + be abe + A -40 +48 -54 +56 -44 +84 -42 +92 +100 B -40 -48 +54 +56 -44 -84 +42 44 84 42 92 C +92 -40 ···48 -54 -56 +44 +84 +42 +92 +28 +64 The main effects are obtained by dividing these contrasts hy 4. A. B = 7. To calculate the contrasts for the interactions. and e. is the coded yield of product. B. B. This data comes from a chemical experiment.2.f. Example 17.I. each corresponding to one of the seven d. and BC (X 2X 3 ). AC (XIX)). They are A = 25. Y. Table 17. and the three two-factor interactions: AB (X 1X 2 ). There arc the three main effects. pressure. He = -4. X I X 3 ' and X 2X 3 ' and then proceed in the same way. An explanation for the apparently large A C interaction can he found either by making an interaction plot or by making a 2 x 2 table of the totals for the four sets of levels of A and C. B. The A.5. and C = 16.S. Their estimates are obtained by dividing their contrasts by 4. The response. temperature.5. acid strength. The interactions are AB == 4/4 =: 1. AC = 20. and C. and ABC = 4. and e contrasts are calculated by multiplying the Y vector hy the appropriate column of the data matrix. The three factors were A. The design matrix and the data are given in table 17.1. we can make columns for XIX". This is shown in tahle ]7.ORS of interest.289 THREE FAC. 2 AS AC (XIX~) < (x IX~) + + BC (X2X. can be tedious and error-prone when the number of factors increases.5.6.) Y --~~-~---< (I) a b ab c ac be abc + + 40 4g 54 56 44 84 42 + + 92 + + + + + + AB AC BC +40 -48 -54 +56 +40 -48 +54 < 56 +40 +48 -54 +44 -44 -84 -42 +92 +g4 -42 192 -44 -84 +42 +92 +4 +80 --16 -56 C A I Iigh Low Total Low lO4 94 High 176 86 180 198 262 460 Total 280 The A C contrast is (176 .10 = 100 o < 17.290 DESIGN OF EXPERIMENTS: FACroRIAL EXPERIMENTS AT TWO LEVELS Table 17.86) .94) = 110 .(104.1) .5. THE REGRESSION MODEL The calculations of section 17. They can be madc more easily on a PC by using a regression program to fit the following model: y:= f30 + {31X 1 + {32 x Z + {31 X 1 + f3J2XIX~ + f313 x l x 3 + f323X2x3 + e. although simple. (17.6. 1. 3 Table 17.202 . We usually make the working Clssumption that interactions involving three or more factors are negligible a priori and omit them from the model. Fitting the Regression Model for a 2 Factorial MTB >set xl in cl DATA> -1 + 1 -1 + 1 -I + 1 -1 + 1 DATA>end MTB > set x2 in c2 DATA> -) -1 + 1 + 1 -1 -1 + 1 + 1 DATA>end MTB > set x3 in c3 DATA> -1 -1 ·-1 -1 +1 + 1 +1 +1 DATA>end MTB > let c4 = c1 *c2 (This puts xlx2 in c4) MTB > let c5 = cI *c3 (This puts xlx3 in c5) MTB > let c6 = c2*c3 (This puts x2x3 in c6) MTB > set y in c10 DATA> 40 4g 54 56 44 84 42 92 DATA>end MTB > name cl 'xl' c2 'x2' c3 'x3' c4 'x Ix2' c5 'xIx3' c6 'x2x3' MTB > regress c 10 6 c1-c6 The regression equation is y = 57. A = 2{3j. ABC. That would make a model with eight terms.0 x Ix3 .5 xl + 3. This is illustrated in tahle 17. corresponding to the three-factor interaction.03 P 0.291 THE REGRESSION MODEL The coefficients in the regression equation are multiplied by 2 to obtain the effects. It is important to note that because of the orthogonality of the design.50 Analysis of Variance SOURCE DF Regression 6 Error I TOTAL 7 xlx2 + 10.1). and eight points. which would force a perfect fit.00 F 14.6.00 2726. flI23X)X2X3' to eljuation (17. For the remainder of this book.50 x2 + g.g. e..OO x3 + 0.5 + 12. which is an edited regression printout from Minitab.00 32.2. including the constant term. We could add another term.6.00 x2x3 SS 2694.6. I. the estimates of the main effects and of the two-factor interactions will be the same whether we include or exclude the higher-order interactions. we will restrict the word interaction to mean only two-factor interactions unless we make a specitic statement to the contrary.00 32.00 MS 449. 9. They contain + 1. by squaring the effect and multiplying by 2"-2). ac . and enter them in column 1 of the work sheet. equivalently.(1). b. and abc . we divide every entry in column 3 by 4.2 A B 2 (1) a b ab AB 3 C 4 + AC 5 Be ABC 6 7 + + + + + + + c (Ie + be abe + + + + + + + + + + + + + + + + + + + + The sum of squares for an effect is obtained by squaring the contrast and dividing by 2" (or.5. Yates (1937) presented an algorithm for calculating all the contrasts simultaneously. We mention this again in section 17. +1 twice. see table 17. except for the total. through column n + 1). Take any pair of columns. the operation is continued for n cycles. In column 5. including those with more than two factors. we notice the orthogonality. and --1. sec tahle 17. ab.6. ab + b. With the model that includes all the interactions. Column 3 is ohtained from column 2 in the same way. ae. and abc. The first step in the algorithm is to writc the observations in the standard order: (1). followed by the differences: a . which appears in the first place. the entries. a. + 1 twice. the total of their sums of squares can provide us with an error term. Column 2 is made up of sums and differences of pairs of observations. are the sums: a + (1). we illustrate the algorithm by applying it to the 2 3 example of section 17. We then go through repetitive cycles of adding and subtracting. -1 twice.292 DESIGN OF EXPERIMENTS: FACTORIAL EXPERIMENTS AT TWO LEVELS Table 17. 17.c. Column 4 contains the contrasts in the standard order. we can rewrite the design matrix to incorporate columns for the higher order terms. together with the grand total. . as we did in section 17.b. and then column 4 from column 3. YATES' ALGORITHM This is another way of calculating the contrasts that docs not require a PC with a regression program. and abe + be. be.6.7.1. It is much faster than computing each contrast separately. -1 twice. Thc differences are always taken in the same way-the second observation minus the first.7. (In a 2" factorial. -1. Again. In this section. When we omit the higher-order interactions from the model.5. ac + c. ab .2. in order.be. c. + 1. 1. B. The output from applying Yates' algorithm is shown in table 17. Yates' Algorithm for 2 3 C 3 41 8X 198 460 110 262 IOO 128 iO 28 90 22 6 -6 80 -16 10 16 2 c 40 48 54 56 44 ac 1:14 134 8 2 be abc 42 92 40 50 (1 ) a b ab 5' 57. With 2" points. we can argue as follows. AN EXAMPLE WITH FOUR FACTORS These data come from an experiment conducted by A.8. An engineer who plans to conduct several 3 4 factorial experiments can make macros for 2 . 17. He was investigating the homogeneity of mixing a vinyl compound used in manufacturing tires. Its variance is (}'2/8 + (}'~/8 = (}'2/4. 2 . and that those two factors arc surely important. the speed of the motor.293 AN EXAMPLE WITH FOUR FACTORS Table 17. C. and D. Column 6 is made by squaring the entries in column 4 and dividing them by 8.8. an effect is the difference between two averages of eight points each. A sample of the compound is placed in a vat and mixed with heads driven by a motor.7.-2. and 2-'. With 16 points. and column 6 the sums of squares. the weight of the sample. C. we should divide the contrasts by 2" -\ and their squares by 2". We can argue that an effect will not be significant unless the absolute value of its estimate is greater than twice its standard deviation: ()' leffectl > V2"-*. column 5 the estimated effects. It is also clear that C is unimportant. whether the sample is preheated (no = low. The four factors are A. which is divided by 8.1.. It is clear that the main effects of Band D are very much bigger than the other effects.5 25 7 4 1 64 16 20 -4 4 61 26450 Mean 1250 A 98 B 2 AB 512 C 800 AC 32 Be 32 ABC . What about A and AB? If we know u. The response recorded is the torque of the motor when the heads are running at a constant speed. Column 1 contains the data.- This is sometimes called the minimum hurdle. yes = high). column 4 the contrasts. . the variance of an effect is u 2/2. Church (1966) at the General Tire and Rubber Company. the temperature of the vat. these are the effects. these are the sums of squares for the effects. In the general case. therefore. This gives a pooled sum of squares 5625 + 900 + 2756 + 16.50 765.00 -26. as a working rule.75 -58.00 ---33.75 15. too. interactions with three or more factors arc negligiblc. A traditional approach is to decide that. must bc due only to noise.00 25. Yates' Algorithm for the Church Data OBS (1) a h ab c ac bc ahc d ad bd abd cd aed bed abed DATA 1460 1345 990 1075 1455 1330 1000 1065 2235 2070 1795 1730 2540 2100 1700 1670 CONTRAST 25560 -790 -3510 900 160 -270 -470 300 6120 -610 -590 120 200 -210 -510 320 EFFECT 3195.937. and come very near to declaring A to be significant. and ABCD. Their sums of squares.00 -98. one could then declare as significant any effect whose sum of squares exceeds 6.75 37.61. we would pool the sums of squares for ABC. and AB to be significant. The 5% value for F( 1. ABO.8. In the absence of such duplicates. whence 52 = 6387.294 OESIGN OF EXPERIMENTS: FACTORIAL EXPERIMENTS AT lWO LEVELS Table 17. we have a problem. This is what happens when we use a regression model that has terms only for thc main cffects and the two-factor interactions. The following alternative procedure is sometimes suggested.1. engineers might set some arbitrary ad hoc level of smallness and pool all the effects that arc less than that standard. it can be estimated by taking duplicate observations at the points of the fraction or by taking scveral center points.75 112.75 40. points where each x.25 -63. .256 + 6400 = 31. and so we can pool them to give us a sum of squares for error. It is fraught with danger and should 1101 be used.9. i. BCD.00 SS 40832100 39006 770006 50625 1600 4556 13806 5625 2340900 23256 21756 900 2500 2756 16256 6400 RANK 16 12 A 14 B 13 AB 2 C 5 8 6 15 D 11 10 1 3 4 9 7 17.61 x 6387 = 42218.50 20.00 -76.75 -438.25 -73.5) is 6.. E. In the example in the previous section. ACD. We would thus declare D.c . is midway hetween -1 and + 1. THE VARIANCE If (T is unknown. ESTIMATIN(. Instead of restricting themselves to using the higher interactions for the error term. This has lately given way to normal plots.59 x 3477 = 19436. 7) is 5. /. when it hecame apparent that it led engineers to declare too many false significant effects. ABC.337 with 7 d. the engineer might decide that W. and BD! The risk in this procedure is that we are using a hiased estimate. CD.000. The bias makes the estimate too low. The 5% value of F( I. D.1. beginning with Cuthhert Daniel's (1959) halfnormal plot. We introduce it when we construct the pooled sum of squares by loading it with the smallest terms that we can find. 17. The list contains six effects: A.2 ** * • * 2 0.4.9 was a standard procedure until 30 years ago. That would give a pool consisting of C.) Since then statisticians have turned more toward graphical procedures. totaling 24.2 * ** * B * -500 -250 o 250 500 Figure 17.0 * * * -1. AB.10. ABD.L and / = 3477.10. it led to the introduction of alternatives to Fisher's least significant difference. AD. Normal plot for Church data (2'). in which he plotted the absolute values of the effects. ACD.295 NORMAL PLOTS In this example. 750 . of the variance in the denominator of the F statistics.59 and we should declare any effect to be significant if its sum of squares exceeds 5.OOO is a fairly low value for a sum of squares in this group and pool all sums of squares that are less than 10. B. sec section 16. NORMAL PLOTS The procedure that was described in the first part of section 17. and ABCD. (A similar phenomenon occurs in comparing all the treatments in the one-way analysis of variance. AC.1 contrasts for the various effects are orthogonal to one another and are linear combinations of D * 1. Thc rationale of the normal plot is that the 2" . either positive or negative. the points ought to lie more or less on a straight line.fraction. It is obvious that there are two extreme contrasts (two significant effects): D on the positive side in the upper right of the plot. We could go ahead and fit a main-effects model for three factors: y = f30 + f3I Xj + f3 zx 1 + f33 x 3 + e. we would have 2" . 17. use those four degrees of freedom for extra factors. The normal scores arc then plotted against the actual contrasts. lie on a straight line. If the effects were all zero. Such designs are called fractional factorials. the plotted points for the others will still lie on a line. The only other way to add a third column that is orthogonal to the first two is to use . The three columns are orthogonal. If there are a few large effects. and abc. at a price. We used three of those degrees of freedom for the main effects of factors and the other four for interactions. each with variance 2n(J2 and zero expectation. THREE FACTORS IN FOUR RUNS The complete two-factor design has four runs with three degrees of freedom. we had eight observations and.X 1X 2 instead of +x\x 2 • Suppose that we add a new factor. The complete 27 factorial calls for 2 x 2 x 2 x 2 x 2 x 2 x 2 = 128 runs and we will run only eight-a one-sixteenth fraction.11.12.b 2 . They were introduced by Finney (1945). as in chapter five. The famous 7 4 industrial statistician Cuthbert Daniel would call it a 2 . 17.1 is a normal plot for the Church data. We added a third column to the design matrix to include the interaction. The estimate of f33 would be half the C contrast: abc + c . seven degrees of freedom. The other contrasts. hence.10. giving it the values in column 3 of J the matrix.296 DESIGN OF EXPERIMENTS: FACTORIAL EXPERIMENTS AT TWO LEVELS normal variables. I<'RACTIONS In the 2 3 experiment that we described earlier.1 independent normal random variables. or write 27// R-to denote a seven-factor experiment in eight runs. Figure 17. C. We get a 2 II4-a half replicate-consisting of the points c. a. b.a . but the points for the large effects will appear as outliers. Normal scores for the 15 contrasts are calculated by computer. including A and AB. Could we. and B on the negative side in the lower left. instead. and thus accommodate seven factors in eight runs? The answer is yes. If we were to plot them on normal probability paper. Since x. we are obtaining a reduction in the number of runs in our expcriment at a price-we cannot distinguish bctwccn main effects and intcractions. ab.:::: -X 1X 2 .. D. FRACTIONS WITH EIGHT RUNS Suppose that we were to take thc 2 3 factorial and add a fourth factor. B.x 2 = 0. Taguchi calls them lattices. == 1 at each point. Two factors in a design are orthogonal if the products of their coordinates sum to zero. we obtain the complcmentary half replicate: (l). or X I X 2X 3 = -]. we must be prepared to assume that it is due to the main effect A and not to the interaction BC. and AD with Be.1 A B C D (1) + + + + + + + + + + + + + + + + ad bd ab cd ae be abed . AC with BD.6. X 1X 2X 3X 4 = +1.l.X 2 = E x IX 3 = E xzX 3 = 0. Notice that the D column is identical with the ABC column in table 17. some say confounded. we can say that this fraction is defined by the relationship X 1X 2X 3 = 1. ac.13. The cost of using the fraction in order to save time and money is that the main effects arc aliased with two-factor intcractions. is orthogonal to A. factors A and B are orthogonal. the two-factor interactions are aliased in pairs. we are better off than we were in the 2 I case. by setting X 4 = X1XZXl or. Thus.13. If we see a large A contrast. E X . 17. However. It follows that XI = X 2X 3 ' and so A is aliased with Be. B is aliased with AC. where main effects were aliased with two-factor interactions. AB with CD. It is a cost that we mayor may not wish to pay.c. The design matrix for this half replicate is shown in table 17. If we set x. The fourth factor.297 FRACTIONS WITH EIGHT RUNS What price have we paid by doing this? We have set X3 = X I X 2 • The main effect of C is completely confused with the AB interaction. equivalently. A design Iikc this in which all the factors are orthogonal to each othcr. Table 17.13. and C. and all four main effects are aliased with highcr-order interactions that we arc preparcd to rc¥ard as negligible. i. In either case. if 2: x .2. In that sense. is callcd an orthogonal array. and be. The technical term for this is that Cis aliased with AB. Similarly. D. are called the defining contrasts. The main effect A is aliased with three interactions. equivalently. SEVEN FACTORS IN EIGHT RUNS We said earlier that we could obtain an orthogonal array for seven factors in eight runs.14. and ignoring the terms with more than two or A = BE "'. Table 17. by the relationship = ACF = BDF = 1== ABCD == ABE == CDE = ADG BCEF = ADEF = BCG = ACEG = BDEG = ABFG = CDFG = Multiplying through by factors.1 A efg adg C -I- + F G + + ·1 + + + + I- + I- + D + + + beg abcdefg E + Mf abe cde acf B + + + + + + + + + + + . F to AC.CF = DG. which define the fraction.298 OESIGN OF EXPERIMENTS: FACTORIAL EXPERIMENTS AT TWO LEVF. E to AB. for new factors. AB. We take the design matrix for the complete 2 3 factorial and use all four interaction columns.14. we have set These products. we see that XI EFG = ABCDEFG . That procedure gives us the design in table 17. Alternatively. and ABC. and G to Be.1.14. Taking all the products of these defining contrasts. BC. we see that we actually have a fraction defined by or. AC.LS 17. We have equated [) to ABC. we can write the four effects in an alias chain: A + BE + CF + DG . and G = . Suppose column 4 is chosen for C.299 THE L(S) 1.DG.ATneR There are six similar chains: B+ AE+ CG + DP C + AF + BG + DE D + AG + BP+ CE E -I- AB + CD + FG F+ AC+ BD+ EG G + AD + BC+ EF 17. He calls the design his L(8) lattice. and one can confirm from the interaction table.BC.15. E = -A8.14. Taguchi changes the order of the columns and writes 1 and 2 where we have written . any factor assigned to column 3 will be aliased with the AB interaction. THE U8) LATTICE If we multiply the columns for E. The entry in the interaction table in row (1) and column 2 is 3. for example.1 by -1. and the BC interaction is in column 6. The engineer who uses this lattice to design. one fraction is as good as the other. chooses four columns of the lattice for the factors. this says that the A B interaction appears in column 3. we maintain orthogonality and get another fraction in the same family.15. we have D = ABC.BE . That leaves column 7 free for the fourth factor.1 A B + + C + + F + + + G + + + + + + E + + + + + + + + + D + + + t + (1) adef bdeg uhfg edfg ueeg beef abed .15. and G in table 17. Table 17. which is shown in table 17.and +. that this Table 17.e. F= -AC.15.CF .2 shows both the design matrix for this lattice and also the interaction table. etc. The AC interaction is in column 5. and so that column should not be used for C. Suppose that column 1 is chosen for A and column 2 for B. The alias chains are changed to A . This time. an experiment with four factors. It is just that some like to have the base point (1) in their design. i. From a statistical point of view.1. F.. 1. No. I A 2 B 4 C 7 D 1 2 1 2 1 2 1 2 1 2 1 1 1 1 1 1 2 1 2 2 1 2 2 2 1 2 2 2 2 1 1 2 (1) cd bd be ad ac ab abed This is the design that we obtained before in table 17. Seven Factors in 8 Runs Design Matrix for L(8) Col. of adding extra factors by equating them to interactions in the basic factorial design. If the engineer goes through with that assignment of factors. No. 17.16.13. No. can be extended .2.14. THE L(l6) LATTICE The method that we used in section 17.J5.300 DESIGN OF EXPERIMENTS: FACTORIAL EXPERIMENTS AT TWO LEVELS Table 17. (1) I (3) 1 (5) 5 2 3 (6) 3 2 1 (7) assignment of factors to columns gives a design with main effects clear of interactions. 2 1 3 4 5 6 7 1 1 2 I 1 2 I 2 2 1 2 1 2 1 2 2 1 2 1 2 1 2 2 1 1 2 2 2 2 2 2 2 2 2 2 1 2 1 2 I t 2 2 I I 2 1 2 Interaction Table for L(8) 2 3 4 5 6 7 3 (2) 2 5 6 7 (4) 4 7 6 7 4 6 5 4 Col. the following design results: Col. 1. The L(16) latticc is shown in tablc 17. or nearly saturated.16.1. 17. I 2 3 (I) 3 (2) 2 1 (3) 4 5 5 4 6 7 7 6 (4) 1 (5) 6 7 7 6 5 5 4 2 3 3 2 (6) 1 (7) 4 8 9 \0 11 12 13 14 15 (8) 9 10 11 11 8 10 9 13 14 12 15 15 12 14 13 1 2 (9) 3 (10) 8 II 13 12 15 14 14 15 12 13 15 14 13 12 8 9 10 11 9 8 11 10 10 11 8 9 11 10 9 8 4 5 6 7 5 4 7 6 6 7 4 5 ( 11) 7 6 5 4 (12) 1 2 3 ( 13) 3 2 (14) I 10 9 8 15 14 13 12 3 2 1 to generate designs for 15 factors in 16 runs. the main effects are all aliased with . the design is said to be saturated. I 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 1 1 1 2 2 2 2 1 I 1 I 2 2 2 2 3 I 1 1 1 2 2 2 2 2 2 2 2 I 4 2 2 1 1 2 2 1 1 2 2 1 I 2 2 5 6 7 1 1 2 2 1 1 2 2 2 2 1 1 1 1 I 2 2 2 2 1 1 2 2 1 2 2 2 2 2 2 2 2 1 1 2 2 1 1 1 I 2 2 1 8 9 I 1 2 1 2 1 2 1 2 2 1 2 1 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 2 10 11 12 I I 2 1 2 2 1 2 1 1 2 1 2 2 1 2 2 1 2 2 1 2 1 2 1 2 2 1 1 2 2 1 1 2 2 1 1 I 2 1 1 2 1 2 \3 2 2 1 1 2 2 1 2 14 15 1 2 2 1 2 2 2 1 2 ] I 1 2 2 1 1 2 I 1 2 1 2 2 1 2 I 1 2 12 13 14 15 1 2 2 I 1 2 2 I 1 2 1 2 2 1 Interactio1l Table for L(16) Col. 31 factors in 32 runs. No. and so on. In designs that are saturated. SCREENING EXPERIMENTS When all seven columns of the L(8) lattice or all 15 columns of the L(16) lattice are used for factors. No.17.16. Fifteen Factors in 16 Runs Design Matrix for L(l6) 2 Col.301 SCREENING EXPERIMENTS Table 17. 7 1.4 8.6 5 6 Eff. etc. we can carry out Yates' al3 gorithm. The engineer assumes that only a few of them will be important.425.15. and wants to spot those important factors to be investigated more closely in a subsequent experiment.17.4 1.7 ]8.425 -I.2 19. In a screening experiment.AB. In this case. Table 17.9 4.725 1. and G = ABC.1.1. Example 17.15 35.BC.17.3 9.5 89.2 -2.6 21.7 7 SS 16.AC.2 19.1.1. 2 (1 ) a(deg) h(dfg) ab(ef) c(efg) ac(df) bc(de) abc( g) 17.) An LU. F = .175 0.] -16.9 4.275 -4.5 5.2 4. the F effect is . As a consequence. and C by setting D = .n5.2 24.3 7. The order of the columns and the assignment of factors to columns is slightly different from the design in table 17. The sums of squares are not affected by the changes of sign.6 21.4 26.6 -13.17.8 23. we have three choices.4 26.' effect is _.31 11.53 0. It is obtained from the 2~ factorial in A.5 2. E = .3 -3.075.70 2.9 43. To calculate the effects.225. AC = -D.) lattice is used in a screening experiment for seven factors. and the G effect is + ABC = 2.05 2.2 24.8 4 84.6 3 43.76 1.AB = +4. (An example of a screening experiment.6 40. we should carry out the algorithm for 2 in ABC and then equate D to .7 2.875 -0. B.7 11.AB. If we know what defining contrasts were used to ontain the fraction. The Data A H C D E + + F G (1) adeg bdfg abef cefg aed! bede abel{ + + + + + + + + + + + + + + + + + + + + + + + + + -+ Y 17.6 . the engineer considers a relatively large number of factors.8 23.225 1.He = -1.302 DESIGN OF EXPERIMENTS: fACTORIAL EXPERIMENTS AT TWO LEVELS several interactions. The engineer argues that the few important factors will stick up through the undergrowth and be detected.76 A B AB C AC BC ABC The D effect is .2 22. they are used mainly for screening experiments.8 45.2 22. the /:. The lattice and the data are shown in table 17. 173.8 -4.075 2.7 18. the engineer can use the method of section 17. The three-factor defining contrasts.18 +4.6 +21.8 23.2 +22.6 -21.8 -23.1 has resolution 1II because .7 18. as in section 17.6 -21.7 -18.1 is also a resolution IV design.7 + 16..4 -26.2 -19.6 -17.28 t2.2 +24.7 + 1.1 + 1l. 22. Calculating the Effects A B C D E F G --17.4 +26. such as X 1X 2X" drop out.2 -22.6 -17.18..2 +24.2 -22.2 -24. D.7 ··HU + \\). because their defining contrasts have at least four letters.9 -4.88 A +4.4 -26.2 122.8 +23.6 +21. This procedure is known as folding over.2 ~ 19.2 shows the calculations for the method of section 17. The combined fractions form a 16-point design.2. The design in table 17. Table 17.17.6 -17.6 +2U! +23.2 "24.4 +26.13.14.5.8 "23. the main effects are aliased only with higher-order interactions and arc "clear" of two-factor interactions.1 can also be obtained from table 17. and the 16 points satisfy the defining relations or 1= ABCD = BCEF = ADEF = ACEG = BDEG = ABFG = CDFG .3 +9.303 FOLDOVER DESIGNS Table 17.6.1 in the following way.7 + \8.2 +19.08 + 2.6 +21.22 D -0. fJ 17.2 -22. Take each point in the first design and change the levels of every factor. as does X 1X ZX\X 4X'X"X 7 .2 -24.4 +26.8 +23. In this larger fraction. Such fractions are said to have resolution IV.2 +19.6 -17.6 -1.72 -1. and G stand out as candidates for further investigation.2 + 24. We see that A.6 -17.9 -2.15.2 .14.2 -24.S Dividing by 4: --0.5 or regression.6 -21.2 -\9.7 +18.6 -17.8 +23.42 G ---------- If the defining contrasts are not known.17. and so on. The design in table 17.2 +22.IiIGNS The array in Table 17.4 -26.4 +26. FOLDOVER DE. Then e!g becomes abed.7 + 18.7 +HU -19. bd. 1946) B C [) E F G H J K 2 1 t 2 1 2 2 1 2 I 1 2 1 2 1 2 I I I 2 I 1 2 2 I I 2 1 2 I I I 2 J 2 2 2 2 1 2 2 2 2 2 1 2 2 1 2 2 2 1 2 2 A 1 2 2 2 I 2 1 2 2 2 2 2 1 2 1 2 2 1 2 1 1 2 2 I 2 2 2 2 I I 2 1 2 2 1 L 1 2 2 2 1 2 2 1 2 2 2 2 1 2 2 1 2 I 2 1 1 2 (1) abde/k bce/Kl acdfgh bdegilj cefhjk dfgjkl aeghkl abfhjl abcgjk bcdhkl acdej/ . abce. each twofactor interaction is aliased only with a three-factor interaction. cd. Plackett and J. acde. ce. abed. J. be:. ac. ae. This fraction can be obtained from the L(16) lattice by assigning A to column 1. D to column 8. C to column 4. RESOLUTION V FRACTIONS Thcre are also fractions with resolution Y. B to column 2.20. and E to column 15. the two parts combine to form a resolution ]V fraction. bcde. R.20.19. be.304 DESIGN OF EXPERIMENTS: FACTORIAL EXPERIMENTS AT TWO LEVELS some of its defining contrasts have only three letters. ad. abde.1. ab. Eleven Factors in 12 runs (PiackeU and Burman. developed an orthogonal Tahle 17. A useful example is the half replicate of the 2~ factorial defined by It consists of 16 points: (1). Each main effect is aliased only with a four-factor interaction. 17. two British statisticians who were engaged in war work. FRACTIONS WITH 12 POINTS During World War II. Burman (1946). P. In resolution III designs. Whenever a resolution III fraction is folded over. in which we can estimate all the main effects and all the two-factor interactions clear of one another. some main effects are aliased with two-factor interactions. de. 17. and E. 56.41.3. 29. Do any of the factors seem to be important? (1).2.1 .2 . bd. 47. abeg.68.2. abc. abd. and what we have said about aliasing does not apply to it. abed. 43.8.20. b. 27.0. e.4 17.4 . abef. bede. 64.0. bcd.8. 59. 43. adeg. 40. F= -CD.2. ad. a.8. (1). a.0. Calculate the contrasts for the main effects in this 24 factorial.4.6. 42 efl!.4 . Note that in some of them you will find that you reach different decisions about which factors are significant depending upon whether you rely upon normal plots or regression! 17.4.2. 20.2.6. 43. be. abe. EXERCISES These exercises involve 2 n -k factorials. 58. 46. b. see table 17. B. 92 c:de 9 beg 90 bdf 65 beef 15 bedg 3 . and setting E = . 47. The fraction was constructed by taking a full factorial in A. 26. cd.0.1. 54. 49.6. 40. This fractional factorial for seven factors in 16 runs was used in investigating a stitch bond process.7. The following data come from a 27 4 fraction. e. Which columns in the L(16) array correspond to the seven factors? (1) 95 def!!. ab. and C contrasts. 23. ae. 42. 26.2. 46. Which effects are important? 17. eefg. 24. bdfg.6 Calculate the A. 20. 48. Match each of the factors with a column in the L(8) array. The derivation of this array is quite different from the derivation of the other fractions that we have discussed.0 .0 17.3.BD. 44. d. Calculate the contrasts. These 12-point arrays should not be used when there are interactions. 49. and G = -ABCD. aedf.2 . The response measured was the percentage of defective stitches. be.5 . 59.6.7.305 EXERCISES array for 11 factors in 12 runs. ab. 67. These are the data from a 2 3 factorial experiment. B. Which effects are important'? (1).6. 46. C.4.1. ae. aed. abc. ahe. C is significant? 17. bc:. Find a 16-point fraction of a 2" experiment (find an assignment of the six factors to the columns of the L(16) lattice) that will enahle you to estimate A and the five interactions that contain A whether or not the other interactions are zero.2.5. 56. lib. 56. B. £lC. ae. a.3 .1. acd. 44. assuming that the only interactions that might exist are AB and AC. cdc. lIC.l . llhd.7 .7. 44. 4R. F = ACD.6. 47. c. b. Find an assignment of factors to columns that achieves this aim. 17.4. C.2. abee meets the requirements. so that the fraction was the half replicate defined by XIX2X~X4X5 = -1 and the experimental points became .306 DESIGN OF EXPERIMENTS: FACTORIAL EXPERIMENTS AT TWO LEVELS ag 94 adef 4 acf 32 acdeg 5 abe 95 abdfg abcefg 24 6 abed 3 Calculate the contrasts and identify the important factors. h. 46.0.6 Was there a significant difference between the two runnings? Which of thc factors A. 47. Which columns correspond to the interactions? Show that the fraction (1 ). 56. hc. (1).8.6 The experiment is repeated later and provided these data. abo 52.9. Which interactions are aliased with A? with f.5. 5t.S.? 17. bde.5 .9. be. (1). 46. 47.ABCD. An engineer carried out a 2 3 factorial and obtained the following set of data: a. If you want to estimate AR and CD instead of AB and AC with an L(S) array can you do it? 17.6. 46. Suppose that a fifth factor had been added to the example in table 17. An engineer wishes to look at five factors in an L(R) array.2. and G = ABD.8. 47.6. An engineer makes a 16-point fraction for seven factors by setting E == . Which points are in the fraction and which interactions are aliased with A? Which columns of the L( 16) lattice give you this fraction? 17. 46. Show that you can also estimate the other five main effects if the other interactions are zero. Will the plot that wc made in the text he changed? How could you handle the analysis using regression? What could you use for an error term? 17.4 14. they constitute factor D.12. He measured two responses: Y1 is the average thickness of the films on four wafers at each set of conditions.8 55. Consider the smaller factorial that consists of the eight points at high A.1 13.6 33.0 d ad bd abd cd aed bed abed 389 765 412 857 457 738 540 880 36. abce. These data are taken from an example that is quoted by Cornell (1981). First the patties were . de.11. A. Which are the important factors? Is there any significant difference between the two materials? 17. Each degree of freedom corresponds to either a main effect or an interaction. B is the temperature of the photoresist.8 50. It is clear in exercise 17.9 16. (1) a b ab c ae be abc 256 767 325 805 389 737 440 942 30. Then the photoresist material is applied. A is the ambient temperature. Suppose that we decide to operate at that level of A. The levels of A were 20.10 that the uniformity is greater at the higher level of the ambient temperature. ab. aede.6 65. t 11. cd. KTI 820 (low) and KTI 825 (high). McBeth (1989) investigates the effect of three temperatures on the thickness of the film of resist that is put on a wafer. bd. The wafer is first placed on a cold platc for 30 seconds to stahilize its temperature. Yz is a measure of the uniformity of the film.0 44. ae.6 15.7 54.5°C. Cis the temperature of the cold plate.10.6 42. ac.9 12. abed with the same data. You can make a normal plot in the usual way.4°C and 23. Three factors were involved. be. be. Arc any of the other factors significant'? Are any of your results shown in interactions? 17. He does not mention the levels of the other factors. abde. ce.307 EXERCISES (1 ). He used two different photoresist materials. Which lines in the Yates table now correspond to the main effect of E and to the interactions that contain E? Notice that you now have a saturated fraction.9 Answer the following questions for each of the two responses.0 32. ad. bede. Fish patties were made at the University of Florida from a fish called mullet. ae. 68. 77. 63. Thc data are a. (1). ab. 28. acd. Y is the coded rate constant as determined by a complex chemical equation. C. abcdefg.6. 17. 26. c. ede. 1. 120. ad. (I). (l).14. 3. 29 . 73. 53 . 61. 4.7. 81. 140. 47. the temperature.4 .13. abd. 86. a.59. bee. 67. abde.04.32. 65.0. and E. be. 66. Which other interactions are aliased with AF? Would you be content to attribute all that sum of squares to AF. 39 . bde. abed. Then they were cooked in an oven at either of two temperatures (factor B). adg. D. acde. ee. aeeg. bdf. each at two levels. hydrogen.4. adef. ahfg. abed. bcde. d.6. 2. edfg. beg. 1.0. b. ace. de. 2. 62. 42. 25 or 40 minutes. Fit a model to the data using Z = In(Y) as the response. II .13. 16.4.0.7 .84. Charles Hendrix of Union Carbide gives the following example of a fraction for seven factors in 16 runs. 32. 64. efg. e. 30. carbon monoxide. abc. b. acf. beef. The factors are A. and not to the others? Make up a two-way table to explain that interaction.7. 17 . cd. steam. The term involving A F is also significant. 66. ae. 29. for a given time (factor C). A petroleum engineer investigated dynamic control of the production of synthesis gas from underground coal gasification and determined an accurate rate equation for new catalysts that are resistant to sulfur poisoning by developing a simple model using five factors. abede. ac. 3. 140 . 78. 23 . 56. be. 32. What interactions are aliased with the main effects of the factors? Show that A and F are significant factors. 15 . abe. bcd. ab. be. ade. abee. The response recorded was a measure of the hardness of the texture of the patty. 36.6 .8. 67.01 . 17. 61 . 62. the design used was a 25 factorial.9. 55. 140. 36.308 DESIGN OF EXPERIMENTS: FACTORIAL EXPERIMENTS AT TWO LEVELS deep fried (factor A) for 25 or 40 seconds. Show that it is a foldover design. bd. ahe. 41. abc. B. e. bdeg. 4. 59.13. 38. carbon dioxide.65. 13.86. cde. Which are the important factors and interactions? Explain.7. 68. 81 . .2. 375 or 425 degrees. . ae.6.S. 29. and C are estimated from the last eight points. X6 == X 2 X 3 · Make two fits. X5::: XIX).1.I. D is estimated by the average of the two estimates from row 1 + row 3. ab.3. be. . e. We can estimate from the 12 points all four main effects and all six interactions. 23. -1.2. b. We cannot use Yates' algorithm for this problem. This design is a three-quarter replicate (John.2. 25. and then including the interactions. which takes the values: -1. + I. and from row 2 + row 3. AC. 20. and CD are estimated from the first and third rows of the data table.6. 24.15. 25. repeat your analysis. 25. -1.3 . D.9.1 . . 1(71). The values of the predictor variables are Xz X3 -I -J +1 +1 +1 -] +1 -1 +1 -1 +1 +1 XI -·1 +1 -1 -1 +1 --1 +] +1 -I -1 -I +1 -1 +1 --1 +1 -1 -1 +1 +1 --1 -} +1 +1 We can add the interactions in the usual way by letting X 4 ::: X 1X 2 . a. ae. and so on.16. first with only the terms for main effects. '--1. letting Xl be the coordinate of A.15 the responses in the third row (the duplicated portion) are higher than the responses in the first row. We can use multiple regression. 26. + I. 27.2. Suppose that the last four points had actually been run a week later using a different batch of raw material. B. BD. (1). + 1. Notice that AB.1 . be. AD. 18. Otto Dykstra (1959) gave an example of partial duplication of a factorial experiment. Include raw material as a fourth factor. abc. -}.309 EXERCISES 17. In the data table in exercise 17. and BC are estimated from only the first eight points.21. ab. + 1. A. -}. Are the interactions orthogonal to the main effects as they were in the L(8) array? Can you estimate the main effect of A by subtracting the average at low A from the average at high A? 17.3. (1). 24. cefhjk. bedg. What are those conditions? What would you estimate the delays to be in the two cases? 17. 4.82 . The engineer is interested in finding the hest case and the worst case. beefgl. 521 . 5. 483. 506. abcdefgh.08 .33 . bdeghj.310 DESIGN Of EXPERIMENTS: FACroRIAL EXPERIMENTS AT TWO LEVELS 17. acdejl.55 . 5. 586. 702. bceh. aceg. Which is it? What would you suggest as a value to replace it? 17. The following data come from an experiment using the Plackett and Burman (1946) L(12) lattice in table 17. cdef. in standard order: 138 145 149 157 154 166 159 158 124 153 151 148 158 150 152 150 One of the observations seems to be a bad observation. In a simulation study an engineer measures the delay time of a device. 484. 5. abdefk.87 . 6. and then folds the lattice over and makes eight more runs degh. abcgjk. 5. aefh. efgh. 6. 399.17. bcdhkl.18. befg.44 . acdfgh. . dfgjkl. 5.50 .20. aeghkl. 561. 490.02 . 7.55. 389 . Short times are desirahle. (1). adfg. 517 .48 . which are the sets of conditions that minimize and maximize the delay. abef~ 568.1. abde.48 .19. A 24 experiment has the following data. abgh. abfhjl. bdfh.475. 522. 452. 562.00 . 7. acdh. Calculate the contrasts for the 11 factors and decide which are important. 4. 5. The engineer begins with seven factors in eight runs using an L(8) lattice: (1). and 3. There is. The linear and quadratic contrasts for A arc given by 311 ..Statistical Methods in Engineering and Quality Assurance Peter W. Suppose. Some statisticians denote the three levels by -. for example. n factors each at three levels.2. and the residual and/or A x B interaction with four d. In the case of factors at three levels. x. 2. again. each with two degrees of freedom. and +.1.f. which may be a complete factorial or a fraction. a difference of opinion about notation. but because they have only two levels for each factor. The model for the 2" contains terms in X" x"x xllx. M. that we denote the sum of the observations at A =0. and 2. and latter cannot be included because each factor := + 1 at every point. Others denote them by 1. Inc CHAPTER EIGHTEEN Design of Experiments: Factorial Experiments at Several Levels 18. TWO FACTORS The two-factor experiment is an ordinary two-way layout with two factors. but not in x. each at three levels. we need at least three levels of each factor. Yates (1937) used 0. We discussed in chapter sixteen how the sum of squares for each factor can be divided into components with single degrees of freedom. and 2 by To. we will see an argument in favor of this approach when we consider fractionation. we turn to the 3" factorials.. 1. John Copyright © 1990 by John Wiley & Sons. they cannot handle quadratic models. 0.x" etc. we can break the sum of squares into linear and quadratic components. The " appears at only two levels. 18.1. The analysis of variance table gives a subdivision into three sums of squares: the main effects of A and B. In order to incorporate quadratic terms in a model or to estimate operating conditions that optimize a response. T l • and 1'2' respectively. and that there are N points in the design. Therefore. FACTORS WITH THREE LEVELS The 2" factorial designs are flexible and powerful. The engineers considered three responses: etch rate in A/min. 125. and D is the flow rate of the etching gas.8.3.3. 1.1. there is no error term and F tests are not available. With all the degrees of freedom used for the factors. and 1. X3 = 2 + 2 = 4. The third and fourth factors are introduced by setting X3 = XI + X z and x 4 = 2x\ + x 2 • The sums are then reduced mod(3). We can compute the averages and the sums of squares for the individual factors in the usual way. They are as Table 18. in nine runs. each at three levels.2. FRACTIONS: FOUR FACTORS IN NINE POINTS The 34 . We look at the first response.1 A B C D 0 0 0 I 1 0 1 2 0 1 2 0 0 1 2 1 2 0 2 I () 0 1 2 2 0 1 1 2 2 I 0 I 2 2 2 . 300.2 cm. Example 18. 160. uniformity. the other two are left as exercises for the rcader.1. 500.3. for example. There are four factors: A is the power applied to the cathode.312 DESIGN OF EXPERIMENTS: FACTORIAl.2 factorial is a one-ninth replicate of the complete design. and 200sccm. The nine points in the fraction are shown in table 18.3.0. 1987).3. EXPERIMENTS AT SEVERAL LEVELS Lin A = T z . C is the gap between the anode and cathode. when XI = x 2 = 2. etch rate.To and and their sums of squares by Lin A 3(Lin A)2 2N and Quad A (Quad A)2 2N and 18. which becomes 1. 275. and selectivity. The data are given in table 18. This example is taken from an experiment to optimize a nitride etch process on a single plasma etcher (Yin and JiIlie. and 325 watts. and 550 mTorr. B is the pressure in the reaction chamber. C Z F 6 . 0. We use the notation 3 4 //9 to denote that it is a fraction for four factors. 450. which means that the actual numbers are replaced by the remainder when they are divided by 3. 0 617.2.5 592.4.0 three factors arc (i) the temperature.5 04.55 .34 25.5 71.34 134.5 79.5 55.4.fXI I138.0 66.2 1787. = 1787.0 59.4.314 DESIGN OF EXPERIMENTS: FAC'TORlAL EXPERIMENTS AT SEVERAL LEVELS Table 18.30 1.0 65. The data are in table 18.5 72. The totals at the various levels of each factor.5 43. Analysis of Variance SOURCE A B C AB AC DF 2 Error TOTAL 8 26 Be 2 2 4 4 4 SS MS 1871.5 606.1.12 2.4 539.5 -11.2.5 72.0 50.5 78. (iii) the level of catalyst.5 75.0 60 gO 70 5 15 25 5 15 25 5 15 25 26. = 107.1 Temp (0C) HzOz (%) WO.81 39.0 278.1 0.66 F 36.5 5555.5 57.5 69.18 648.0 83.5 1.11 8.0 107.2.0 77.5 79. (%) 0.62 893.0 37.0 590. The analysis of variance table is table 18. The sums of squares for the three main effects are S" = 1871.74 5. 51.4 203.5 440.6. (ii) the concentration of hydrogen peroxide. W0 1 .4.0 77.O 248.79 222.0 43.6 889.06 1431.79 35.0 JOl:I. are as follows: A 503.5 67.5 68.5 54.5 R9.0 686. 5. together with the contrasts and the sums of squares for the linear and quadratic components.13 2.48 53.5 C 574.0 B 498.3 157.5 84.0.2 935.45 x 0:00 x=1 x=2 Lin Quad SS Lin SS Quad Table 18.5 g4. which is tungstic acid.96 105.43 213.0 62.5 676. Table 18.TORS IN 27 RUNS The linear and quadratic contrasts for a factor were introduced in section 16.43 = 4. We notice that the F test for the main effect of C is not significant at a = 5%. R) = 3. higher levels of acid arc better. in both cases. Temperature vs. 574. The spread in the totals for C is less than the spread for A or for B.0 199. if we fitted a parabola. which is F*( I. 0.0. Quad: Q = (sum at x = 0)" 2(sum at x = 1) + (sum at x = 2) = (-}.5 200.0 617.4.f. At 60 and 70 degrees.1 %. +2. we should find.5 503.3. the lowest level of acid is best.315 THREE PA(. and AuCo' The Lin A Lin C contrast represents a change in the linear trend in Lin C as we change levels of A. Table 18. (sum at x = 0) == (-. It seems at first that. There arc clearly quadratic trends in A and B. but nothing to indicate any curvature in the response.0 59.5 Another way to say this is to note that the linear contrast for C is positive at low temperature and negative at high temperature.3.0 -14.4. In this case. I t is the difference between the Lin C at high A and Lin C at low A.5 163.12/ 25.0 188.0 -1.46. -}). 1.0 590.5 -11. and yet there is a significant AC interaction. although not significant at a = 5%. Since there are nine observations at each level.13. the sums of squares are L 2/(2 x lJ) and Q2/(6 x lJ).0 229.0 1781. 590. on the average.5 43.0 -18. they are the following: linear: L == (sum at x == 2) _.5 574. which.5 2. the linear component alone has an F value of 105.L 0. However.5 3.0 686.0 230.5 592.5%. At 80 degrees.5 227. There is thus some evidence for a linear trend. exceeds the 10% value.5 \97. totals for the levels of A and C.f. that the maximum response occurs somewhere between x = 0 and x = 2.1 0.: A/C" ALCU' AUCL . For a factor with three levels. shows the interaction. .0 .0%. The totals for the three levels are 0. 617. How can we evaluate the significance of this switch in sign'! One way is to divide the four d.7. I). Tungstic Acid Totals Temperature W0 3 (%) 0. for the AC interaction into single d.5 1.5. there is no difference between the levels of tungstic acid.0 Sum Lin C Quad C 60 70 80 Sum 140. which is analogous to a resolution IV fraction. and is -59.84 6.23 5.0.0) = -36. with SS 484. 0 18.!H 5.45 5. Table IS. The Lin A Quad C and Quad A Quad C terms arc obtained in similar fashion from the quadratic contrasts at the three levels of C.44 5.84 5.0) + 2(3. 539. which was done in the following example.s.98 5.72 4.7 .27 A () J 2 2 2 2 0 1 2 A B C D Y A 0 0 1 2 0 I 2 0 1 2 1 2 0 2 0 1 0 1 2 6.5) .0. The sums of squares for the four components add up to the sum of squares for AC interaction. 2.63 6.(--18.31 5.46 5.0/(3 x 6 x 6) = 4.0)2/(3 x 2 x 6) Quad A Quad C = -( -14.1 B C D Y 0 0 0 0 0 0 0 0 0 0 0 0 1 2 0 0 1 2 1 2 0 2 0 1 5.85 5.24 7.12 5.5.18 6. FOUR FACTORS IN 27 RUNS 3 We can add a fourth factor to the complete 3 factorial and still have a design that has main effects clear of interactions.0) .316 DESIGN OF EXPERIMENTS: FACTORIAL EXPERIMENTS AT SEVERAL LEVELS Since there are three observations at each combination of levels of A and C.08 .75 2 2 2 2 2 2 2 2 2 () () 1 1 1 1 1 2 2 2 B C 0 () 0 I 1 1 2 2 2 D Y 0 2 1 2 U 0 1 () 2 2 6.(-14.3.3 x 2 x_') = 494. The lion's share is accounted for by the linear by linear component. We can.0 + 2(2. = 4.0) = 22.42 5. SS = (13.0.50 1 1 () J 1 2 0 2 . The Quad A Lin C contrast is obtained by applying ( -1. set X 4 =X 1 +X 2 +X 3 ' reduced (modulo 3).64 5.5) .5.82 6. Its sum of squares is (36. for example.83 5.0. -1) to the Lin C terms. the corresponding sum of squares is (77.0)2/(3 x 2 x 6) = 36.09 4.17 6.0)2 .69 5. Their values are the following: Lin A Quad C = (-1.29 5.01 6.(-1.0) = 13.1. 164 To minimize the propagation delay. all of the linear terms are.421 5.567 0.2.018 5.478 0.S. Analysis of Variance SS OF 2 2 2 2 0.896 9. It is not necessary to have a computer program that handles four-factor analyses of variance for this problem.525 5.114.158 SOURCE A B C 0 Lin 0.748 S2 Quad 0.028 0.567 0. and D from the total. on the second pass. F*(l. None of the quadratic terms is significant.158 MS 0. Because the factors arc orthogonal.089 0. we ignore C and D and do a two-way ANOYA for A and B.317 FOUR FACTORS IN 27 RUNS Table IS.S.2. The first pass gives the following table: SOURCE A B interaction error TOTAL OF 2 2 4 18 26 SS 0. The sum of squares for error is then obtained by subtracting the sums of squares for A. B.5. and so a contrast will be significant if its sum of squares exceeds 4.1.781 2 6. The averages for the various levels of the factors are as follows: A B C D 0 5.748 5.031 = 0.779 Error 18 0.043 0. the best conditions are the lowest level of A.783 0.773 1. This gives us the sums of squares for A and for B and the total.773 1.686 6. we can make two passes on a two-factor ANOYA program.623 5. C. the response was the propagation delay in nanoseconds in an integrated-circuit device.514 TOTAL 26 9. C. In this experiment.170 2.5.161 .541 I 5. and D and the highest level of B. On the first pass.826 5.262 2.783 5. we ignore A and B and obtain the sums of squares for C and D.525 5.318 6.41.0286 Example IS. and we are only interested in main effects. The analysis of variance table is table 18.038 6. The data are shown in table 18. 18) = 4.41s2 = 0.000 0.497 5.l. THIRTEEN FACTORS IN 27 RUNS 4 Earlier.1 1 2 2 () 2 0 1 (J 2 0 1 1 2 0 2 0 1 0 2 2 0 1 2 {) U 0 1 2 2 2 1 2 () 0 1 t 2 1 2 1 2 () (J 1 2 0 (J 2 0 I 2 0 0 1 2 1 () 2 1 0 2 1 0 1 2 0 0 2 0 1 2 2 0 1 2 2 0 1 1 0 2 0 1 1 2 2 0 1 1 2 2 0 1 1 2 2 (J 1 0 0 1 2 I 2 0 0 I 2 2 0 1 . The actual derivation of the array.1 can be obtained from this lattice by assigning A to column I. A similar procedure enables us to derive an orthogonal array for 13 factors.6. in 27 runs.1 I 2 1 1 0 1 I 1 1 2 0 1 {) {) 0 1 2 {) (J 0 1 2 () 0 0 2 2 2 1 1 2 2 2 2 1 2 0 1 2 1 2 2 2 {) (J () () t 1 0 1 0 I 2 2 2 1 2 1 2 0 1 2 () () 1 2 () U {) 0 2 2 () 2 2 2 2 1 1 () (I 1 1 0 t () t 1 1 1 t 2 2 1 1 2 2 2 2 0 0 2 2 2 2 2 0 0 0 2 2 2 2 2 1 2 2 2 2 2 () 0 0 () 1 I I 2 2 0 () 0 1 I t t 1 1 2 2 2 0 0 0 1 2 2 1 2 0 1 2 I) () 1 1 2 2 0 I 2 0 1 2 0 1 2 0 1 2 0 2 0 1 2 0 0 1 2 I () 2 1 () I 2 0 I 2 0 1 I 2 0 1 2 0 1 2 0 2 I 2 0 2 I) 1 1. and so is the error tcrm. and D to column 9.1.318 DESIGN OF EXPERIMENTS: FACT0l>IAL EXPERIMENTS AT SEVERAL LEVELS The interaction sum of squares is spurious. we obtained a 2 7 . The design used in example 18.1. which goes back to Fisher. B to column 2. 0 18. with three levels each. The interactions in the three-level Table 18.fraction hy starting with a 2' factorial and adding the four extra factors hy equating them to interactions. is more complicated than it was in the two-level case and will not be given here.6. The IA27) Lattice 2 3 4 5 6 7 R 9 10 11 12 13 U I) 0 0 0 () {) {) 1 1 0 1 1 1 0 2 t (J (J 1. C to column 5.5.6. This array is shown in table 18. but the sums of squares for A and B are what are needed together with the total. at two levels by running the first three of the triples at the low level of H and the other three at the high level of H. If. An example of the use of this lattice is given in the next chapter. See table 18.7. The first 12 columns of the lattice form an orthogonal 12 array for 3 . THE L(36) LATTICE The last of the standard lattices that we present is the L(36) lattice.1. and this lattice can be used for main effects only.8. We can say nothing about the interactions. In the table. which is given in table 18.8. We can add a thirteenth factor with three levels to the experiment by assigning four groups to level 0 of the new factor. We can add an eighth factor.1.1.7. we have also added two more factors at two levels. The L(l8) Lattice A 0 0 0 1 1 I 2 II C 0 I 0 1 2 0 I 2 1 2 2 0 2 0 2 0 I 2 () () 0 1 2 0 1 0 I 2 I 0 2 1 1 1 0 2 2 2 0 t t 2 2 0 1 2 2 D 0 1 2 2 E F G H 0 () 0 I () 1 2 2 2 2 1 2 0 0 I J 0 I () "'- 0 1 2 2 2 0 2 0 () I 2 1 2 2 0 I 1 2 () 2 '") () 1 1 2 0 0 0 0 0 0 0 0 1 1 0 I 2 0 0 2 0 0 1 2 (I 2 0 1 1 1 2 0 1 2 0 IS. THE L(l8) LATTICE This is a lattice for 37 /118. and four to level 2. . and there are no interaction tahles for them. IS. Table IS. The 36 points in that array form 12 groups of three points each. four to level I. The eighteen points fall into triples.319 THE L(36) LA'rnCE lattices form a complicated web.7. and thus obtained an orthogonal array for 2 3 3 13 1/36. 8. .2.1. One row of the L(12) lattice is added to each group of three points. The resulting array is an orthogonal design for 2113 12 / /36.8. 2 3 tJ lf36 0 0 0 1 2 1 2 I 0 I 0 I 0 I 0 1 0 1 0 1 2 2 2 2 2 2 2 0 1 2 0 0 1 2 0 1 1 2 2 2 1 2 2 2 0 0 0 0 0 1 0 1 0 2 1 2 0 2 1 0 1 2 0 1 2 1 0 1 2 1 2 0 1 2 t 2 0 1 0 0 2 0 1 2 0 0 0 0 t 2 2 0 1 1 2 0 0 1 I 2 2 2 0 0 1 2 2 0 I t 1 0 1 2 0 1 2 t 2 0 I 2 t 0 2 1 2 0 2 0 I 2 0 I 2 2 0 0 I 0 1 2 0 I 2 2 t 0 t 0 2 0 2 2 2 0 1 I 2 0 2 2 0 0 I 2 0 2 1 2 0 1 I () 0 1 2 0 1 1 2 0 1 2 0 2 2 0 1 I 0 () 2 2 0 1 1 2 2 0 2 0 2 ) I J 2 0 () 2 () 0 0 0 0 1 t J 2 0 2 0 1 2 2 0 1 2 0 0 1 2 2 0 0 1 2 0 2 2 0 1 2 2 I 0 1 1 1 0 1 2 2 1 0 1 2 0 2 0 2 2 0 0 I 2 0 I 0 1 2 1 2 0 2 0 1 2 0 1 2 2 0 I 1 2 0 2 0 2 0 1 1 0 1 2 0 1 2 0 1 1 2 2 0 1 0 1 0 2 1 2 () 1 2 1 2 1 2 1 0 0 0 0 1 2 2 0 2 0 2 0 1 2 () 0 2 1 2 0 0 1 1 2 1 2 0 1 2 0 I 2 0 0 ) 2 2 2 I 0 1 2 1 2 0 0 0 1 2 2 1 2 0 2 0 0 1 () 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 2 0 2 0 1 0 0 2 1 0 0 1 2 0 0 1 2 1 1 I 1 2 I 0 I 2 0 0 1 1 1 1 1 I () 0 0 () () 0 2 2 2 t 2 0 0 0 0 0 1 2 2 2 0 1 2 2 I 2 2 0 0 0 0 1 2 1 I 0 0 0 1 2 t 0 0 0 0 0 1 0 0 0 2 2 2 I 0 0 0 2 0 0 0 0 0 0 2 2 2 We could also start with the first 12 columns and add a Plackett and Burman design.320 DESIGN OF EXPERIMENTS: FACTORIAL EXPERIMENTS AT SEVERAL LEVELS 1 Table 18. It is shown in table 18. In this section.S.9. A NONORTHOGONAI. we are focusing on fractional factorials that are orthogonal arrays. 2" 3 12 / /36 000 0 0 0 0 0 0 0 0 0 111111111111 2 2 2 2 2 222 2 222 o o 0 0 0 0 0 0 0 000 0 0 0 0 0 0 0 000 0 0 0 II 0 0 0 0 0 0 0 o o 000 0 2 222 1 1 1 1 2 2 2 2 000 0 222 2 0 0 0 0 1 00120220112 1 120 1 200 220 220 I 201 1 200 1 o o o o 202 022101 120102 1 0 0 2 ) 2 o 1 0 2) 0 o 0 1 2 1 0 0 2 1 221 0 120 2 0 2 0 0 2 20 022101102 o o o 2 0 o 2 0 2 1 () 2 2 2 0 o o o o o o 0 0 2 I 0 2 1 2 I 021 1 0 2 1 0 2 021 0 2 2 2 1 021 0 1 021 0 lJ o o () (l I) I) () 0 () o 202 1 1 1 2 0 () 2 0 000 () () 1 1 1 o o o 1 1 1 00010 o 0 0 1 0 o 0 0 ] 0 o 2 I ) () 0 2 1 2 0 2 102221102010 1 1 2 1 0222 21 0 00 1 0 0 0 2 () 2 2 1 2 t 1 2 1 1 I 0 I 002 022 o o 1 o ) o 020 ) 2 ) 2 0 1 2 () 1 0 1 2 0 2 0 ) 220 ) 22010120012 1 o o o (J () o o o o o () 0 0 () 0 0 000 0 o o o o o 0 o () o o () () 0 () 000 000 o o 1 1 1 I) () 2 1 0 0 0 2 2 1 () 000 0 () 000 I) 021 0 2 2 020 1 1 0 2 1 2 0 0 1 () 122 210201 21200 0 o o o () 000 000 1 1 220 1 0 () 2 2 1 1 2 2 () () 1 2 1 I 0 0 2 200112022 )0 1 o o 101 2 000 1 2 2 1 2 0 o J 0 000 1 2 1 000 o o o o o o 000 000 000 o o o o () 0 (J 1 (J (J (J 1 (J 0 0 o 1 000101 o 0 () 1 () 1 11000101 18. FRACTION Throughout the last three chapters of this book.321 A NONORTHOGONAL FRACTION Table IS. we give an example of a nonorthogonal fraction. We can only .2. a Box-Behnken design. parabolas.1. + 1.4 0 --1 0 +1 -1 0 -1 0 3R. Example 18. or minimizes. + 4. Several responses are observed.57x 2 + 1O.l 37.5 34.57x2X3 .0 45. This example comes from an etching study.9.5 () -) 13_9 () -1 -) 0 0 . The design and the observed responses are shown in table 18. y. The levels of the factors are taken as -).3. The contours for given values of yean be a set of concentric quadratic surfaces. Table J8.1.0 22.- 2.50X 1X3 = 10% are x 2 ' - 8.1 Ratio 0 -\ +1 Pulse Power Oxide Uniformity +1 0 +1 -1 21. the response.5 28.t-! +1 -I 0 -1 +1 -1 0 0 +1 +1 0 -1 0 0 +1 0 0 12.4.5x 3 . The three factors arc A.94xi .6 28.1.29x 1 . and the engineer fits a complete quadratic model: 222 Y = 130 + 13I X l + 132 X 2 + {33 X 3 + {3lJX J + (3n x 2 + f333 X 3 + 1312 X IX 2 + f3I3 X IX) + {323 X 2X 3 • The rcader who remembers analytic geometry will recall that this equation is a three-dimensional generalization of conic sections (ellipses.24x. and C. pulse.1. The center of the system is the set of conditions that maximizes. power. the ratio of gases used.0 0 (1 +I +1 0 The initial regression fit of the model to the data is y = 30.{.4 1l. and hyperbolas).O.27x J x 2 The only significant terms at a . which is called response surface methodology.2 12. we will look at oxide uniformity.0 43. 3" factorials without the vertices of the cube and with several centcr points. The most commonly used designs handle three or four factors. The Box-Behnken designs are. By fitting the . B. X3' and X2X]. 0.9.67x~ .0 25. in essence.1 25.9.1 .322 DESIGN Of EXPERIMENTS: FACTORIAL EXPERIMENTS AT SEVERAL LEVELS touch on the area represented by this example. The predicted responses at the four "corners.323 GRAECO-LATIN SQUARES reduced model.10. b. following tradition. Thus. IS.55 24. and the model reduces to a quadratic in x~ and x" which is a hyperbola. LATIN SQUARES A Latin S4uare is an arrangement of p letters in a p x p square in such a way that each letter appears exactly once in each row and each column.58x 2 + 1O.31. We denote the factors by P. and R instead of our usual A.1. and C because we are using a.3. They are the points of the L(9) array. The rows represent the levels of the first factor. when we put one on top of the . Q. and so on.8.4. GRAECO-LATIN SQUARES Suppose that we take two Latin squares of side p. and. the regression equation is y = 27. o 2 1 1.55 50. or a saddle surface.01 . the points in the first column are at level 0 of Q. It is an orthogonal array for three factors. 2 0 2.63 20. 2 0. Thc model clearly indicates interaction betwecn the two remaining factors. o IS. The points in the first row are at level 0 of P. table 18. the columns represent the levels of the second factor." all of which arc outside thc area in which the observations were taken. B. For this set of data. The two squares are said to be orthogonal if.46x 3 . arc as follows: XI X2 -1 -1 +1 +1 -1 +1 -\ +1 Estimate 12. and the letters the levels of the third factor.11. o 1. 2. each at p levels. we have a square a h c b C £l cab 3 This is a fraction for 3 in nine points. o 2 2 2 2. the nine points are o 0 O. usc Latin letters for the first s4uare and Greek letters for the second. the gas ratio is not important. For p = 3. I 0. with the last factor omitted.58x 2x) . 1. and c as the letters in the square. ).1. 5.3. which is the lattice of table 18. Algebraists have proved that one can always find a set of p . each at four levels. the Latin letters to the third. and of an array for 5/J in 25 runs. This.13.13. 2 1 0 2. We 5 will give an example of a square of side four that can be used for a 4 factorial in 16 runs.12. .13. 3. 0 1. . and the Greek letters to the fourth. 0 0 I 2 2 2 2 2 2. we can superimpose them to form a hyper-Graeco-Latin square. each at p levels. 3 (4:= 2-..1 mutually orthogonal squares when p is a prime number (2.2 is the orthogonal array for five facto~s. 1 1 1.. 1 2.. each Latin letter appears exactly once with each Greek letter. we can take the two squares abc b c a cab and u y f3 a f3 y y {3 a Superimposing the squares. On the other hand. I]. and so no Graeco-Latin square of side six exists. 1 0. For example. Table 18. Table 18.1 shows four mutually orthogonal Latin squares of side four. 2 1. I.13. we get the Graeco-Latin square. The square 4 that we have just shown provides a fraction for 3 in nine points: 0 0 1 0 2 0 0 0. Orthogonal Squares with h a d c a ~ l) b a d c 'Y {j d a b c Y d c b a a /3 p =4 y a ~ {j 0 ~ a 'Y A D B C B C D C A B D D A A C B .324 DESIGN OF EXPERIMENTS: l'AC'TORIAL EXPERIMENTS AT SEVERAL LEVELS other. in turn. 18. 18. ca a{3 c{3 ay ba We can now make an orthogonal array for four factors by letting the rows correspond to the levels of the first factor. in 16 runs that we derive from the three squares. the columns to the second. ) or a power of a prime number . 7..8 = 2. au bf3 cy b. 1 1 2 0. one cannot even find two orthogonal Latin squares of side six. provides an orthogonal array for five or more factors. HYPER-GRAECO-LATL"J SQUARES If we can find three or more squares that are orthogonal to each other. FIVE }'ACTORS AT FOUR LEVELS IN 16 RUNS Table 18. with p = 3.. 16 24.10 Table 18. the purpose of the experiment is merely to detect the one or two factors that clearly stand out as important.4. 4 5 in 16 Runs A B C J) E A B C D E A B C D E A 0 1 2 3 0 1 2 3 0 1 2 3 2 2 2 2 2 3 0 1 2 0 1 3 2 J 3 () 0 0 0 0 0 0 1 2 3 0 2 3 1 0 3 1 2 1 1 1 1 1 0 3 2 1 3 2 0 1 2 0 3 B C 3 1 3 2 3 3 3 J) E 3 3 3 2 1 0 1 0 2 0 2 1 The data in the example are shown in table 18.43 22.4.58 23.2.54 23.15 26. Analysis of Variance SOURCE A B C D E DF 3 3 3 3 3 SS 7.87 21. so that they may be the targets of further investigation.37 0. 18.08 MS 2. some statisticians suggest that this design should be used with only three factors so that there remain six d.t. The observations are written in the same order as table 18.13.68 0.2.29 1. and there is no error term for F tests. Those who use all five factors argue that at the screening stage.80 23. they are associated with two of the . FOUR-LEVEL FACTORS AND TWO-LEVEL FACTORS A factor with four levels ean he accommodated in a 2" factorial.325 FOUR-LEVEl.09 5. for the error term in the analysis of variance.3.86 3. This is a matter of choice. FACTORS AND lWO-LEVEL FACTORS Table 18.13 21.98 23. The analysis of variance table is table 18.03 It is obvious that D is a major factor and that A is also worthy of attention. are used for the factors.97 22. Data 20.13.13. Table 18.10 21.13. and that no formal test of significance is necessary at this stage.89 0. for the four-level factor.27 22.05 23. f.3.13. There are three d.14.28 15. For those reasons. The price of accommodating five factors is that all1S d.09 22.74 22.f.11 2.13. 2. 2 1 becomes 1. we write the levels of the four-level factor by 1.1. respectively. To be consistent with the notation used in the L(8) lattice in table 17.14. [J Example 18. 9. 13. Table 18. CD. but we can handle five four-level factors in 16 runs.2. The array is shown in table 18. We can follow chapter sixteen and associate the three degrees of freedom with orthogonal contrasts. II becomes the A contrast. and 2 2 becomes 3. 5.2. 7. We cannot extc!ld this idea to handle two factors at four levels in eight runs. 4. . AB. 3. 15. Contrasts for a }'our-Level Factor 2 3 1 11 -1 +1 -1 -J -1 +1 III t) -1 -1 4 tl +1 +1 two-level factors and their interaction. as shown in table 1S. we assign A. ABCD.326 DESIGN OF EXPERIMENTS: FACTORIAL EXPERIMENTS AT SEVERAL LEVELS Tahle 18. and D to columns 1. and 4. C. 1 2 becomes 2. ACD. ABC.2. and III the AB contrast. Example 18. The other columns arc available for the two-level factors. fl. If. in the L(16) lattice. 3. consequently. The first example is an orthogonal main-effects array for five factors with four at two levels each and the fifth at four levels. The combination 1 1 in the first two columns of L(S) (and. 2. We take the L(8) array and assign A and B to columns 1 and 2. the remaining columns will represent: 3. 6. their interaction is in column 3. 12. We demonstrate the rationale in the next paragraph and follow that by two examples.1. BD. BCD. AC. If we replace 1. BC.2 with the four-level factor in the first column.2. AD. 10. ABD. 2441 in Eight Runs I 1 2 2 3 3 4 4 1 2 1 1 2 2 I 2 1 2 2 1 2 L 2 1 1 2 2 1 1 2 2 1 2 2 1 2 1 1 2 11. and 4 by (1) b. 1 in column 3) becomes level 0.14. ab. respectively. a. contrast I becomes the B contrast.14.14.14. 14.1. as we saw in the previous section.14. and 8. 2. 8.8. 2.1 1.14. 0 Table IS. AB) 4. CD) R S T 5. 4 1 21 22. 14 (ABC. 17. 20.327 FOlJR-LEVEL FACTORS ANn TWO-LEVEL FACTORS Table 18. B. 15 (AC. 5 2 2 5 2 4 5 2 1 12. 14. 3. 1. 4 s in the l. 11. 10. ABD.13. ACD) P The array is shown in table 18.3.3 (A. Changing the levels to 0. 4 2 5 1 3 2 4 1 3 5 2 3 4 5 S 2 1 4 3 I 3 2 5 3 4 4 5 2 1 2 3 4 1 5 5 4 4 4 1 4 2 4 3 4 2 5 4 I 3 6. and 3 gives the array in table 18. lJ. 1 3 2 3 5 3 2 2 4 4 4 5 3 3 3 4 3 1 3 2 3 5 1 3 2 5 4 .IS. 16.) . ABCD) 7. 12 (c. I 1 2 3 5 4 2 2 5 S 5 4 5 1 5 3 5 2 13. 9. 4. 2 5 3 4 1 2 1 2 4 3 3 5 5 4 2 5 1 5 3 2 2 4 4 I 5 3 2 3 4 5 4 1 2 3 3 11. 25. 18. BCD) I. 2.3. 3 23.2.( 16) Array p Q R S T I I 1 2 2 3 4 I 4 3 3 4 2 3 4 3 4 1 2 4 3 I 2 3 4 4 3 J 2 2 4 3 2 3 4 I 1 1 2 2 2 2 2 3 4 2 3 3 3 3 4 4 4 4 I 2 3 4 2 3 4 2 1 2 1 4 3 3 4 1 2 4 3 I 2 We assign the five factors at four levels to the columns in the following way: Q 1. 9. 5.14. AD. 24. BD.2. 15. 11. It is left to the reader to fill in the details of the derivation. 10. 7. 13 (BC. EXERCISES IS. The first column gives the approximate value of the stress.1. The levels of the three factors. the second column gives the value from a more detailed analysis.0 201.8 230. Such an array is shown in table 18.13 can also be used to construct an orthogonal array for six factors.5 201.5 294.4 206.6 235.5 232.8 250.7 198. each at five levels. B. T.1 217.7 281.6 369.0 152.8 187.7 216.8 177. Carry out the analysis of variance.5 269.5 295. 1.1 235. Oka (1987) report the use of a 3 x 3 x 3 factorial to investigate differences between two methods of estimating membrane stress.3 153. Oikawa and 1'.6 293.9 250.1 245. uniformity and selectivity. The authors were interested in the ratio of Col.6 231.0 286.1. An application appears in the exercises.5 234.3 236.1.15. FACTORS AT FIVE LEVELS The method of section 18.6 154.6 278. are given in the last three columns.15.3.3 175.328 DESIGN OF EXPERIMENTS: FACTORIAL EXPERIMENTS AT SEVERAL LEVELS 18. Repeat the analysis for the other two responses. in 25 runs.0 208.6 304.8 170.2.6 159.1 369.3 158.0 174.7 171. and C.8 267.5 349. A.5 186.5 185.8 227.2 1 1 1 1 1 1 1 1 1 1 1 1 I 2 2 2 3 3 3 1 1 1 2 2 2 2 2 2 2 2 2 2 3 2 3 2 3 3 1 3 1 3 I 3 2 3 2 3 2 3 3 3 3 3 3 2 3 1 2 3 1 2 3 1 2 3 J 2 3 t 2 3 J 2 3 1 2 3 1 2 3 . In a paper.3 176. 2/Co!.5 278. IS. 191.8 167.0 169. we analyzed the response etch rate in the experiment by Yin and lillie. In example 18.3 154.7 213.1 361.1 246.1 251.9 204. be due to a high level of noise (note the value of s).x. + {32x 2 + {3Jx. This is overkill. Show that Xp X 2 . XI -·1 13. 15. 10)/10. 0 0 12.3. 10. Xl z: (power _. 7. +1 -1 14. 0 0 II. XI 0 -1 +1 5. there are several terms whose coefficients have small I-values. X2 X3 Y +1 -} 0 -1 0 0 0 +1 +1 16 48 43 68 0 30 Xl X3 Y -1 +1 +1 +1 0 7 20 +1 -1 0 0 -I 0 +1 68 0 0 35 26 XI 6. 3. and X3 are uncorrclated.3. I. -1 -1 0 0 21 28 9. .) Fit the complete quadratic model to the data in exercise 18. X2 X3 Y +1 +1 0 0 -I 36 12 -1 14 8. the first factor is the power in Mtorr at levels 10. 18. It is adapted from an experiment carried out in the development of semiconductors. Print out the observed values. and residuals. 15. () The values taken by the X variables are coded values for the levels taken by the factors.329 EXERCISES 18. Fit the model y = {3o + {3. (Continuation of exercise 18. 20. predicted values.3. This an important point about the design. Drop one of the two terms with the smallest t-value and rerun the regression.1 . Continue this procedure until you have a model in which every coefficient has a significant t-value. There arc a few residuals that seem to be large-but that could. This is an example of a Box-Benkhen design with three factors and three center points for a total of 15. which might lead you to conclude that the model gave a very good fit to the data. perhaps. 2.. There arc substantial I-values for each of the predictors and a large value of R2. Thus.4. 4. 471 0. (Continuation of exercise J8. (1987) also report the data for an L(9) array for three factors. The response is the outlet temperature minus 360 degrees. They are related to components of the circuit.326 4. Notice that there is a marked reduction in the size of the residuals in the second fit. 18.7. Y. Notice that the estimated standard deviation.330 DESIGN OF EXPERIMENTS: FACTORIAL EXPERIMENTS AT SEVERAL LEVELS 18.14.164 1.076 3.464 Which are the important factors? Which levels of those factors give the largest voltage? 18.03 85.254. The observations in the order that is given in the table are as follows: 0.153 1.82 Which are the important factors? Do they exhibit linear or quadratic trends? Are these conclusions dominated by the single large response.150 0.189 2.3.90 125. 18.97 A 1 2 B C 2 1 1 1 2 3 2 2 3 2 Y 58.005 5.18 67.4). Fit the model to the data of exercise 18.560 0. Wu et al.177 3. The response.248 1.3.64 81.850 0. A 1 1 2 3 B C 1 3 I 2 3 3 3 3 3 I 2 Y 54.260 0. The array that was used is the first five columns of the array in table 18. The following data set comes from a simulation study of an output transformerless (OTL) push-pull circuit reported by Wu et a!.78 14.71 19. The experiment involves a heat exchanger.1.685 0. (1987). s. is the midpoint voltage of the system.423 1. There arc five factors.522 7.5. has dropped to less than half its earlier value.066 8.704 6.391 2.917 1. each at five levels.64? Would your conclusions be changed if you used as the response In( Y) or VY? If the objective was to minimize Y.326 0. particularly for the two points with the highest responses (y = 68).3 and compare the fit to the results of exercise 18.6. 2. would the engineer be justified in recommending that A should be at level 3 and that the levels of the other factors are immaterial? .085 3. Work out the details of the derivation of the lattice array for 4~ in table 18. Compare the predicted values and the residuals in the two fits. 125.15.8.085 1.25 19. or can we improve our husbandry to coax a higher yield from our land? They have had enormous success in achieving their aims. Inc CHAPTER NINETEEN Taguchi and Orthogonal Arrays 19. One of the leaders in the field is Japanese engineer and statistician G. Sequences of well-designed experiments lead the engineer to the operating conditions that optimize the yield and also map the region around that maximum so that the operator can adjust for such departures from optimum as thc gradual loss of potency of a catalyst. Taguchi.1. worked hard at solving it. An example of these designs was given in the previous chapter. In earlier chapters. Many engineers arc now using these methods. which were developed by Box and others in the 1950s and 1960s. The agronomist wants to compare different varieties of wheat and different levels of fertilizer with the idea of optimizing the yield: can we develop a hybrid that will yield more bushels per acre. attention has usually been focused on a response such as the yield. He has been mentioned several times in previous chapters. 331 . This is the topic of response surface methods. H. This is another area in which statistics has played a major role. our attention has been attracted to a third fieldmanufacturing engineering-in which the objective is not to squeeze the extra bushel or the extra octane number. The Japanese.Statistical Methods in Engineering and Quality Assurance Peter W. In the last decade. INTRODUCTION In the traditional strategy of experimentation in engineering. Much of this chapter is based on his contributions. especially the series of experimental designs introduced by Box and his student D. and other sciences. but to produce thousands and thousands of widgets that are as dose to some specifications as possible. we painted a picture of the old days when the engineer had a process rolling along smoothly in a state of statistical control. agriculture. John Copyright © 1990 by John Wiley & Sons. Behnken. however. This is an area that was not of major concern in the West until recently. under the guidance of Deming. recognized the problem after World War II and. M. The chemical engineer wants to know which levels of pressure and temperature will optimize the amount of product coming out of the reactor. Although we cannot improve quality by inspection. her mean square error hecomcs smaller. what can you do? You can begin by rejecting the old simplistic notion that to be outside specification is bad.J-to)2. Taguchi calls this the system design phase.332 TAGUCHI AND ORTHOGONAL ARRAYS between the engineer and the factory gate was a squad of inspectors who. or she misses the green and is out of specification. the inspectors cannot alter that figure. in which case she is ready to go into the next stage of production. but to be within specification. When we talk about designing quality into the process. The sample median can be preferred to the sample mean as an estimator of E( Y) because it is robust against extreme observations (outliers). or robust design. we can. Wilcoxon's nonparametric rank test can be preferred to the . J-Lo. even by only a whisker. Two things must be said at this stage. those limits may bear little relationship to the actual specifications. as her game improves. who design new items and perhaps build a single prototype model. doing rework in the rough. within its natural limits. One might think of two golfers playing iron shots from the fairway.e. So. One aims for the green: she either meets the specification of being on the green. Replace it by the new ideas of process capability that were discussed in section 9. The other player is not content with the criterion of being either on or off the green. we would like to refine our manufacturing process to have zero bias and the smallest possible variance. and she works continually to reduce it even further. i. We arc more concerned with the next step. we mean carrying out research-and-development experiments to find the best operating conditions for manufacturing not just one item but thousands. they can reduce the percentage of substandard widgets that leave the plant. E(y . The second is that the inspectors can spot most of the bad batches and turn them back for rework or for scrap.15 in connection with the criteria C" and Cpk ' One of the major contributions of the Japanese approach is to make the engineer think in terms of aiming for a target value. and the target value. J-Lo' and work to reduce the mean square error. build quality by designing it into the production process. You cannot inspect quality into a process.. which is to take their prototype and make it manufacturable. Statisticians use the term robust to mean insensitive to departures from set conditions. putting. We want to control both of them. is satisfactory.. By their diligence. Too often the word design is used in a restricted sense to mean the work of a group of specialist engineers. and the variance. All they can do is decrease the fraction of that 20% that goes out to the customer. spotted all the bad widgets and allowed only good ones to leave the plant. sometimes far away physically and intellectually from the shop floor. The first is to remind the reader that even though a process can be rolling along smoothly under control. This is called off-line quality control. The mean square error is the sum of two parts: the square of the bias. but they cannot directly improve quality. and must. in theory. She takes the hole itself for her target and aims for the flag with the criterion of getting as close to the hole as possible. p. If you are producing 20% defective widgets. which is the difference between the actual process mean. In the parameter design stage. for location. In the present context. we conduct experiments to determine the "best" nominal settings (or design parameters) for these factors. Therefore we may save cost by using in the second group cheaper parts than we had originally projected. tolerance design. and the sample variance. i. the engineer finds the tolerances around the nominal settings that were determined by the parameter design. One of the varia hies that controls the output is the input ratio at a certain point in the circuit. and some other statistics. We may also use transformations of them to measure the dispersion. We may learn that some of the parts used in the process require very close tolerances. parameter design. We discuss parameter design in the next sections. which we mentioned earlier. THE THREE STAGES OF PROCESS DESIGN Taguchi suggests that there are three stages of design for a process: system design. called signal-to-noise ratios. . He advocates a systematic approach to achieve that aim. The Taguchi approach to robust design makes the engineer focus on reducing the variability of the process. which we call the design factors.4. The engineer talks about y as the response. The process involves several factors. A STRATEGY FOR PARAMETER DESIGN Suppose that we are manufacturing an electric circuit and that we are interested in the output of a certain stage of the circuit. TERMINOLOGY The generic observation has different labels in different fields. such as In(/). Taguchi has led the way in making engineers think of carrying out experiments in the manufacturing process to make the product robust to variation both in the parts that are used in the manufacture and in the environmental conditions in which the item is to be employed. the variable that is being observed is often called a performance characteristic. while others arc not so demanding.3.A STRATEGY FOR PARAMETER DESIGN 333 two-sample I-test because it is insensitive to departures from the assumption of a normal distribution.2. 19. which are advocated by Taguchi. 19. The agronomist usually talks about the yield of a plot. and tolerance design. for dispersion. The two most commonly used performance characteristics for a sample of data are the sample mean. 19. y. In the final stage. + 54 -+-- -------- I 52 __ r-----·-I I -------r-- 25 Ratio Figure 19. We will perform a factorial experiment to investigate the influence of these factors. 58 --~---------~-----I I I I . the output is increased to Y = 60. corresponds to a ratio. As part of this strategy. The target value.334 TAGUCHI AND ORTHOGONAL ARRAYS Figure 19. and then usc the others to adjust the performance characteristic to its target valuc. From it. We should use the latter group to reduce the variance as much as we can. If the ratio is increased to x=: 20. but the sensitivity of Y to variation in the ratio is decreased because of the nonlinearity of the response curve. the slope of the curve is guite steep at that point. . Yo = 55. 60 . but not the variance. Perhaps we can find another part in the circuit. Xu = 13. Taguchi makes extensive use of orthogonal arrays. changing the levels of some other factors affects the variance.. ratio. that we can vary to adjust the output back to Yo without losing the advantage of lower sensitivity to small changes in the ratio that we discovered at x = 20. such as a capacitor. As the ratio varies about the value x o' the output varies about Yn.5.1. 56 .4.4. we will learn that changing the levels of some factors affects the process mean. This suggests a design strategy for parameter design experiments.1 shows a plot of output against the ratio. Plot of output vs. Increasing A decreases .0 6\. or dispersion.. e.0 65.0 92.1:1 50. We run the complete factorial (all eight points) and we make four runs at each of the eight sets of conditions.5. the best levels of those factors. is modest.n 71. In this example.3 49. 19.. Many statisticians prefer to use the natural logarithm of S2. because it is more nearly normally distributed. we used the sample variance as the measure of variability.7 76. The combined analyses suggest that we should run the process at the low level of C to cut down on the variability.g. Suppose that we have three factors in our process and that we decide to look at two levels of each.0 65. We can analyze the data in two ways.5 8H.2 49. the average output voltage. we take as the response the means of the sets of five observations.5. That tells us which factors have a major influence on the location of the process.5. we take as the performance characteristic the estimated variance at each of the eight sets of conditions.2 76. or parameters. PARAMETER DESIGN EXPERIMENTS The purpose of the parameter design experiment is to determine which of the factors are important in the manufacturing process and to find the optimum set of working conditions.7 n.6 90.5 46. We find that A and B are the important factors.5 91.1 64.4 . SIGNAL-TO-NOISE RA TlOS In the example of the previous section.6 70. First. though positive.1 71. which is greater at the high level of C.l. changing the level of C produces a significant change in the variance.e. i.7 66. on the other hand.1 A B C -1 +1 -1 -1 -1 +1 +1 -1 -1 -1 -1 +1 +1 +1 +1 +1 -1 +1 -1 +1 -I -1 +1 +1 60. rather than S2 itself. We learn that changing the levels of A or B has little effect on the variance.6.0 71. The data are shown in table 19.3 60. We start with a simple example that illustrates the basic principles.5 91. and then adjust the levels of A and B to get the desired value of the process mean.y.3 9\.5 78.5 47. In the second analysis.7 46. we obtain the same result with either choice.0 60.5 76.335 SIGNAL-TO-NOISE RATIOS 19. increasing B increases f The effect of changing the level of C.2 91.8 47. and ji as the measure of the process Table 19.2 50. " and "on target is best. y it is often expressed as a percentage. s = -::. It can be shown that this ratio is mathematically equivalent to the logarithm of the variance of log( y). Pushing that a bit further. where n is the number of points involved. In all cases. Taguchi advocates several other measures. which is the ratio of the standard deviation of the response to its mean. The last one. It has been said that he has developed over 70 different such signal-to-noise measures! The three most commonly used of Taguchi's signal-to-noise ratios are these: (i) L y. is a transformation of the mean square error when the target value is zero.!] n' S when y is to attain a specific target value.336 TAGUCHI AND ORTHOGONAL ARRAYS mean. the objective of the engineer is to optimize the process by choosing the set of working conditions-the levels of the experimental factors-that makes the signal-to-noise ratio as large as possible. Why we should be considering the variance of log( y) rather than looking at y or log(y) is not clear. That is why some of them are prefaced with minus signs." The fact that the ratios are expressed in terms of 10gJO rather than natural logarithms is not important. It is clearly a transformation of the well-known coefficient of variation. which he calls signal-tonoise ratios. SNs = -10 loglO ( -n-) . (ii) when the objective is to make y as large as possible. It may well be more appropriate in some cases. For a set of data C. but not as a general procedure. They are given the slogans "smaller is better. the second is related to the mean square error for lly when its target value is zero.v." "larger is better. It can be argued that the first of them. SN s . however. when the objective is to make y as small as possible. . (iii) SN N = 10 log 10 [(i)2 . The use of the signal-to-noise ratios has been a matter of dispute. is more difficult to justify. perhaps the L( 18) lattice for six factors in 18 runs. In a letter to the editor of Quality Progress. The first step in planning the experiment is to select the design factors and to choose their levels. However. Suppose that there are N vertices in the inner array and n observations in each outer array. We can also calculate one or more signal-to-noise ratios. The design factors are factors that we can control-more or less. 19. SN N contains no mention of the target value. We consider two types of factor: the design factors that we have already discussed and noise factors. or do not choose to. Hunter (1987) has presentcd a strongly written and convincing argument against using this criterion. the total number of possibilities is only eight. the total number of possibilities is 729. Therefore. which makes it hard to justify as a measure of meeting that target. INNER AND OUTER ARRAYS The example in section 19. . We can decide that factor A will be a certain resistor with three levels: 80 ohms. we choose a fractional factorial-one of the orthogonal arrays. J. The overall experiment consists of running the outer array once at each point (vertex) of the inner array. The outer array is another orthogonal array. S. and 120 ohms. we want small values of s. y" and the overall variance of the If points. The analysis proceeds in two steps. and the pollen count.5 was simple in two ways. each at three levels. the humidity. the experimental design is more complicated. We could also treat as a noise factor the resistance of the component that was mentioned earlier by perturbing it by a few ohms above and below its nominal values. Examples of these are the temperature in the work place. which exceeds most budgets. we calculate two performance characteristics of the outcr array: the average observation. We obtained our values of the variance by making four observations at each of the eight vertices of the cube.337 INNER AND OUTER ARRAYS Furthermore. 100 ohms. There were only three design factors. Maximizing the quantity y2// places a premium on points with large values of y and small values of s. This lattice for the design factors is called the inner array. but we do not necessarily want to have y as large as possible. representing environmental noise. with six design factors. They arc the knobs on the black box that is our process. With more factors and more levels. These arc the noise factors. each at two levels. That is the sort of thing that might happen if we used cheaper parts. and it has also been the subject of articles hy Box and others. this time in the noise factors.7. control. Certainly. giving a total of Nn observations in the complete experiment. and so we were able to run the complete factorial experiment. This can turn out to be an awful lot of observations! At each vertex. With three design factors at two levels each. There are other factors that affect the performance of the final product that we either cannot.. s. 21 45. we will put the dispersion factors at the levels that minimize the variability.77 42.37 39. and we make an analysis for each.30 42.1 shows the design of the inner array and the summary statistics.1.4237 8.R222 9.1985 9. We note that the linear contrasts for A and both the linear and quadratic contrasts for Bare Table 19.6670 9.1668 11. If that is so.65 41. The outer arrays consist of nine points each. we can use analysis of variance.2546 22.82 43. S2. but do not affect the location.3092 22. whieh we can call the signal factors.8275 7.6S59 11.7165 24.9051 23.S0 45.1639 9.8.5.759R 23. We begin with an analysis of the data for the response y. s~.28 46. that affect y" but do not alter the dispersion.1. The factors are assigned to the first five columns of the lattice.9526 22.3032 7. Table 19.86~ 10.8727 23. 73S2 23.4675 21.404S 22. y.5696 12. If nature is kind.8.5976 21. the second with C and D.91 45. It is decided that these factors be considered at three levels each and that the inner array consist of the L( 18) lattice described in section 18.825S 2 2 0 0 0 1 1 1 2 2 2 () 2 0 I 2 1 2 0 1 2 I () (I 2 0 1 I 2 0 1 2 2 2 0 2 0 1 0 1 2 1 2 0 1 1 2 J 2 0 2 1 () () 2 2 n I 2 0 1 7. we will find some factors. and some other factors that control the dispersion. and use the adjustment factors to get y back on target.07 46. Because the array is orthogonal.338 TAGUCHl AND ORTHOGONAL ARRAYS Now we have the inner array with two performance characteristics observed at each vertex.6673 8.3950 9.67 10.7.1767 21. and the third with E and one of the others.1:1870 JO.3150 22. The easiest way is to make three runs on a two-way analysis of variance program.47 46.5.3533 lO.67 39.7558 23. just as in section 19.08 41.15 46.3058 23. following the pattern of example 18.51 43.l:!302 1:1.9163 22. AN EXAMPLE OF A PARAMETER DESIGN EXPERIMENT We consider an example in which there are five factors that are possible control factors.82 39.8245 22.1--one with factors A and B. 19. The L(18) Lattice for the Inner Array A B C D E Y S2 SNN 0 0 0 1 1 I 2 0 I 2 0 1 2 0 1 2 0 1 2 0 0 0 1 2 2 0 1 39.9026 22.8.05 46. and SN N' for the outer arrays carried out at each of the 18 vertices.3564 . We already saw that in the more obvious cases. namely.77 9.8. and leaves the choice of the level of C to other considerations.8.7 42.1 significant at the 5% level. and to adjust the level of y by altering A and B.9 42. to a certain extent.5 45.47 9. and.9 43. The reason for this is that the use of the factorial experiment. It is clear that one should try to operate at the middle level of D. It is not always really necessary to carry out the analysis of variance procedure.0.7H 9. the linear component of E arc important for controlling the variability. Averages for s 2 2 (] A B C D E 9.09 9. This tells us that A. E have a big influence upon y.R9 9. Table 19.16 9. shows that factor D and. or its orthogonal fractions.73 9. The usc of a signal-to-noise ratio Ii ke SN N makes life simpler for the engineer because there is only one criterion to be considered. the decision process is easier.7 2 43.7 4l. /. enahles us to estimate effects by averages. and that the linear component of E would be significant if we took a = 10%. Averages for j o A B C [) E 40.2 that factors A and B. table 19.10 9.2. we should certainly move to the middle level of E because that brings us ncar to the target with a further reduction in variability. One chooses the "best" values for each of the factors A and B. It is clear from an examination of table 19.2. Analyzing the variance response.70 8.71 9.4. The averages are shown in table 19. points in the same direction.3.65 9. but that Table 19. together with D = E = 1.5 43. If the target value for JL is about 43.5 43. to a lesser extent. The table of averages of In(i).8 42.7 43. E arc signal factors. one can come to the same conclusions from looking at a table of the averages at the various levels.8. and.8.8 43.3.24 9. If the objective is to maximize y. A = 2 and B = 1.8.32 11.8 43. possibly.8 44.8.339 AN EXAMPLE OF A PARAMETER DESIGN EXPERIMENT Table 19.92 9.H5 . B.9 44. A reasonable set of recommendations would be to run the process at the middle level of D. 88 23. We could decide that the optimum conditions for our process are those that maximize SN N for each factor.266 2.66 2 22.202 2.4. we relied on an orthogonal array-an orthogonal main-effects plan. we hope.263 2.287 2.266 2. they are A = 2 and B=C=D=E=1. In this cxample.286 2 2.340 TAGUeHl AND ORTHOGONAL ARRAYS Table 19. The averages for the ratio SN N are given in table 19.8.18 23.202 2.37 22.282 simplification may not always be realistic. the result is a set of operating conditions that appear to be the best.21 22. In deciding which of the factors should be varied to minimize the variance and which should be varied to adjust the response.74 23.279 2.5.94 22.9. we will.37 23.09 22.37 22. Perhaps we need to expand our inner array to make it a resolution IV fraction.40S 2. We should run the outer array at the set of values for the factors in the inner array that we are going to recommend.96 .8. We need some more data. manufacture product that is close to the target with a minimum of variability.241 2.219 2. In the analysis. The results could be disastrous. Averages of In(sl) ------ 0 A B C [) E 2. we assumed that there were no two-factor interactions.65 22.8. They are the best in the sense that if we run the process at those conditions. or a resolution V Table 19.06 22. But there may be a pitfall. 19. in which case we have to stop and think (and be thankful that we did not make a rash rccommendation that would stand as a memorial to us).5. Averages for SNN () A B C D £ 22. But we could have been wrong! This leads us to the last stage of thc experimental plan for parameter design-the confirmation experiment.227 2.163 2. THE CONFIRMATION EXPERIMENT When all this has been said and done.98 23.262 2.55 22. 3. In this chapter. in the noise array.11. and more costly. 19. but we might just as well continue to use low-cost parts for A. FINALE In the last three chapters. If the variability is too high. by the nature of their prices. we are going to have to use better parts in some places. we turn to tolerances. we have given a short introduction to the design of experiments. This time the noise factors are noise in the parts themselves. and save a lot of money. resistors for that item. We may find that A is critical and that the best way to reduce variability is to use very precise. relatively speaking. That happened in the example discussed in section 18. and calls for some serious thinking and statistical planning. This classical approach to experimentation has two . 19.FINALE 341 fraction. we find how much of the variability is associated with each factor. a factor with levels. Until about 60 years ago. we have focused on the approach of Taguchi. experimenters were taught to change one variable at a time. We run a second outer array centered on the optimum parameter conditions. If the outer array on our confirmation experiment has shown a satisfactory level of variability. where we have to make our further improvements. Taguchi recommends that the parameter design experiment be done with low-price components that have. which has its strengths and its weaknesses. The results could confirm our earlier conclusions. we can go into production with the low-cost parts and a corresponding savings in production costs. some variability. and not with precision parts in a white coat laboratory setting. it becomes.10. But which parts need to be improved? A tolerance design experiment is an experiment to find out where the variability occurs and where the adjustments should be made. we may find that "wiggling" the level of A does not contribute much to the variability of the final product. say. If factor A is a resistor at a nominal level of 100 ohms. Not only will improving the precision of A not help us much. When the array has been run. 100 ± 2. hence. and. on the other hand. We have our optimum conditions. We also have a basis for the next stage of the investigationtolerance analysis. This can become costly in terms of both money and time. TOLERANCE DESIGN Having settled on the parameter values of our process. and the outer array at the confirmation run provides some data about the area around that set of conditions where we will be working. the engineer runs the base point. and of the gains that Japanese have made through intensive use of statistical methods. and c. With three factors at two levels and four runs.342 TAGUeHI AND ORTHOGONAL ARRAYS drawbacks: inefficiency and the inability to detect interactions. ac. particularly George Box. It is still very good reading for both engineers and statisticians. Furthermore. was developed at Rothamsted Agricultural Station by Fisher and Yates in the 1930s. Fractional factorials were developed by Finney. and be. At the end of World War II. and then investigates A by keeping Band C fixed and varying A alone. (1). we arc able to compare averages-the average of all the points at high A minus the average of all the points at low A. several factors simultaneously. It is still difficult to find room for statistics courses in most engineering curricula. or doubling of efficiency. b. Because the designs arc orthogonal. the estimates in the one-at-a-time experiment are correlated.(I) with variance 2u 2• The 2 3 1 half replicate also has four points. The halving of the variance. The main reason has been that. mainly in-plant short courses. Many of them are hearing about these things for the first time in the framework of the Taguchi methodology. American engineers arc learning about designed experiments through programs arranged by their companies. This property is sometimes called the hidden replication property of factorial designs. The growth in the usc of these ideas in engineering in the United States was slow. also at Rothamsted. it was the standard book from which most industrial statisticians learned about design of experiments. The ideas of multifactor experiments took a while to move from agriculture to engineering. ab. and the statistical independence of the estimates were achieved by considering. The methods that they were using were gathered in a book (Design and Analysis of Industrial Experiments. particularly in the semiconductor industry. edited by O. Davies) first published in 1954. some of the more zealous disciples of Taguchi get overheated and proclaim that all the ideas of multifactor experimentation arc Japanese. and C in the factorial experiment arc uncorrelated. Multifactor experimentation. Unfortunately.be -. Now that practicing engineers are aware. For a long time. unlike the Japanese. of the importance of statistics. during World War 11. American engineering education has not included mueh emphasis on statistical methodology. were llsing them in chemical engineering research. The strength of these designs lies in their efficiency and their ability to detect interactions. and were invented by Taguchi . The estimate of the A effect is a .(1)] /2 with variance (T2. scientists at the laboratories of Imperial Chemicals Industries (ICI) in Great Britain. a. we have two choices. as we know it. we have the average of two at high A and two at low A. B. but now the A effect is estimated by [ab + ae . TIle efficiency comes from the balance. the estimates of A. it followed and built upon Fisher's earlier introduction of the procedure for the analysis of variance. instead of one point at high A minus one point at low A. and varying. L. The classical one-at-a-time design consists of four points: (1). 5 9. But some caveats must be mentioned again. newspapers conveyed the absurd allegation that the western statisticians are still advocating one-at-atime experiments and are unaware of the principles of multifactor experimentation.0 3. we have outlined a systematic approach to experimentation. 19.0 8.343 EXERCISES himself.5 4.5 3.5.5 5. and you need to watch out for them. This data set comes from Karlin (1987).0 0. That is one of the reasons for the confirmation experiment in section 19. If you do it properly.S. -1 +1 -I -} -1 !-l -} -] +} -1 4. Because they are orthogonal factorial experiments. but you will have no good handle on the variability. Another caveat concerns variability. One of the important uses of the analysis of variance is to obtain a valid estimate of the error. Carry nut the details of the analysis of the data set in tahJc 19.5 9. Because of the orthogonality. the results are easy to present and easy to understand. It is tempting to load up an L(32) lattice with 31 factors.1. An article in one of the major U. but remember: not only will the interactions (if any) he aliascd all over the place.2. It can too easily get out of hand and result in engineers blindly following the instructions in the handbook while their brains and engineering judgment are left hanging on the hook outside the door. you can hardly avoid learning something! Good luck! ORTHOGONALITY = BALANCE = SIMPLICITY = EFFICIENCY EXERCISES 19. He is reviewing a software package and he includes this example of a design with five factors in an L(8) lattice and five observations at each point. Some rabid disciples also go so far as to deny the existence of interactions. Setting up a multifactor experiment in an orthogonal array is a straightforward matter. including the calculation of all three signal-to-noise ratios. In the last three chapters. There really are interactions. It is an example for a Taguchi parameter design.5 . But there is more than that. which it claimed were developed outside the West.0 7. A policy of picking the winner among the means of three levels of it factor may get you into trouble if there is no significant difference. No matter what language you speak.9. and use orthogonal arrays. the important thing about welldesigned experiments is that they arc efficient and simple. the calculations are straightforward. Fisher had to discourage some of his agronomist colleagues from going off half-cocked and claiming that effects were significant when they were not. Such blind partisanship by zealots is regrettable.1. arguing that they do not exist in the real world. 44 7.88 7.44 7. Pigniatello and Ramberg (1985) reported an experiment on the heat treatment of leaf springs on trucks.56 7.81 7.25 7.78 8. There were three runs at each set of conditions.4.5 4.5 8.69 7.69 7. and E.09 7. Kackar and Shoemaker (1986) used designed experiments to improve an integrated-circuit fabrication process.18 7.56 7.59 7.44 7.5 7.. They were prepared to accept that BC.oo? 19.6 19.0 0.5 1.50 7.88 8.5 16. C.5 7.78 8.5 6. Y. The data follow: B 1 2 I 2 1 2 1 2 C BC D BD CD E I 2 1 2 2 1 1 1 1 1 2 2 1 2 2 J 1 2 2 2 1 1 I 1 1 2 2 1 1 2 1 1 2 2 1 1 2 1 2 1 2 1 2 2 2 2 2 . The engineers considered five factors in the manufacturing process: B.56 8.5 5. they measured the thickness of the epitaxial layer at five places on each of 14 wafers (70 measurements in all). The design of the experiment was an L(8) array in factors 8. susceptor rotation method. 7.0 4.00 8. heating time.18 7. is the free height of the spring.69 7. quench oil temperature. and 0. The response.15 7.(16 7. transfer time. Its target value is eight inches.0 5.62 7.' Which would you regard as the important factors? 19.0 4.50 7.0 17.0 9.12 7. hold-down time. which was repeated at both levels of O.56 7.25 7. The factors werc: A.56 7.50 7.62 7.5 2.5 Suppose that you were to use the signal-to-noise ratio (SN) I. C. For each run.5 4.50 7. and CD were the only interactions of interest.0 9. E. They investigated eight factors in an L(16) array. 10 10glO( ~:) . What levels of the factors would you recommend? Which factors would you adjust to minimize the variance? Having done that.50 7.to-noise ratio Z. which factors would you adjust to move the response to y = 8.59 They took as their signal.18 7.81 7.5 1. D. code .63 7. BD.5 9.94 7.0 13.88 7.5 15.56 7.81 Low 0 7.44 7.0 6. D.75 7.56 High 0 7.0 10.0 2.32 7.5 7.1 +1 0.3..50 7.5 1. B.5 11..5 7..88 7.75 7.344 lAGUCHI AND ORTHOGONAL ARRAYS -I +1 -) +1 -1 +1 -1 1 +1 +1 +1 +1 +1 +1 -1 -1 +1 +1 -1 +1 +1 -1 -1 +1 +1 -1 -1 +1 . high heat temperature. 878 14.880 2 2 111 212 14.037 13. deposition time.4369 -0.3131 --0.888 211 212 1 2 2 1 2 1 14.914 -0.fi270 1 2 I 2 2 1 I 2 1 1 I 1 2 2 2 2 1 2 14.4307 -0.757 -0. nozzle position. The data follow. 14 of the 32 bits failed during the first cycle. G. for example.821 14. at what levels of the four-level factors should you run your process? Does it matter? A 1 1 1 1 1 1 !JCDl:. HCI flow rate.5 . Which factors should you control to minimize the variance? Which factors should you then adjust to obtain a mean thickness of 14. deposition temperature.345 EXERCISES of wafers.3267 2 1 2 1 2 1 1 1 2 221 14. He had eight factors at two levels each and two factors at four levels. Phadke (1986) used an L(32) lattice in an experiment to compare the lives of router bits.5 17.907 13.843 14.5 0.5. He assigned the four-level factors to three columns of the lattice. see chapter 18. F HCI etch temp.1)21 13. D.5. If the bit failed during the eighteenth eydc. E.5 1 1 122 134 221 0. Which were the important factors? Given that large values of Yare desirable..3467 -0.8625 19.5 0. Y is the number of cycles to failure of the bit.972 -0. and H.932 13.165 1 1 2 13JW) 2 2 2 2 1 2 1 221 -1.5 2.032 1 2 1 1 1 1 212 2 2 2 B 1 I 1 1 1 1 1 2 2 2 2 222 1 2 1 122 2 1 14.8563 -0.4230 -0.5 0.5 0. Y was recorded as 17.6505 -1.5 JLm? A 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 log (s 2 ) -0.1989 -1.5 I 1 1 1 I 1 I I 1 1 4 3 223 1 I 1 224 2 2 1 222 2 4 3 2 1 1 2 4 122 212 2 1 2 1 J 1 2 I 2 1 2 2 122 17.4425 C D E F G H y 1 1 2 2 1 2 1 2 1 2 1 2 14.4969 14.415 14.6154 -0.1190 -0. The orthogonal array follows together with the values of y and i at each point. arsenic gas flow rate.: 1 1 1 1 F G 11 l Y 1 1 1 1 3. C.2292 -0. 5 n.5 3.s 17.5 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 0.5 0.5 n.5 1 3..5 0.5 2 1 3.5 0.5 2.5 17.5 .5 0.") 2 2 1 2 1 1 1 2 1 2 2 2 2 I 2 1 2 1 1 2 J 2 2 1 ") .5 2 2 2 2 3.5 3.5 0. 2 2 1 1 2 1 2 2 1 1 1 2 1 1 1 2 2 1 1 2 2 2 1 1 2..5 17..5 1 1 0.5 17.5 17.5 14.5 17.:.346 TAGUCHI AND ORTHOGONAL ARRAYS 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 1 2 2 2 2 1 1 1 1 2 2 2 1 3 2 1 2 1 1 1 1 4 1 1 1 I 2 3 1 2 4 3 1 2 4 3 2 1 2 3 4 4 1 3 1 2 4 2 4 3 I 2 3 2 2 2 4 2 2 2 2 2 2 2 1 2 1 1 I 3 4 2 2 2 2 1 1 1 4 3 1 2 4 2 1 3 3 1 2 2 1 2 2 1 2 1 1 1 2 1 . 1. OUo (1959). 5. Dodge. C (1959). New York. Homewood. (1943). N. vol." Industrial Quality Control. New York." Annals of Eugenics.. W.) (1956). 3-5. "The fractional replication of factorial arrangements. vol. H. (J9H6). and Torrey. Dodge. (Ed." Industrial Quality Control. A. vol. "A sampling inspection plan for continuous production. Alonzo C. Finney. Church. J. Draper. G. Hunter. and Wood. E. H. F. Hafner. vol. H. Cornell. 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"Identifying useful differences in a multiplehead machine. Hunter. 198-206. Ott. J. 1-12. T. 47-57. Lucas. UNIPUB. H. issue 2. 54-57. (1986). series B. (1973). M. New York. J. 7-8. R. 203-210.{7). E. D.. J. (195'». Introductioll 10 Stalistical Quality CUlilrol. E. vol. Plackett. D. Reading. (1984). W. W.22-25. Tukey. J. S." Techllometrics. Don W." Impcrial Bureau of Soil Science. (19tn). A. 119-130. "The design and analysis of factorial experiments. Massachusetts. Madison. Exploratory Data Analysis. "Developing blending models for gasoline and other mixtures. S. Yates. Wald. R. J." QUfllity Progress. Wu. vol. "Orthogonal design for process optimization amI its application in plasma etching. A.. Economic Control of Quality of Manufactured Product. 1. G." Techllometrics. S. (1931). (1977). and Jillie. Van Nostrand. 17. G. May. Snee. vol. Yin. Sequential Analysis. Dekker.REFERENCES 349 Schilling. "A quick compact two sample test to Duckworth's specifications. New York. F. Center for Quality and Productivity Improvement. Gerald Z. (1982). (1987). (1987).. Tukey. "An overview of acceptance control. C. Mao. vol." Solid State Technology. Harpenden. E. and Ma. Wiley. (1959). 23. 24. Acceptance Sampling in Quality Control." Report No. England. New York. 31-48. 127-132. (1947). F.. . "An investigation of OA-based methods for parameter design optimization. J. Shewhart. E. Schilling.. University of Wisconsin. (1937). New York. Addison-Wesley. F. W. 6368 .6026 .0:: en c.7549 .8810 .94179 .8686 .95728 .95637 .92647 .6 .8790 .94950 .92922 .5793 ..90658 .5636 .8508 .6064 .1 .7257 .91774 .04 .7486 .6772 .92364 . 1.6628 .93319 .7123 .7291 .5557 .90824 .8051 .5040 ..95449 .96638 .7157 .6915 ...7995 .5832 .5478 .92073 ..4 .8962 .~ ~ cr o Table I.96327 0.5 .97062 .5359 ..6591 .5319 .5279 .6985 .0 1.7764 .7054 .8599 .7454 .6217 . I::l 3"::J'"~ () ::::l ~ .91466 .5596 .6700 .5 1.7 1.7734 .2 1.8577 .96712 .7422 .00 .90147 .5120 .7673 .7324 .6879 .6736 .8888 .96164 .5987 .8389 1.3 .6 ...8980 .6406 ...94062 ..5517 .02 .59\0 .93189 1.8708 .97441 .6443 .5239 .97128 .97615 .95053 .96856 .4 .8729 .93056 .8133 .97193 .8023 .96485 .7190 .93448 .8485 .97500 . ~ (0 (0 ~ S· o IT ~ "< I::l c.6255 .97558 .. .96080 .94845 .8212 .' i::" ".8106 .93574 .8925 ..761 I .7852 .90988 .9\ 149 .5675 .95352 .95994 .06 .96995 .05 .7703 .5199 .96562 .95254 .5000 .5753 .7967 .8869 .8621 .92785 ..97320 .8289 .91621 .6141 .8 1.8531 .09 .7881 .91309 .8907 .97257 .5948 .2 1.96246 .1 1. c.8944 .5438 .8078 .5714 .6331 .0 .91924 .79\0 .93699 .95907 .7823 .8264 . :: S ~ a "'- :::2: .8186 ..7088 .8413 .03 .6950 .8665 .07 .6480 .6808 .92220 ..97670 ..6103 .6179 .7794 .: :::: ::::l • ..08 .846\ .7357 .9 ~ .)I:: ro@-~ "<~~ R<>:::2::'" (fl' ~ .94738 .6293 .8438 .96926 .8365 .90320 .8554 .8238 .9 .94630 . The Normal Distribution*·t z .8749 .92507 .01 .7224 .94408 .6517 .8159 .96407 .7019 ..8997 .95154 .7580 .5160 .8315 .8849 .94295 . ~ ~ ~ s-~ '"3.5398 .7939 .8830 .3 1.96784 .7642 .6554 .7517 .o ~ ~ ~ @ :.5080 .8 .5871 .8643 .7 .6664 .7389 .93943 .95543 .97381 . UI ~ s (D rJ:J VJ S' .93822 .8770 .90490 .8340 .6844 .94520 .95818 . 93 8834 .9 4 1838 .9 3 4230 .98382 .9 2 7814 .92 0358 .9 2 2656 .9 3 8347 .9 4 2765 .9 2 8694 .9 4 2159 .98778 .94 6554 .94 7748 .9 3 2112 .9 2 7110 .9 .5 2.92 4766 .9 2 8511 .9 4 7090 .9 3 7493 .9 2 1802 .94 4988 . John Wiley and Sons.98169 .9 2 5855 .9 2 3053 .94 6833 .02 .9 3 6752 .9 30324 .94 3848 .9 3 7398 .9 24132 .9)5811 .97932 .98077 .9 3 2378 .01 .9 3 6505 .9·3327 .9 1 7365 .9 3 7842 .9 2 2240 .9 2 8893 .98214 .97882 .9 3 8282 .9 4 6092 .9·0039 .9 3 6982 .97831 .9 1 1576 .9 3 2886 .9 2 4915 .9 1 3431 .9 3 8527 .4 .9 3 4991 .00 .07 .9)5166 .9 2 7948 .92 1106 .9 2 7445 .9 2 7744 .0 2.9 27020 .9 1 8605 3.9 2 8650 . Hald.97778 .9 2 2857 .9 1 6093 .9 2 8193 .9 3 6103 . by A.9 .9 4 6964 .920863 .93 1553 . .9 4 0426 .9 3 4429 .92 7882 .9 3 8583 .9 2 4457 .9 3 8637 .08 .9 3 0957 .9 2 5604 .9 3 7299 .94 5926 .9 20097 .6 3. reproduced by permission of Professor A.9 2 4297 .Jbt -~ (!-~'Il .91 8559 .(N (II Table I (continued) N :z I .9 28134 .91 8012 .9 20613 .9 4 4094 .9 2 2451 .98870 .9 2 1344 .9 1 8462 .9 2 5060 .98300 .93 8215 .9 2 8736 .9 2 8930 .9 2 5339 .98537 .94 7843 • cJl(zl = - If .98809 .9 3 4623 .9 2 7599 .9 3 8879 4.9 1 1260 .9 1 3244 .9 3 6242 .9 2 8777 .9 3 8409 .97982 .98713 .9 3 8469 .0 3.9 2 8856 .2 3.98956 .94 6253 .1 3.9 4 5385 .9 3 7585 3.98574 .9 27282 .93 2636 .06 .1 2.9 3 5959 .9 2 2024 .9 3 8922 .94 6406 .9 2 3613 2.8 2.9 2 3790 .97725 .98679 .9 3 8146 .9 3 7759 . Hald and the pUblishers.9 3 7922 .9·1158 .9 3 8787 .2 2.9 3 7197 .9 3 7999 .98500 .94 7211 .9 2 7197 .7 2.98257 .9 2 8999 .9 4 3593 .9 3 6631 .8 3.98422 . New York (1952).9 3 5658 .9 2 5975 .~2468 .9 2 6928 .9 3 6376 .9·0799 .9 2 8817 .9 3 3810 .9)5499 .7 3.9 3 6869 .9 26636 .98341 .98745 .9 38689 .9 38074 .9 2 8411 .0 .9 2 8074 .9·1504 .9 1 5201 .98840 .9 4 7439 .9 3 7091 .9 3 5335 .9 4 5190 .9 3 8739 .9 1 8965 .9 3 1836 .5 3.9 3 7674 .4 .94 4558 .9 3 4810 .05 .9 2 8305 .9 2 5473 .9 2 6427 .9 30646 .3 2.92 3963 .94 7327 .9 4 5753 .9 1 8359 .94 6696 dx t From Statistical Tables and Formulas.94 4331 .9 38964 .98899 .9 2 6736 .9 28250 .9 2 7523 .94 4777 .03 .9 3 3590 .9 2 6207 .9 2 6833 .9 2 6533 .98645 .09 2.98124 .9 2 6319 .6 2.94 7546 .98030 .98610 98928 .9 4 5573 .9 34024 .94 3052 .9 2 5731 .9 3 3363 .9 24614 .98461 .9 4 7649 .98983 .3 3.9 3 3129 .9 2 7673 .04 . 694 .333 1.485 2.921 2. edited by E.282 1.677 .686 . values of t .776 31.717 1.052 2.05 .787 2.182 2.539 2.699 2.764 2.314 1.000 .690 .779 2.093 2.689 .998 2.925 5.602 2.687 ..895 1. where m equals degrees of freedom and l" t• • r[(m + 1)/2] ( I + -tlrC.604 .363 1.012 2.372 1.734 1.021 2.325 1.440 1.571 2.658 1.771 2.326 2.688 .703 1.701 1.045 2.201 2.679 .341 1.358 2.750 2.064 2.056 2.878 2.457 2.617 2.684 1.685 1.624 3.390 2.106 3. lltl dl :.316 1.729 2.706 \.756 30 40 60 .746 1.708 \.310 1.397 1.727 .725 1.143 2.812 1.383 2..697 .765 .552 2. ~nm/2) m t From Biomf!lrika Tables for Statisticians.086 2.797 25 26 27 28 29 .841 4. I (2nd edition) Cambridge University Press (1958).833 2.821 6.000 1.476 1.816 .296 1. .069 2.415 1. Hartley.943 1.819 2.807 2.671 1.685 .423 2.250 14 .355 3.473 2.960 2.977 15 16 17 18 19 .657 9.703 1. CI.318 1.684 .533 6.740 1.330 1.315 1.721 1.782 1. O.365 3.576 I 2 3 4 5 6 7 8 9 10 II 12 13 120 00 .860 1.132 12.078 1.323 1. Vol.861 20 21 22 23 24 .131 2.160 2.042 2.319 1.303 3.947 2.700 .718 2.683 .683 .1 . interpolatton should be carried out using the reciprocals of the degrees of freedom.796 1.681 ..365 2.697 1. Where necessary.650 2.704 2.688 \.980 1.345 1.350 1. reproduced by permission of the publishers.683 1.714 1.492 2.761 2.314 2.706 4.032 3.353 2.356 1.674 1.228 2.645 2.306 2.965 4.074 2..447 2. • That is.060 2.499 3.681 2.500 2.01 .289 1.692 1.262 3.110 2.179 2.462 2.541 3.0 1m.711 .660 2.528 2.015 1.321 1.771 1.567 2.055 3. .508 2.691 .005 1.896 2.311 1.080 2.101 2.337 1.169 3.353 TABLES ~ Table II.718 .845 2..920 2.707 3.831 2..886 1.328 1.025 .684 .313 1.695 .467 2.145 2.120 2.638 1.303 1.747 63..821 4.518 2.753 1.706 . Pearson and H.583 2.686 .898 2. S.711 2.25 .684 .479 2.741 3.048 2. The t·Distribution··t·:j: .763 2. and for this the function l20/m is convenient.. 831211 1.60094 5.0100251 .2093 24.5966 12.872085 \.26214 6.9200 23.5346 19.1908 32.81575 4.7496 18.087912 .81221 6.60321 3.5264 32.0717212 .5073 16.2170 27.710721 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 .344419 1.63490 9.4087 34.9999 33.4884 28. The Chi-Square Distribution*·t ~ .7250 26. .3621 23.8 .1170 24.8119 18.63273 6.24697 3.975 .68987 2.99147 7.114832 .995 .6848 20.010 5.5916 14.1433 6.34840 11.8013 34.7569 28.96164 8.0863 16.21034 11.1190 23.1910 31.206990 157088 X 10.90655 7.0228 15.1435 27.07382 3.990 .3367 24.3070 19.1882 26.90766 7.734926 .2767 7.16735 2.56418 8.66043 3.32511 11.6883 29.2777 21.89186 6.6751 21.10 .9550 23.37776 9.0128 17.989265 1.646482 2..1564 38.215795 .14224 5.9958 26.2962 27.40379 5.15585 2.00874 5.62872 3.8693 30.63539 2.484419 393214 X 10.0902 21.26481 6.102587 .005 .81473 9.5779 31.2995 29.01491 7.23075 8.94030 4.9 .9 .05347 3.237347 \.5476 20.8194 31.0506356 .70039 1.9190 12.351846 .57063 18.73264 3.69724 6.17973 2.3193 4.40776 7.297110 982069 x 10 .3449 13.239043 1.675727 .10691 4.8523 30.0261 22.8325 14.c".554300 .7356 26.5893 2.22603 5.57481 5.015 .1413 25.5871 28.0201007 .55821 3.22935 5.02389 7.145476 1.57056 4.4831 21.26094 7.8381 14.84398 5.8602 .0671 15.2672 35.84146 5.8454 30.8053 36..411740 .950 1 2 3 392704 x 10.~ 3.39046 10.4753 20.0705 12.67176 9.5822 .56503 4.6660 16.7185 37.07468 2.48773 x!" .4494 16.87944 10.I (II "" ~ 0 Table III. 7393 60.0471 14. For rn> 100.8564 9.6114 15.299 140.l36 129.1981 12.8852 40.5733 15.879 113.6907 76.1539 88.1963 67.807 104.1 and unit variance.169 .4104 32.329 124.( .9798 39.5879 46.19567 10.1879 43.9823 11.4613 13.5585 10.Table III (continued) m: 20 21 22 23 24 25 26 27 28 29 30 40 50 60 70 80 90 100 1 .215 116.3276 45.3791 16.4331 32.5093 34.5240 12.4848 16.1211 11.6466 74.88623 8.7867 20. e-.9968 41.7642 43.8439 14.9278 48.1603 11.7908 24.9279 17.0905 13.9792 59. Cambridge University Press (19S8).8076 12.3380 13.6465 41.010 .7065 27. 2=0: ••• ' 2r(rn/2) 2 .6885 12.28293 10..43386 8.975 .t.7576 57.2565 13.561 • That is.5569 40.2782 49.025 .3141 45.0757 39.59083 10.050 .64272 9.03366 8.6525 38.6705 33. O.9295 90.0231 106.7659 79.2899 49.3079 16.3641 37.4607 45.5648 14.950 .116 135.6449 50.7956 44.5346 14.:.7585 67.6384 42.8922 63.4151 34.1813 45.425 112. S.7083 37. linear interpolation is adequate.89720 9.1197 13. 100.• ' where rn represents degrees of freedom and r X2 -1.1720 59.3574 40.3794 53.5197 11. edited by E.26042 9.'!2 d1. is approximately normally distributed with mean j2rn .4202 83.8484 31.321 128. Hartley.7729 55.342 95. reproduced by permission of the publishers.5662 38.8786 13.2. J22 For rn < 100.7067 37.4926 26.4418 53.9907 35.26040 8..9535 22. values of 1.3915 69. I (2nd edition).9933 52.2894 41.3356 13.4011 10.0648 48. Vol..1643 29.6417 46.54249 10.9244 35.7541 70.4010 42.005 7.1696 35.4900 91.4789 36.5312 101.6720 66.-. '.1133 41.5400 61.9517 43.995 . Pearson and H.8508 11.9630 48.7807 38.1532 65.2219 51.0819 46.3372 42.7222 44.629 118.1260 77.It til t From Biometrika Tables for Stalisticians.1725 36.990 . I.2976 50.3417 71.5048 79.4817 18.145 124.5913 12.9321 40.1513 16.2752 51.9232 43.1944 44. so that percentage points may be obtained from Table 11. 0880 2.3668 5.2426 2.9830 2.9826 1.9357 5 6 7 8 9 4.9238 2.9985 1.1745 2.6683 2.9189 1.8652 1.1774 2.5202 3.0045 1.0785 2.0481 3.204 9.2850 3.593 9.3491 5.1300 20838 2.5613 2.0862 2.0397 2.9407 1.2400 3.8841 1.9001 1.1655 2.3805 5.4955 2.1279 2.8064 2.0922 2.5222 2.3092 4.3512 2.0909 2.9959 1.1362 3.3970 2.0262 3.9605 2.9245 2.8871 2.8068 2.0145 2.3932 2.9271 2.2333 2.1783 2.9899 2.8833 2.9327 1.6053 2.2730 2.9449 2.7265 2.2735 2.9480 \.2135 2.3649 2.6056 2.2438 2.9790 59. The F Distribution*·t IX =.241 9.1280 2.7265 2.9747 2.9819 \.9091 2.3255 5.0084 1.4374 2.3170 2.4801 2.833 9.5383 4.8939 2.4404 2.9836 20 21 22 23 24 2.9374 2.8473 \.1843 2.9177 2.6682 2.6055 2.9577 2.7167 1.2987 2.7675 1.4624 4.1524 2.8947 1.2261 2.2858 2.8129 3.9878 1.4579 3.1075 2.0171 2.3427 4.5894 3.3473 2.7632 2.5263 5.8289 1.2830 2.7380 1.0492 \.3026 2.1149 2.0741 2.4512 2.8841 1.1571 2.3603 3.7748 1.858 9.2663 2.10 2 3 4 5 6 7 8 9 2 3 4 39.3077 2.0613 2.8918 1.9457 1.9549 59.3274 2.1931 2.9269 1.8238 \.8560 30 40 60 120 2.9531 1.5493 2.5612 2.0098 58.1539 2.2926 5.7516 2.1618 5.9427 1.1808 2.7797 3.6446 2.2888 3.5893 2.0822 2.1296 2.2517 3.0410 1.2574 3.8595 2.4530 3.1073 57.4045 3.5893 2.2081 2.1220 15 16 17 18 19 3.9091 1.6843 1.3473 2.1017 2.7914 2.5362 2.7849 2.2193 2.9063 25 26 27 28 29 2.1582 2.4887 2.0506 58.0604 3.9714 \.8743 1.7055 2.864 8.1131 3.4140 2.9345 \.3467 2.1908 55.3246 53.439 9.500 9.7760 3.9012 2.1953 2.8194 1.9269 1.6602 2.4337 2.0065 3.1094 2.6241 2.3310 2.1765 3.3075 2.2906 2.1949 2.9923 1.0913 2.1494 2.2828 2.9803 1.3040 2.1185 2.2831 2.3772 2.1582 2.2434 5.3069 2.4633 3.0351 2.0605 2.4403 10 11 12 13 14 3.0645 2.9649 1.9188 1.6106 3.7220 \.2065 2.8807 2.1022 2.3416 2.7929 \.2252 3.3801 2.8498 1.3940 2.9486 2.0047 1.0546 2.8354 2.0284 2.3614 2.2183 2.3393 2.7247 2.7478 2.0232 2.2489 2.9609 2.8725 1.5053 3.0379 2.9949 1.5028 2.0472 2.9515 1.2761 2.4898 2.4618 2.5383 2.5106 2.0751 2.4160 2.5746 2.5216 2.1638 2.7277 2.0139 2.6239 2.7741 1.6195 3.1030 2.1958 2.3327 2.0732 3.5191 2.6952 2.3387 2.9610 1.3908 4.6702 1.2662 3.5283 2.5509 3.8274 2.5603 2.0553 2.4606 2.9668 \.3163 2.0241 2.1760 2.5448 49.356 TABLES Table IV.0566 2.9292 1.1423 2.8959 1.2847 4.4694 3.6315 I 00 .1422 2.0000 5.9968 1.906 9.6927 3.0070 2.3679 3.2446 2.3947 2.0730 2.0580 2.3891 2.2341 2.8747 1. 8200 1.6252 1.5198 1.76()7 3.7756 1.5450 1.5555 2.1926 1.6074 1.6856 1.8090 1.8572 2.6589 1.2646 1.1686 1.5313 1.2983 3.8388 1.0360 1.8272 1.6377 J.4663 5.0597 2.7395 \.0095 2.5732 1.328 9.3789 3.4829 5.3476 1.9997 1.2380 2.6042 1.9117 1.740 9.8775 1.7938 1.9369 2.6308 1.6554 1.8656 1.9827 1.4989 1.8903 1.1554 2.3226 2.6407 1.6662 1.6356 1.1474 2.1337 3.2435 2.8451 1.7182 1.6890 1.4373 1.7808 1.7873 1.1597 3.2087 2.7537 1.6864 1.7021 1.4871 1.2067 28363 2.9770 1.2926 2.7439 1.7223 1.4041 2.0762 2.2951 1.5142 2.7805 1.9043 1.6224 1.265 9.3952 1.9399 1.4094 1.8728 1.8454 1.2547 3.6034 1.7175 1.7827 1.7159 1.6012 1.7302 1.3916 5.9036 1.5570 1.5036 1.5458 1.8658 1.1764 3.6052 1.9243 1.5056 1.8712 2.8000 2.8647 1.7025 2.7667 1.0070 1.7193 1.S825 1.5871 1.2974 2.7964 1.8759 1.002 9.5925 J.6032 1.6910 1.6951 1.6852 1.7627 1.6890 1.195 9.8750 1.7867 1.1845 3.2156 3.8462 1.9047 2.6574 1.9625 2.8359 1.5934 1.8624 1.7973 2.3396 3.7070 1.6468 1.5753 2.1681 3.S741 1.8103 1.3830 2.1402 27620 2.4413 5.8142 1.8274 1.7146 1.8852 2.1376 2.6567 1.5435 1.1072 20261 1.0281 2.4906 1.0115 1.9315 1.7481 1.7708 1.0537 2.8450 1.220 9.0554 1.0171 1.6465 1.8111 1.6228 1.7191 1.8183 2.2007 2.2915 1.8990 1.2304 3.1878 2.1784 2.3832 1.1741 2.8174 62529 9.0966 2.8036 62794 9.1512 3.9333 \.8195 1.8955 61.9323 1.2400 1.1230 2.1425 3.4746 5.8550 1.10 10 12 15 20 24 30 40 60 120 co 60.7596 1.6322 2.7831 1.4206 1.5020 2.4246 2. =.6560 1.0532 2.5327 1.4928 2.2768 3.1671 2.0009 1.4755 1.4821 1.8913 1.6771 1.4472 1.7895 1.4081 5.9861 1.7551 1.6272 \.9199 60.5490 .8868 1.6673 1.7083 1.2320 3.3419 1.7896 63.8362 1.5107 1.9043 1.4784 1..6681 2.9119 2.5715 1.8689 61.7059 1.4496 5.9557 2.357 TABLES Table IV (continued) ~ 0 (I F""''''2.1317 2.6155 1.8310 62.5987 1.9854 1.7753 63.6768 1.9377 2.4579 5.5668 1.7506 1.8593 1.3676 1.7816 1.4913 5.3614 2.6524 1.3391 2.1050 2.9576 1.7507 27222 2.6433 1.4708 2.6721 1.3769 1.7019 1.4642 2.061 9.5380 2.6759 1.0818 1.7232 1.2482 2.6389 1.7989 1.705 9.0000 1.8108 1.0593 2.8280 2.7312 1.4163 3.4247 5.1573 2.6988 1.5805 1.2085 3.7298 1.7592 1.9577 1.4672 1.0954 2.6659 1.5351 2.5376 1.8368 \.1592 1.3203 1.5437 1.4670 1.8168 1.5411 1.3162 2.1843 3.1905 2.8319 1.5090 1.7306 1.4564 1.5947 2.1000 2.6147 1.0516 1.8090 \.7812 2.2841 2.6065 1.8449 1.6624 1.6569 1.7727 1.7382 1.9367 1.1228 27423 2.9197 1.8924 1.5176 1.2682 2.5686 1.9721 1.5575 1.9597 1.6073 1.7492 1.7590 1.5472 1.5862 1.4248 1.6714 1.9722 1.1049 2.8443 62.2003 3.9117 1. 6331 5.3719 2.3513 4.6987 2.9340 2.4139 4.3881 3.77 238.3541 3.4453 2.9823 3.8477 3.2296 4.7144 2.9384 2.3419 2.6672 4.3404 3.0135 8.7083 3.0123 2.6408 2.9874 5.6049 2.7905 2.5431 4.4205 2.5435 2.2317 3.9467 2.4028 3.5915 3.4638 2.5082 2.9134 2.5377 2.3690 3.9588 1.1631 6.3852 3.6613 2.3883 6.3009 4.2172 3.128 9.4904 2.2501 2.5005 3.7413 2.371 19.0012 3.7729 2.3883 23732 2.2360 2.2821 2.5337 4.7587 2.4768 2.8056 3.2597 3.6572 2.0867 2.8167 2.6458 15 16 17 18 19 4.9582 3.7109 2.4471 2.3248 4.7472 4.7374 4.50 215.4940 4.4590 4.3463 2.1274 3.1789 10 11 12 14 4.6030 2.7278 2.4817 4.0717 2.8321 2.99 236.4780 3.5454 2.2107 2.3201 2.1791 3.6802 2.0280 3.0725 3.6875 3.5491 2.4495 2.16 233.2100 4.4422 2.6613 2.8401 2.8951 2.7086 6.1504 3.3928 2.7725 4.3683 2.3439 3.4047 2.6896 2.5252 2.1922 4.1174 5.5990 2.3277 2.0254 2.8661 2.3468 4.5914 6.9480 2.3053 2.1355 3.2793 4.4513 4.9201 3.6400 2.5806 3.6207 2.3593 2.0990 3.9443 6.5876 2.1203 3.3874 3.1968 3.1750 2.7389 3.3371 2.7955 2.3258 3.1665 2.4741 2.8853 3.0088 2.0503 4.2490 2.5219 3.7870 3.5277 2.2066 3.8660 3.05 5 6 7 8 9 161.3661 2.7426 2.6060 2.4943 2.7669 2.2540 2.000 19.5336 2.5102 2.9406 8.2592 3.2655 2.0401 1.353 19.247 19.4105 3.4381 3.0410 5.0096 2.9153 2.2900 2.5868 2.3551 2.8799 13 00 .4668 3.7964 2.3177 5.9013 2.0946 2.8868 8.164 19.7014 2.3205 2.6987 3.8378 3.4591 2.2662 21802 2'0970 2.4226 2.0718 2.8443 4.2417 4.5914 5.2252 4.0848 4.1468 3.1240 2.5911 2.5581 2.9957 2.3748 2.0984 3.0556 3.4903 3.5719 2.3965 2.2766 9.9223 2.6337 3.2141 2.8759 4.71 224.3808 3.5546 3.4434 3.1709 4.8452 8.2565 5.0662 3.5480 2.6143 2.8415 3.6079 5.296 19.4205 2.5140 2.7571 4.2039 3.2782 2.9277 2.7257 3.3738 4.358 TABLES Table IV (continued) il '" 2 I 2 3 4 3 4 .7141 2.58 230.1122 3.1599 3.3343 2.8387 2.9646 4.9751 2.7861 5.0204 2.0164 1.9715 3.5767 2.9912 2.5521 9.9647 2.2927 4.6283 2.8100 2.8626 5.5874 3.45 199.9961 2.4472 2.9604 2.5727 2.330 19.54 18.2560 6.9988 5 6 7 8 9 6.7642 3.7401 2.0986 2.6001 4.3567 3.2839 3.1059 3.7581 2.4876 2.2229 30 40 60 120 4.3158 3.2389 3.4227 20 21 22 23 24 4.1960 4.4928 3.2874 3.9503 4.1028 3.4324 2.8123 7.7763 2.88 240.6823 3.0491 3.1433 4.6848 2.1830 3.3002 25 26 27 28 29 4.7066 2.385 10.4095 4.6767 3.2913 2.8183 4.0942 6.4221 3.8524 2.1172 9.8962 2.513 19.3359 2.0069 2.4563 2.8486 2. 9643 1.0629 2.7855 8.1323 2.4445 3.8082 3.9898 1.5089 1.0825 2.9139 1.8536 2.9276 2.8895 1.6866 2.2214 1.1507 2.9586 1.479 19.8027 1.9010 1.496 19.3522 2.7744 5.2043 2.3082 2.0102 1.7537 1.1601 2.5437 2.8117 1.7929 1.5342 2.5944 8.7743 3.0729 4.3180 1.0061 4.9299 1.9245 1.0264 2.7475 4.8450 2.2962 2.6223 1.2033 2.2756 2.7307 1.9605 1.7897 1.2747 2.6688 3.3210 2.9669 2.9446 1.7459 5.1152 2.6928 1.5943 1.9168 1.5494 8.05 30 40 60 ]20 00 241.9926 1.7138 1.8718 1.7851 1.8874 1.3275 2.7331 2.7029 8.6981 1.3410 2.6541 1.5801 2.9005 4.5265 5.4753 2.9005 1.3043 3.8533 1.8025 5.2298 2.6609 2.413 19.2184 3.1058 2.396 19.9604 1.487 19.0267 2.9645 1.0558 2.2547 2.3842 2.9317 1.454 19.6581 5.0889 2.7067 2.4901 2.7351 4.8424 1.5108 3.3472 3.558\ 3.2354 2.8364 1.6084 1.8687 1.6021 2.7444 1.6169 2.7740 2.4630 2.4638 3.1900 2.6377 2.8649 1.0035 1.0540 2.0391 2.7372 2.3519 1.0716 2.1497 2.8259 4.1045 2.8307 2.2036 2.1071 2.0053 2.20 253.3487 2.1242 2.5331 2.7570 1.4314 3.09 252.0166 1.8842 1.5173 1.25 19.4952 1.7398 3.2966 2.9782 2.4480 2.0050 1.9130 2.4673 1.6188 3.01 251.1757 2.1141 2.471 19.6373 1.9192 1.0921 2.7505 1.2304 2.1373 4.7186 2.2840 3.1649 2.6777 3.6664 1.7534 2.9302 2.2258 2.8361 1.4259 2.9174 1.6385 8.4935 2.8217 1.0148 1.3080 2.1077 2.3779 2.7396 1.1646 2.5436 2.7480 1.8963 1.9365 4.3879 2.6710 2.2967 2.2504 2.6906 1.446 19.7831 1.8543 1.0707 2.6037 2.6464 2.2468 2.4591 1.5747 3.5543 1.1938 2.2197 2.9390 1.1555 2.7488 1.2341 2.0960 2.7872 4.4035 2.4202 2.3404 3.6835 1.1834 2.0096 1.1477 2.5720 8.1508 2.429 19.6602 8.0428 2.1768 2.5705 1.14 254.32 249.9464 1.2365 2.7876 2.9681 1.2524 2.6717 1.6090 2.0476 2.8055 1.8742 3.9838 2.3392 2.4117 2.0275 2.4589 2.0075 1.1503 2.0589 2.4499 2.1906 2.8637 4.9117 5.9796 2.2539 1.0600 3.9736 1.4105 3.1282 2.7170 5.9380 1.0411 2.2043 2.4957 3.6878 5.7001 1.8415 3.0283 2.0712 2.5379 2.4663 2.05 250.5705 2.95 248.4247 2.5272 3.9381 3.6365 3.3758 3.9938 1.9644 5.3984 3.4045 2.3421 2.5055 2.9842 1.0584 2.7684 1.0772 1.9147 1.6491 1.8920 1.2664 2.2674 2.7110 1.2230 2.8409 1.3893 1.8389 1.6211 2.91 245.8657 1.1040 2.1479 2.4290 1.3807 2.2064 2.2776 2.5766 1.8380 1.9999 3.8780 2.5343 1.1179 2.7446 8.1141 2.3803 2.6996 2.7918 1.0000 .1898 2.7047 3.5309 2.359 TABLES Table IV (continued) 2 10 12 15 20 24 = .7522 2.9165 1.3077 2.2686 2.462 8.9105 1.1778 2.88 243.8578 5.0107 1.0794 2.9464 1.1307 2.7689 1.8432 1.8337 1.0658 2.6281 4.2878 2.8128 1.3479 2.3650 3.8203 1.3940 1.6587 1.6166 8. 360 TABLES Table IV (continued) II - 2 I 2 3 4 647.79 38.506 17.443 12.218 799.50 39.000 16.044 10.649 .025 3 4 S 6 7 8 9 864.16 39.165 15.439 9.9792 m.S8 39.248 15.101 9.6045 921.85 39.298 14.885 9.3645 937.11 39.331 14.735 9.1973 948.22 39.355 14.624 9.0741 956.66 39.373 14.540 8.9796 963.28 39.387 14.473 8.9047 5 6 7 8 9 10.007 8.8131 8.0727 7.5709 7.2093 8.4336 7.2598 6.5415 6.0595 5.7147 7.7636 6.5988 5.8898 5.4160 5.0781 7.3879 6.2272 5.5226 5.0526 4.7181 7.1464 5.9876 5.2852 4.8173 4.4844 6.9777 5.8197 5.1186 4.6517 4.3197 6.8531 5.6955 4.9949 4.5286 4.1971 6.7572 5.S996 4.8994 4.4332 4.1020 6.6810 5.5234 4.8232 4.3572 4.0260 10 11 12 13 14 . 6.9367 6.7241 6.5538 6.4143 6.2979 5.4564 5.2559 5.0959 4.9653 4.8567 4.8256 4.6300 4.4742 4.3472 4.2417 4.4683 4.2751 4.1212 3.9959 3.8919 4.2361 4.0440 3.8911 3.7667 3.6634 4.0721 3.8807 3.7283 3.6043 3.5014 3.9498 3.7S86 3.6065 3.4827 3.3799 3.8549 3.6638 3.5118 3.3880 3.2853 3.7790 3.5879 3.43S8 3.3120 3.2093 15 16 17 18 19 6.1995 6.1151 6.0420 5.9781 5.9216 4.7650 4.6867 4.6189 4.5597 4.5075 4.1528 4.0768 4.0112 3.9539 3.9034 3.8043 3.7294 3.6648 3.6083 3.5587 3.5764 3.5021 3.4379 3.3820 3.3327 3.4147 3.3406 3.2767 3.2209 3.1718 3.2934 3.2194 3.1556 3.0999 3.0509 3.1987 3.1248 3.0610 3.0053 2.9563 3.1227 3.0488 2.9849 2.9291 2.8800 20 21 22 23 24 5.8715 5.8266 5.7863 5.7498 5.7167 4.4613 4.4199 4.3828 4.3492 4.3187 3.8587 3.8188 3.7829 3.7505 3.7211 3.5147 3.4754 3.4401 3.4083 3.3794 3.2891 3.2501 3.2151 3.1835 3.1548 3.1283 3.0895 3.0546 3.0232 2.9946 3.0074 2.9686 2.9338 2.9024 2.8738 2.9128 2.8740 2.8392 2.8077 2.7791 2.8365 2.7977 2.7628 2.7313 2.7027 25 26 27 28 29 5.6864 5.6586 5.6331 5.6096 5.5878 4.2909 4.2655 4.2421 4.2205 4.2006 3.6943 3.6697 3.6472 3.6264 3.6072 3.3530 3.3289 3.3067 3.2863 3.2674 3.1287 3.1048 3.0828 3.0625 3.0438 2.9685 2.9447 2.9228 2.9027 2.8840 2.8478 2.8240 2.8021 2.7820 2.7633 2.7531 2.7293 2.7074 2.6872 2.6686 2.6766 2.6528 2.6309 2.6106 2.5919 30 40 60 120 co 5.S675 5.4239 5.2857 5.1524 5.0239 4.1821 4.0510 3.9253 3.8046 3.6889 3.5894 3.4633 3.3425 3.2270 3.1161 3.2499 3.1261 3.0077 2.8943 2.7858 3.0265 2.9037 2.7863 2.6740 2.5665 2.8667 2.7444 2.6274 2.5154 2.4082 2.7460 2.6238 2.S068 2.3948 2.2875 2.6513 2.5289 2.4117 2.2994 2.1918 2.5746 2.4519 2.3344 2.2217 2.1136 361 TABLES Table IV (continued) !X ., 10 12 15 20 24 .025 30 40 60 120 00 984.87 1005.6 968.63 976.71 993.10 997.25 1001.4 1009.8 1014.0 1018.3 39.398 39.415 39.431 39.448 39.456 39.473 39.481 39.490 39.498 39.456 14.419 14.037 13.947 14.337 14.253 14.167 14.124 J4.081 13.902 13.992 8.8439 8.6565 8.5599 8.4111 8.3604 8.2573 8.5109 8.4613 8.7512 8.3092 6.6192 5.4613 4.7611 4.2951 3.9639 6.5246 5.3662 4.6658 4.1997 3.8682 6.4277 5.2687 4.5678 4.1012 3.7694 6.3285 5.1684 4.4667 3.9995 3.6669 6.2780 5.1172 4.4150 3.9472 3.6142 6.2269 5.0652 4.3624 3.8940 3.5604 6.1751 5.0125 4.3089 3.8398 3.5055 6.1225 5.9589 4.2544 3.7844 3.4493 6.0693 4.9045 4.1989 3.7279 3.3918 6.0153 4.8491 4.1423 3.6702 3.3329 3.7168 3.5257 3.3736 3.2497 3.1469 3.6209 3.4296 3.2773 3.1532 3.0501 3.5217 3.3299 3.1772 3.0527 2.9493 3.4186 3.2261 3.0728 2.9477 2.8437 3.3654 3.1725 3.0187 2.8932 2.7888 3.3110 3.1176 2.9633 2.8373 2.7324 3.2554 3.0613 2.9063 2.7797 2.6742 3.1984 3.0035 2.8478 2.7204 2.6142 3.\399 2.9441 2.7874 2.6590 2.5519 3.0798 2.8828 2.7249 2.5955 2.4872 3.0602 2.9862 2.9222 2.8664 2.8173 2.9633 2.8890 2.8249 2.7689 2.7196 2.8621 2.7875 2.7230 2.6667 2.6171 2.7559 2.6808 2.6158 2.5590 2.5089 2.7006 2.6252 2.5598 2.5027 2.4523 2.6437 2.5678 2.5021 2.4445 2.3937 2.5850 2.5085 2.4422 2.3842 2.3329 2.5242 2.4471 2.3801 2.3214 2.2695 2.4611 2.3831 2.3153 2.2558 2.2032 2.3953 2.3163 2.2474 2.1869 2.1333 2.7737 2.7348 2.6998 2.6682 2.6396 2.6758 2.6368 2.6017 2.5699 2.5412 2.5731 2.5338 2.4984 2.4665 2.4374 2.4645 2.4247 2.3890 2.3567 2.3273 2.4076 2.3675 2.3315 2.2989 2.2693 2.3486 2.3082 2.2718 2.2389 2.2090 2.2873 2.2465 2.2097 2.1763 2.1460 2.2234 2.1819 2.1446 2.1107 2.0799 2.1562 2.1141 2.0760 2.0415 2.0099 2.0853 2.0422 2.0032 \.9677 \.9353 2.6135 2.5895 2.5676 2.5473 2.5286 2.5149 2.4909 2.4688 2.4484 2.4295 2.4110 2.3867 2.3644 2.3438 2.3248 2.3005 2.2759 2.2533 2.2324 2.2131 2.2422 2.2174 2.1946 2.1735 2.1540 2.1816 2.1565 2.\334 2.1121 2.0923 2.1183 2.0928 2.0693 2.0477 2.0276 2.0517 2.0257 2.0018 \.9796 1.9591 1.9811 1.9545 1.9299 1.9072 1.8861 1.9055 1.8781 1.8527 1.8291 1.8072 2.5112 2.3882 2.2702 2.1570 t0483 2.4120 2.2882 2.1692 2.0548 1.9447 2.3072 2.1819 2.0613 1.9450 1.8326 2.1952 2.0677 1.9445 1.8249 1.7085 2.1359 2.0069 1.8817 1.7597 1.6402 2.0739 1.9429 1.8152 1.6899 1.5660 2.0089 1.8752 1.7440 1.6141 1.4835 1.9400 1.8028 1.6668 1.5299 1.3883 1.8664 1.7242 1.5810 1.4327 1.2684 1.7867 1.6371 1.4822 1.3104 1.0000 TABLES 362 Table IV (continued) ac 2 1 2 3 4 3 = .01 4 5 6 7 8 9 5763.7 4052.2 5624.6 5859.0 5928.3 5981.6 6022.5 4999.5 5403.3 99.299 99.388 98.503 99.332 99.356 99.374 99.249 99.000 99.166 34.116 29.457 28.710 28.237 27.911 27.672 27.489 30.817 27.345 15.207 14.976 14.799 15.977 21.198 16.694 15.522 14.659 18.000 5 6 7 8 9 16.258 13.745 12.246 11.259 10.561 13.274 10.925 9.5466 8.6491 8.0215 12.060 9.7795 8.4513 7.5910 6.9919 11.392 9.1483 7.8467 7.0060 6.4221 10.967 8.7459 7.4604 6.6318 6.0569 10.672 8.4661 7.1914 6.3707 5.8018 10.456 8.2600 6.9928 6.1776 5.6129 10.289 8.1016 6.8401 6.0289 5.4671 10.158 7.9761 6.7188 5.9106 5.3511 10 11 12 13 14 10.044 9.6460 9.3302 9.0738 8.8616 7.5594 7.2057 6.9266 6.70\0 6.5149 6.5523 6.2167 5.9526 5.7394 5.5639 5.9943 5.6683 5.4119 5.2053 5.0354 5.6363 5.3160 5.0643 4.8616 4.6950 5.3858 5.0692 4.8206 4.6204 4.4558 5.2001 4.8861 4.6395 4.4410 4.2779 5.0567 4.7445 4.4994 4.3021 4.1399 4.9424 4.6315 4.3875 4.1911 4.0297 15 16 17 18 19 8.6831 8.5310 8.3997 8.2854 8.1850 6.3589 6.2262 6.1121 6.0129 5.9259 5.4170 5.2922 5.1850 5.0919 5.0103 4.8932 4.7726 4.6690 4.5790 4.5003 4.5556 4.4374 4.3359 4.2479 4.1708 4.3183 4.2016 4.1015 4.0146 3.9386 4.1415 4.0259 3.9267 3.8406 3.7653 4.0045 3.8896 3.7910 3.7054 3.6305 3.8948 3.7804 3.6822 3.5971 3.5225 20 21 22 23 24 8.0960 8.0166 7.9454 7.8811 7.8229 5.8489 5.7804 5.7190 5.6637 5.6136 4.9382 4.8740 4.8166 4.7649 4.7181 4.4307 4.3688 4.3134 4.2635 4.2184 4.1027 4.0421 3.9880 3.9392 3.8951 3.8714 3.8117 3.7583 3.7102 3.6667 3.6987 3.6396 3.5867 3.5390 3.4959 3.5644 3.5056 3.4530 3.4057 3.3629 3.4567 3.9381 3.3458 3.2986 3.2560 25 26 27 28 29 7.7698 7.7213 7.6767 7.6356 7.5976 5.5680 5.5263 5.4881 5.4529 5.4205 4.6755 4.6366 4.6009 4.5681 4.5378 4.1774 4.1400 4.1056 4.0740 4.0449 3.8550 3.8183 3.7848 3.7539 3.7254 3.6272 3.5911 3.5580 3.5276 3.4995 3.4568 3.4210 3.3882 3.3581 3.3302 3.3239 3.2884 3.2558 3.2259 3.1982 3.2172 3.1818 3.1494 3.1195 3.0920 30 40 60 120 7.5625 7.3141 7.0771 6.8510 6.6349 5.3904 5.1785 4.9774 4.7865 4.6052 4.5097 4.3126 4.1259 3.9493 3.7816 4.0179 3.8283 3.6491 3.4796 3.3192 3.6990 3.5138 3.3389 3.1735 3.0173 3.4735 3.2910 3.1187 2.9559 2.8020 3.3045 3.1238 2.9530 2.7918 2.6393 3.1726 2.9930 2.8233 2.6629 2.5113 3.0665 2.8876 2.718' 2.558"6 2.4073 'X.l 363 TABLES Table IV (continued) IX 10 12 15 20 = .01 24 30 40 60 120 00 6366.0 6260.7 6286.8 6313.0 6339.4 6055.8 6106.3 6157.3 6208.7 6234.6 99.483 99.491 99.501 99.474 99.399 99.416 99.432 99.449 99.458 99.466 26.125 27.229 26.316 26.221 26.690 26.598 27.052 26.505 26.411 26.872 13.463 13.745 13.558 14.546 14.374 13.652 13.838 14.198 14.020 13.929 10.051 7.8741 6.6201 5.8143 5.2565 9.8883 7.7183 6.4691 5.6668 5.1114 9.7222 7.5590 6.3143 5.5151 4.9621 9.5527 7.3958 6.1554 5.3591 4.8080 9.4665 7.3127 6.0743 5.2793 4.7290 9.3793 7.2285 5.9921 5.1981 4.6486 9.2912 7.1432 5.9084 5.1156 4.5667 9.2020 7.0568 5.8236 5.0316 4.4831 9.1118 6.9690 5.7372 4.9460 4.3978 9.0204 6.8801 5.6495 4.8588 4.3105 4.8492 4.5393 4.2961 4.1003 3.9394 4.7059 4.3974 4.1553 3.9603 3.8001 4.5582 4.2509 4.0096 3.8154 3.6557 4.4054 4.0990 3.8584 3.6646 3.5052 4.3269 4.0209 3.7805 3.5868 3.4274 4.2469 3.941\ 3.7008 3.5070 3.3476 4.1653 3.8596 3.6192 3.4253 3.2656 4.0819 3.7761 3.5355 3.3413 3.1813 3.9965 3.6904 3.4494 3.2548 3.0942 3.9090 3.6025 3.3608 3.1654 3.0040 3.8049 3.6909 3.5931 3.5082 3.4338 3.6662 3.5527 3.4552 3.3706 3.2965 3.5222 3.4089 3.3117 3.2273 3.1533 3.3719 3.2588 3.1615 3.0771 3.0031 3.2940 3.1808 3.0835 2.9990 2.9249 3.2141 3.1007 3.0032 2.9185 2.8442 3.1319 3.0182 2.9205 2.8354 2.7608 3.0471 2.9330 2.8348 2.7493 2.6742 2.9595 2.8447 2.7459 2.6597 2.5839 2.8684 2.7528 2.6530 2.5660 2.4893 3.3682 3.3098 3.2576 3.2106 3.1681 3.2311 3.\ 729 3.1209 3.0740 3.0316 3.0880 3.0299 2.9780 2.9311 2.8887 2.9377 2.8796 2.8274 2.7805 2.7380 2.8594 2.8011 2.7488 2.7017 2.6591 2.7785 2.7200 2.6675 2.6202 2.5773 2.6947 2.6359 2.5831 2.5355 2.4923 2.6077 2.5484 2.495\ 2.4471 2.4035 2.5168 2.4568 2.4029 2.3542 2.3099 2.4212 2.3603 2.3055 2.2559 2.2107 3.1294 3.0941 3.0618 3.0320 3.0045 2.9931 2.9579 2.9256 2.8959 2.8685 2.8502 2.8150 2.7827 2.7530 2.7256 2.6993 2.6640 2.6316 2.6017 2.5742 2.6203 2.5848 2.5522 2.5223 2.4946 2.5383 2.5026 2.4699 2.4397 2.4118 2.4530 2.4170 2.3840 2.3535 2.3253 2.3637 2.3273 2.2938 2.2629 2.2344 2.2695 2.2325 2.1984 2.1670 2.1378 2.1694 2.1315 2.0965 2.0642 2.0342 2.9791 2.8005 2.631K 2.4721 2.3209 2.11431 2.6648 2.4961 2.3363 2.1848 2.7002 2.5216 2.3523 2.1915 2.0385 2.5487 2.3689 2.1978 2.0346 1.8783 2.4689 2.2880 2.1154 1.9500 1.7908 2.3860 2.2034 2.0285 1.8600 1.6964 2.2992 2.1142 1.9360 1.7628 1.5923 2.2079 2.0194 1.8363 1.6557 1.4730 2.1107 1.9172 1.7263 1.5330 1.3246 2.0062 1.8047 1.6006 1.3805 \.0000 364 TABLES Table IV (continued) 01- .005 2 3 4 5 6 7 8 9 1 2 3 4 16211 198.50 55.552 31.333 20000 199.00 49.799 26.284 21615 199.17 47.467 24.259 22500 199.25 46.195 23.155 23056 199.30 45.392 22.456 23437 199.33 44.838 21.975 23715 199.36 44.434 21.622 23925 199.37 44.126 21.352 24091 199.39 43.882 21.139 5 6 7 8 9 22.785 18.635 16.236 14.688 13.614 18.314 14.544 12.404 11.042 10.107 16.530 12.917 10.882 9.5965 8.7171 15.556 12.028 10.050 8.8051 7.9559 14.940 11.464 9.5221 8.3018 7.4711 14.513 11.073 9.\554 7.9520 7.1338 14.200 10.786 8.8854 7.6942 6.8849 13.961 10.566 8.6781 7.4960 6.6933 13.772 10.391 8.5138 7.3386 6.4511 10 11 12 13 14 12.826 12.226 11.754 11.374 11.060 9.4270 8.9122 8.5096 8.1865 7.9217 8.0807 7.6004 7.2258 6.9257 6.6803 7.3428 6.8809 6.5211 6.2335 5.9984 6.8723 6.4217 6.0711 5.7910 5.5623 6.5446 6.1015 5.7570 5.4819 5.2574 6.3025 5.8648 5.5245 5.2529 5.0313 6.1159 5.6821 5.3451 5.0761 4.8566 5.9676 5.5368 5.2021 4.9351 4.7173 15 16 17 18 19 10.798 10.575 10.384 10.218 10.073 7.7008 7.5138 7.3536 7.2148 7.0935 6.4760 6.3034 6.1556 6.0277 5.9161 5.8029 5.6378 5.4967 5.3746 5.2681 5.3721 5.2117 5.0746 4.9560 4.8526 5.0708 4.9134 4.7789 4.6627 4.5614 4.8473 4.6920 4.5594 4.4448 4.3448 4.6743 4.5207 4.3893 4.2759 4.1770 4.5364 4.3838 4.2535 4.1410 4.0428 20 21 22 23 24 9.9439 9.8295 9.7271 9.6348 9.5513 6.9865 6.8914 6.8064 6.7300 6.6610 5.8177 5.7304 5.6524 5.5823 5.5190 5.1743 5.0911 5.0168 4.9500 4.8898 4.7616 4.6808 4.6088 4.5441 4.4857 4.4721 4.3931 4.3225 4.2591 4.2019 4.2569 4.1789 4.1094 4.0469 3.9905 4.0900 4.0128 3.9440 3.8822 3.8264 3.9564 3.8799 3.8116 3.7502 3.6949 25 26 27 28 29 9.4753 9.4059 9.3423 9.2838 9.2297 6.5982 6.5409 6.4885 6.4403 6.3958 5.4615 5.4091 5.361 I 5.3170 5.2764 4.8351 4.7852 4.7396 4.6977 4.6591 4.4327 4.3844 4.3402 4.2996 4.2622 4.1500 4.1027 4.0594 4.0197 3.9830 3.9394 3.8928 3.8501 3.8110 3.7749 3.7758 3.7297 3.6875 3.6487 3.6130 3.6447 3.5989 3.5571 3.5186 3.4832 30 40 60 120 9.1797 8.8278 8.4946 8.1790 7.8794 6.3547 6.0664 5.7950 5.5393 5.2983 5.2388 4.9759 4.7290 4.4973 4.2794 4.6233 4.3738 4.1399 3.9207 3.7151 4.2276 3.9860 3.7600 3.5482 3.3499 3.9492 3.7129 3.4918 3.2849 3.0913 3.7416 3.5088 3.2<;11 3.0874 2.8968 3.5801 3.3498 3.1344 2.9330 2.7444 3.4505 3.2220 3.0083 2.8083 2.6210 ro 365 TABLES Table IV (continued) a - .005 10 12 15 20 24 30 40 60 120 24224 199.40 43.686 20.967 24426 199.42 43.387 20.705 24630 199.43 43.085 20.438 24836 199.45 42.778 20.167 24940 199.46 42.622 20.030 25044 199.47 42.466 19.892 25148 199.47 42.308 19.752 25253 199.48 42.149 19.611 25359 199.49 41.989 19.468 2S465 199.51 41.829 19.325 13.618 10.250 8.3803 7.2107 6.4171 13.384 10.034 8.1764 7.0149 6.2274 13.146 9.8140 7.9678 6.8143 6.0325 12.903 9.5888 7.7540 6.6082 5.8318 12.780 9.4741 7.6450 6.5029 5.7292 12.656 9.3583 7.5345 6.3961 5.6248 12.530 9.2408 7.4225 6.2875 5.5186 12.402 9.1219 7.3088 6.1772 5.4104 12.274 9.0015 7.1933 6.0649 5.3001 12.144 8.8793 7.0760 5.9505 5.1875 5.8467 5.4182 5.0855 4.8199 4.6034 5.6613 5.2363 4.9063 4.6429 4.4281 5.4707 5.0489 4.7214 4.4600 4.2468 5.2740 4.8552 4.5299 4.2703 4.0585 5.1732 4.7557 4.4315 4.1726 3.9614 5.0705 4.6543 4.3309 4.0727 3.8619 4.9659 4.5508 4.2282 3.9704 3.7600 4.1J592 4.4450 4.1229 3.8655 3.6553 4.7501 4.3367 4.0149 3.7577 3.5473 4.6385 4.2256 3.9039 3.6465 3.4359 4.4236 4.2719 4.1423 4.0305 3.9329 4.2498 4.0994 3.9709 3.8599 3.7631 4.0698 3.9205 3.7929 3.6827 3.5866 3.8826 3.7342 3.6073 3.4977 3.4020 3.7859 3.6378 3.5112 3.4017 3.3062 3.6867 3.5388 3.4124 3.3030 3.2075 3.5850 3.4372 3.3107 3.2014 3.1058 3.4803 3.3324 3.2058 3.0962 3.0004 3.3722 3.2240 3.0971 2.9871 2.8908 3.2602 3.1115 2.9839 2.8732 2.7762 3.8470 3.7709 3.7030 3.6420 3.5870 3.6779 3.6024 3.5350 3.4745 3.4199 3.5020 3.4270 3.3600 3.2999 3.2456 3.3178 3.2431 3.1764 3.1165 3.0624 3.2220 3.1474 3.0807 3.0208 2.9667 3.1234 3.0488 2.9821 2.9221 2.8679 3.0215 2.9467 2.8799 2.8198 2.7654 2.9159 2.8408 2.7736 2. 7132 2.6585 2.8058 2.7302 2.6625 2.6016 2.5463 2.6904 2.6140 2.5455 2.4837 2.4276 3.5370 3.4916 3.4499 3.4117 3.3765 3.3704 3.3252 3.2839 3.2460 3.2111 3.1963 3.1 51 5 3.1104 3.0727 3.0379 3.0133 2.9685 2.9275 2.8899 2.8551 2.9176 2.8728 2.8318 2.7941 2.7594 2.8187 2.7738 2.7327 2.6949 2.6601 2.7160 2.6709 2.6296 2.5916 2.5565 2.6088 2.5633 25217 2.4834 2.4479 2.4960 2.4501 2.4078 2.3689 2.3330 2.3765 2.3297 2.2867 2.2469 2.2102 3.3440 3.1167 2.9042 2.7052 2.5188 3.1787 2.9531 2.7419 2.5439 2.3583 3.0057 2.7811 2.5705 2.3727 2.1868 2.8230 2.5984 2.3872 2.1881 1.9998 2.7272 2.5020 2.2898 2.0890 1.8983 2.6278 2.4015 2.1874 1.9839 1.7891 2.5241 2.2958 2.0789 1.8709 1.6691 24151 21838 1.9622 1.7469 1.5325 2.2997 2.0635 1.8341 1.6055 1.3637 2.1760 1.9318 1.6885 1.4311 1.0000 00 • That is. values of F., .•, .•. where (m,. ma) is the pair of degrees orrrc:edom in F., .•, and f,'" rllm, + mz)/2) (~) ... 11 f1 .. ,Ill-. r(m,/2)r( ml/2) mz '., .,.' (I + ~ F) -\til. ·.,111 dF '" (I. mz t From MTables of percentage points of the Inverted Beta (F) Distribution," Biometrika, Vol. 33 (1943). pp. 73-88. by Maxine Merrington and Catherine M. Thompson; reproduced by permission of E. S. Pearson. If necessary, interpolation should be carried out using the reciprocals of the degrees of freedom. Statistical Methods in Engineering and Quality Assurance Peter W. M. John Copyright © 1990 by John Wiley & Sons, Inc Index A A2.147 Ac. see Acceptance number Acceptable quality level (AQL), 180. 183. 198 Acct:ptance number. 179, 199 Acceptance region, 113 Acceptance sampling, 1, 7 Algorithm. Yates', 292 Alias. 297-298 Alpha risk, See Hypothesis testi ng Alternative, see Hypothesis testing American National Standards Institute 197 ' American Society for Quality Control, 197 Analysis of covariance, 267-271 Analysis of variance, 228, 251, 264 ANOVA., see Analysis of variance ANSI, see American National Standards Institute ANSIIASQC Z1.4, 197 ANSIIASQC Z1.9, 197,201 AOQ, see Average outgoing quality (AOQ) AOQL, see Average outgoing quality limit (AOQL) Approximations to probability densities: binomial approximation to hypergeometric, 40 normal approximation to binomial 72 ' normal approximation to rank sum 135 ' Poisson approximation to binomial, 182 AQL, see Acceptable quality level (AQL) Arcsin, 171 Arcsine transformation, 171 Arithmetic moving average chart, 203, 207 ARL, 154. See also Average run lengths (ARL) Array, see Orthogonal arrays ASQC, see American Society for Quality Control Assignable cause, 113, 146, 169, 173 ATI, see Average total inspection (ATI) Attributes. 29-51 control charts for, 164- I75 Average outgoing quality (AOQ), 184 Average outgoing quality limit (AOQL), 184-186 Average TUn lengths (ARL), 154,204, 206 Average total inspection (ATI), 186, 189 B B 3, B 4 • 156, 157 Back-solution, 231 Balance, 274, 342 Bayes theorem, 42 Behnken, D.H., 321-322. 331 Behrens-Fisher problem, 132 Bernoulli trials. 36-37 Beta distribution, see Distributions Beta risk., see Hypothesis testing 367 7. 156. WI.8. 218.E. 172-175 Central limit theorem.332 Capability of a process. 303 Degrees of freedom.4. 44. 41 Confidence intervals. J. 203-207 attributes. 331. 160-161 np charts. see Distrihutions Bivariate distributions.248 Box. 197 Cumulative distribution function. Jr. 113 arithmetic average. 336 Coin tossing model. 187 Characteristic life. 207. 106. 113.279-280 Demerit..326.. 98. 60 Charts. 342 Deciles. see Distributions Church. 77. 165-166 p charts. see Distributions Chain. 159 CPU (upper capability level). 209-213 EWMA. 203. 67-69. A. 98. 213-214 I charts. 10. see Analysis of covariance Covariate. see Cumulative sum control chart D d1• d2 (for double sampling). see Distributions Continuous variahles. 158-160 Causality. 321-322. 146-156 Chi-square distribution. 164-171 Binomial distrihution. 104. 138. 268 Conditional density functions. 203. 230. 267 CPL (lower capability level). 67 Conditional probabilities. 284. 52-69 Contrast. 2-5. assignable. 54 Cumulative sum control chart. Cpk. 107 Binomial charts. See also Defining contrast Control lines. 189 148. 207. 31 I.E. 29. 276 BLUE (best linear unbiased estimate). 178 Con1ounded. 148-149 Daniel.C..P .• see Plackett. 164 Defective. W.7. 156 x-bar charts.332 C4. 174 Deming. 218 analysis of. 342 Box-Behnken designs. 203. See also Standard deviation dJ. 297. see Assignable cause Cause-and-effect diagram. 94 Density functions. 228. C3. C. 251 Covariance. 164-175 binomial. 147 D3• D4 . 164-171 c-charts.• 7. 298. 148-151 s charts.159. 292. 33 Concomitant variahles. alias.368 INDEX Biased estimates. G. 157 Gp.232 Confirmation experiment. 104. 1. O. 214-215 geometric average. 331 DeMoivre. 159. 293 CoctIicient of variation. see Box-and-whisker plots Box-and-whisker plots. 114. RS.L. 340 Conforming. 126..13 Defect. 101 Burman.P .248. 297 Constant failure rate. 164. 295-296 Davies. 321-322 Boxplot. 22-25. 295. 288. 314-315.90-97. 13-14 c-charts. 122. 209-213 Cusum chart.. see Hypothesis testing CSP (continuous sampling plans). 159 Critical region. 72.80 CFR (constant failure rate) model. 190 d2 (for control charts).• see Degrees of freedom Discrete variahles. 81. 29-5\ . A. 217. 203. 66 Blocking. c CI. 178 Defining contrast.146-147 Correlation coefficient. see Probability density functions Deviation. 88.I72-175 cumulative sum. 298 Chain sampling plans.317.218 Cause. Q.274. 167-170 R charts. 284 Hurdle.25 Homoscedasticity. \08. 214-215 Exponential model.182 Student's 1.. 197 Dodge-Romig sampling plans.265 Poisson.201 E EtIect. 30 WeibulJ. 280 Farrell. 112 uniform (discrete). 102. F. 207. fitting by regression. 235. 203.. 3. 7. 177. 30.S . 55. 238 Duckworth test.72-87. 78. 61-(l4. 235-237 Hypergeomelric distribution. N.94. 116 heta risk..264. 255 F Factor. see Exponentially weighted moving average chart Expectation. 115-116 . 181 critical region. 318.63-65. 181 alternative hypothesis. 190 Draper.189. 37-38. 217.R.256-257 Gosset. 52. 5.25. 73. 227 Horizontal V-mask. 113. see Distributions. 303 four-level factors. H. 81. 14 Fisher. 45-46 hypergeomctric. 238 exponential. 284 Hunter. 296 Fishbone diagram. 1. 75.267.140. 139 . 113-127. 141. see Distributions Hyper-Graeco-Latin squares. 59 Hinge. 285-289. see Distributions Gauss-Markov theorem. 285 G Galton. 125. 186. 19. 89. E. 186. see lnterquartile range Hunter. 285. 60-61 Dodge. 237-238 Extrapolation. 136 Duncan. 124. 52. 82-83 multinomial. 46-47.• 214-215.113 Exponential distribution.213-214 Goodness-of-fit. 100. 186-187. 15.. set' Distributions Gaussian distribution. 62. 23 Finney.. 71. 18. DJ. 123.90. W. 129. 197. 293 Hyperhola. J. 100 Graeco-Latin squares. 85. see Distributions Geometric moving average chart 203. 30 normal. See also Tests acceptance region. 96. 125-127. I Fence. 324-325 Fractional factorial experiments. 33-37. lOt. 71. Sir Ronald A.G. 120.69.102. 115.. Il8. 80. 292 EWMA. AJ. 56-60. normal geometric..342-343 foldover designs.369 INDEX Distrihutions: beta.R. 13-21. 112 binomial. 66. ("Student"). W. 82.83.l37 gamma. 100 uniform (continuous).129. 130 Double sampling. 220 Gamma distribution.182 chi-square. fitting by regression. 71. 77. 273 Factorial experiments. 81-82. 248 Geometric distribution. 189 Dotplot. 40. ]]2 Gaussian. 212 Houghton. 114. 13 H Half-normal plOL 295 Hazard rate.172-176.. minimum. Il5. 324 Hypothesis testing. 271. 207.S. see Distributions Exponentially weighted moving average chart. 115-116 alpha risk. 46--49. 22-23 Histogram. 72. 147.F. Sir Francis. 85. 31. 89 lognormal. 113. 6 H-spread. 323-324 Graphs. J. 332 MIL-STD·105D.101 Mean square. 177. see Mean square . 24 K Kaizen. 284-285 Joint density function. 64-66.78. t 16-117 power.58. 76-7'd Moving average charts. 287. 199 reduced. Wilcoxon's test Marginal probabilities. 256 Mean square error. 75 estimators. I. 323 Lattices.319 INDEX LeL (lower control limit). 12. 303 Interaction plot. 18.71. 332 Median. 183 LTPD. 118. 64-66.337-338 L(27). 251. 276. 63-65. K. 83.. 77.• 2.. 189-201 MIL·STD-414. 327 L(18). 209 M Mann. see Stem-and-Ieaf diagrams Least squares.. 120 I I charts. 245 Lifetime. 318-319 L(8). 132.299 L(9). lM.. 201 tightened. see Confidence intervals Ishikawa. 67 Juran. 199-200 Interaction. 105 Lot. 177 disposition action. 108-110 Mean. 106-107 generating function. 56-57.319. 179. 59-61.81 Indifference quality. 174 Linear estimator. 4 Kurtosis.312 L(2).218.M. 121 type I error. 320 L(l6). 76-78 Moments. 300-301.201 MIL-STD-1235B. see Tests. 80. 203 MS. PW. 115 type 11 error. 70. 71 Mathematical expectation. see Limiting quality (LQ) LSD (least significant difference).. 76-77. 115-116 sample size. 337 Inspection level.22-23. 179 sentencing. 68. 299. 74 Linear combinations of random variables. 156.M . 106-107 Moment generating function. 304. 78. 159. 120 rejection region. 134. 45.71-73. 249 Modal interval.75.370 Hypothesis testing (Cofllinued) null hypothesis. 13 J John. 6. 41 Independent variables. 122. 71.TPD) Lucas. 22 Interval estimates. 223. H. 87.318 L(36). 199-200 switching rules. 312. 304.18. 287 Interquarti1e range. 289. Lord. see Expectation Maximum likelihood estimates. 89 Likelihood ratio. 160-161 Independent events. 187. 87 L Latin squares. 180 LQ. 14-15. 67-69. 115. 297. t70-t71 Leaf. 198 Inner arrays. 197. 15 Moment estimators. 15. 179 Lot tolerance percent defective (LTPO). 9-10. 243. 198 Limits. 4 Kelvin. 166. 261.72. 70. 300-301. 191 Limiting quality (LQ). see Lot tolerance percent defective (I. method of. 267 LSL (lower specification limit). 158. II. 197 Minitab. 45. 199 normal. 101. 209 Nelson. 295-296 Normal scores.371 INDEX Multinomial distribution. 171 Random samples. 342 One-sided intervals. 220. see Distrihutions Multiple sampling plans. 117 One-way layout. 271 Quartiles. 271 Quantitative factor. 287 Point estimates. K. 18. 277 Nonconforming. 297. E. 125 Percentiles.. 146-147 R charts. 213 . 74 Power. 209 Parameter. 179. 274 Ott. 285 Nonparametric tests.S .• 171. 277 Outer arrays. 79 Random variable.. 93 One-sided tests. S. 88. 271 Pooled estimate of variance. 209 Paired !-test. 129 Normal equations. &3-85 np-charts. 13 Performance characteristic. 177 MVUE (minimum variance unbiased estimate). 304. 19& Orthogonal arrays. 113 Parameter design. see Rejection number Rectification. 167-170 Pearson.58 R Radians. see Hypothesis testing Reliability function. 178 Nonconformity.. 255. 184. 13. 52 conditional density functions. 337 Outliers. 147. 333 Plackett. 88-90. 178 Nonorthogonal. R. E. 73. 10. 271 orthogonal. 223. 24 P charts. 204. 320 Plot: as an experimental unit. See also Interquartile range Rank tests.296 p Page. see Distributions Normal plots. 104.R. see Operating Characteristic curve Octane blending. 164.W. 264 interaction. E. 165-166 Null hypothesis. 121 One-at-a-time design. 258-259 Ogive. see Hypothesis testing o DC.90 Poisson distribution. 172 Nondefective. P. see Tests Rational subh'fOUP.. 303-304. 59 Residual plots. 247 Normal distribution. 233 Residuals. 83. 67 Process capability. 169. 148-151 Re. 290 Rejection number. 231 Resolution of a fractional factorial. 16. 186 Regression.S . J. See also Lattices Orthogonality. 265 Operating Characteristic cu rve. 253 Qualitative factor. 179 Rejection region. 7. see Distributions Polynomials: filting by regression. 270.S. 224.. 11-12. 105 N Nelson. 120. 91 Nonadditivity. 253. lSI! Neyman. filting by regression. L. see Capability of a process Q Quadratic blending model..340 Roberts.• 91 Pearson.R. 333-340 Pareto diagrams.S . 164. 259 Quadratic model. 133 Parabolic masks.22. 30 Range.. 131. 53.294 Population.. see Hypothesis testing Probability density functions. I01 Tolerance design. 20-22. see Hypothesis testing u ueL (upper cOlltrollimit). 187.203. 126 Tics in rank tests. 122. 8. 335 Single sampling plans. S. 15 Statistic. 158. 104.'iQC Z1.229 Runs: experimental points. 273 Type 1 error. 177-178. 333. 284-286. 154.. 187 Smith. see Normal scores Screening experiments. 229. 235.133.G.206.201 MIL-STD-1235B. 197. 197 Transformations. see Tests Tukey.230. 137.177. 204 s Sam pIc.203 Stratification.12 t-tcst. 174. 301-302 Sequential probability ratio lest.. 264 Trials. l.CJ. stopping.. 76. IS6.179.. E. see Bernoulli trials Trimmed mean.. 197 ANSIIASQC Z1.G..264 Wilcoxon's test. 197 skip-lot sampling plans. 131. see Hypothesis testing Type II crror.N. 39 Sampling plans: ANSI/A. H .4. 123. 332-333 Robust statistics. Wilcoxon's test paired [-test. 74. 187. 18 TRMEAN.214 Signal-ta-noise ratios. 31 chi-squarc test. 1.332 Tests. H. Wilcoxon's test sequential probability ratio test.191-197 Seven tools. 264 Mann-Whitney test. 134-136. 17. IS8.. 177.D . 12.189-201 MIL-STD-414. 13 Target value. normal. 98. see Tests. 190 Sampling: without replacement. see Test~. 129. 258 SchiIling. 171 Treatment. 119-120. 145. 238 Snee. 187 Scatter plots.136 2 X 2 tahles. 165. 187 CSP (continuous sampling plans). 134 TINT. 10 Sample size.. see Stopping rules for control charts Rules for switching between levels of inspection.251. 21-22. Ishikawa's. W.9. 129.4. 12 Stem-and-Ieaf diagrams. 183.W. H. 258. 331-343 'lally sheets.332 z-test. 9-10. IS6 Scheffe. 13 Student's distribution. M. 187.. 13. 159.218 ~ charts. 129 Duckworth test.372 Robust design. 340 lorrey. 191-193. 197. 87 Skip-lot sampling plans. II. chi-square test for. see Distributions Sum of squares. 190 Skewness. 133 rank tests.171 . 189 Rule. 136 F-test. 165. 277 Standard deviation. see Tests Sequcntial sampling. J3 Shewhart. 101. 265 of points Oil quality control charts. 264 Symmetric.201 chain sampling plans. 131 Two-sided intervals.7. 18 Romig. 92 Two-way layout. 197 Dodge-Romig sampling plans. 332-333 T Taguchi.76 System design. 179-187.209 t-test. 9. 114 INOEX Slatgraphics. 186 Scores. 125-127. 122. 39 with replacemenl. R.228.297. 186 189 • MIL-STD-105D.• 217. 25 Stopping rules for control charts. 141 Tho-sample [-test. 201 Rule of thumb. 373 INUEX Unbiased estimators. see Distributions Whiskers. 132 Behrens-Fisher test. 123.71. 32-69 discrete. 268 continuous.203 Weibull distribution. 292 z Zero-defects. 217 x x-bar. 159 v Variables: concomitant. Fo. 147 x-barchart~. Wilcoxon's test Wilcoxon's test. 97-102. coefficient of.38.. see Tests . 103 USL (upper specification limit). 97-102.294 estimation of..~ee Distributions Upper confidence intervals. 336 V-mask. 90 x-bar-har. 131. .342 Yates' algorithm.3 11. 131.64. 10. 29.294 Variation. 7. F. 93. 284. 75. 146-156 y Yates. see Box-and-whisker plots Whitney. 105 Unequal variances.R. 9. 132 F-tests for. 189 Warning lines. see Tests Wood. 154. 29-51 Variance.292. see Tests. 137 Uniform distribution. 137. 336 use as SN ratio. A. 209 W Wald. 78. D. 178 z-test. 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