Solution Manual for Design and Analysis of Experiments, 10th Edition

Solution Manual for Design and Analysis of Experiments, 10th Edition Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 1-1 Chapter 1 Introduction Solutions 1.1S. Suppose that you want to design an experiment to study the proportion of unpopped kernels of popcorn. Complete steps 1-3 of the guidelines for designing experiments in Section 1.4. Are there any major sources of variation that would be difficult to control? Step 1 – Recognition of and statement of the problem. Possible problem statement would be – find the best combination of inputs that maximizes yield on popcorn – minimize unpopped kernels. Step 2 – Selection of the response variable. Possible responses are number of unpopped kernels per 100 kernals in experiment, weight of unpopped kernels versus the total weight of kernels cooked. Step 3 – Choice of factors, levels and range. Possible factors and levels are brand of popcorn (levels: cheap, expensive), age of popcorn (levels: fresh, old), type of cooking method (levels: stovetop, microwave), temperature (levels: 150C, 250C), cooking time (levels: 3 minutes, 5 minutes), amount of cooking oil (levels, 1 oz, 3 oz), etc. 1.2. Suppose that you want to investigate the factors that potentially affect cooked rice. (a) What would you use as a response variable in this experiment? How would you measure the response? (b) List all of the potential sources of variability that could impact the response. (c) Complete the first three steps of the guidelines for designing experiments in Section 1.4. Step 1 – Recognition of and statement of the problem. Step 2 – Selection of the response variable. Step 3 – Choice of factors, levels and range. 1.3. Suppose that you want to compare the growth of garden flowers with different conditions of sunlight, water, fertilizer and soil conditions. Complete steps 1-3 of the guidelines for designing experiments in Section 1.4. Step 1 – Recognition of and statement of the problem. Step 2 – Selection of the response variable. Step 3 – Choice of factors, levels and range. 1.4. Select an experiment of interest to you. Complete steps 1-3 of the guidelines for designing experiments in Section 1.4. Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 1-2 1.5. Search the World Wide Web for information about Sir Ronald A. Fisher and his work on experimental design in agricultural science at the Rothamsted Experimental Station. Sample searches could include the following: 1.6. Find a Web Site for a business that you are interested in. Develop a list of factors that you would use in an experimental design to improve the effectiveness of this Web Site. 1.7. Almost everyone is concerned about the rising price of gasoline. Construct a cause and effect diagram identifying the factors that potentially influence the gasoline mileage that you get in your car. How would you go about conducting an experiment to determine any of these factors actually affect your gasoline mileage? 1.8. What is replication? Why do we need replication in an experiment? Present an example that illustrates the differences between replication and repeated measures. Repetition of the experimental runs. Replication enables the experimenter to estimate the experimental error, and provides more precise estimate of the mean for the response variable. 1.9S. Why is randomization important in an experiment? To assure the observations, or errors, are independently distributed randome variables as required by statistical methods. Also, to “average out” the effects of extraneous factors that might occur while running the experiment. 1.10S. What are the potential risks of a single, large, comprehensive experiment in contrast to a sequential approach? The important factors and levels are not always known at the beginning of the experimental process. Even new response variables might be discovered during the experimental process. By running a large comprehensive experiment, valuable information learned early in the experimental process can not likely be incorporated in the remaining experimental runs. Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 1-3 Experimental runs can be expensive and time consuming. If an error were to occur while running the experiment, the cost of redoing the experiment is much more manageable with one of the small sequential experiments than the large comprehensive experiment. Reserved Problems 1.1R. Have you received an offer to obtain a credit card in the mail? What “factors” were associated with the offer, such as introductory interest rate? Do you think the credit card company is conducting experiments to investigate which facors product the highest positive response rate to their offer? What potential factors in the experiment can you identify? Interest rate, credit limit, old credit card pay-off amount, interest free period, gift points, others. 1.2R. What factors do you think an e-commerce company could use in an experiment involving their web page to encourage more people to “click-through” into their site? Font size, font type, images/icons, color, spacing, animation, sound/music, speed, others. 1.3R. Two of the leading contributors to design of experiments over the last 60 years were George E. P. Box and J. Stuart Hunter. Search the World Wide Web for information on these two individuals and briefly summarize their contributions. 1.4R. Suppose that you want to make brownies. You plan to use a brownie mix, but there are a number of factors that could impact the results. a. What would you use as a response variable? Could there be more than one response? Taste would probably be the primary factor. Other factors could be texture and aroma. Possibly a combination of all three. b. Identify the factors that might impact the results. Amount of oil, number of eggs, amount of water, cost of mix – expensive or cheap, baking temperature, baking time, type of pan – glass or metal. c. Complete the first three steps of the guidelines for designing experiments in Section 1-4. 1. Problem Statement – To maximize brownie tastiness. Or to make the best brownie while minimizing cost (Can a cheap mix give the results of an expensive one?) 2. Response Variable – Tastiness is the primary response. Secondary responses could also be texture and aroma. 3. Choice of factors, levels and range – for the maximize brownie tastiness: Mix cost (cheap, expensive), number of eggs (2,3), amount of oil (1/2 cup, ¾ cup), pan type (glass, metal), oven temp (350, 375), bake time (35 min, 45 min). One might want to reduce the number of factors from 6 to 3-4 to reduce the number of experimental runs. Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-1 Chapter 2 Simple Comparative Experiments Solutions 2.1. Computer output for a random sample of data is shown below. Some of the quantities are missing. Compute the values of the missing quantities. Variable N Mean SE Mean Std. Dev. Variance Minimum Maximum Y 9 19.96 ? 3.12 ? 15.94 27.16 SE Mean = 1.04 Variance = 9.73 2.2S. Computer output for a random sample of data is shown below. Some of the quantities are missing. Compute the values of the missing quantities. Variable N Mean SE Mean Std. Dev. Sum Y 16 ? 0.159 ? 399.851 Mean = 24.991 Std. Dev. = 0.636 2.3. Suppose that we are testing H0: µ = µ0 versus H1: µ ≠ µ0. Calculate the P-value for the following observed values of the test statistic: (a) Z0 = 2.25 P-value = 0.02445 (b) Z0 = 1.55 P-value = 0.12114 (c) Z0 = 2.10 P-value = 0.03573 (d) Z0 = 1.95 P-value = 0.05118 (e) Z0 = -0.10 P-value = 0.92034 2.4S. Suppose that we are testing H0: µ = µ0 versus H1: µ > µ0. Calculate the P-value for the following observed values of the test statistic: (a) Z0 = 2.45 P-value = 0.00714 (b) Z0 = -1.53 P-value = 0.93699 Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-2 (c) Z0 = 2.15 P-value = 0.01578 (d) Z0 = 1.95 P-value = 0.02559 (e) Z0 = -0.25 P-value = 0.59871 2.5. Suppose that we are testing H0: µ1 = µ2 versus H1: µ1 = µ2 with a sample size of n1 = n2 = 12. Both sample variances are unknown but assumed equal. Find bounds on the P-value for the following observed values of the test statistic: (a) t0 = 2.30 Table P-value = 0.02, 0.05 Computer P-value = 0.0313 (b) t0 = 3.41 Table P-value = 0.002, 0.005 Computer P-value = 0.0025 (c) t0 = 1.95 Table P-value = 0.1, 0.05 Computer P-value = 0.0640 (d) t0 = -2.45 Table P-value = 0.05, 0.02 Computer P-value = 0.0227 Note that the degrees of freedom is (12 +12) – 2 = 22. This is a two-sided test 2.6. Suppose that we are testing H0: µ1 = µ2 versus H1: µ1 > µ2 with a sample size of n1 = n2 = 10. Both sample variances are unknown but assumed equal. Find bounds on the P-value for the following observed values of the test statistic: (a) t0 = 2.31 Table P-value = 0.01, 0.025 Computer P-value = 0.01648 (b) t0 = 3.60 Table P-value = 0.001, 0.0005 Computer P-value = 0.00102 (c) t0 = 1.95 Table P-value = 0.05, 0.025 Computer P-value = 0.03346 (d) t0 = 2.19 Table P-value = 0.01, 0.025 Computer P-value = 0.02097 Note that the degrees of freedom is (10 +10) – 2 = 18. This is a one-sided test. 2.7. Consider the following sample data: 9.37, 13.04, 11.69, 8.21, 11.18, 10.41, 13.15, 11.51, 13.21, and 7.75. Is it reasonable to assume that this data is from a normal distribution? Is there evidence to support a claim that the mean of the population is 10? Minitab Output Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-3 13 Median Mean 1st Q uartile 9.080 Median 11.345 3rd Q uartile 13.067 Maximum 13.210 9.526 12.378 8.973 13.078 1.371 3.639 A -Squared 0.33 P-V alue 0.435 Mean 10.952 StDev 1.993 V ariance 3.974 Skewness -0.45131 Kurtosis -1.06746 N 10 Minimum 7.750 A nderson-Darling Normality Test 95% C onfidence Interv al for Mean 95% C onfidence Interv al for Median 95% C onfidence Interv al for StDev 95% Confidence Intervals Summary for Sample Data According to the output, the Anderson-Darling Normality Test has a P-Value of 0.435. The data can be considered normal. The 95% confidence interval on the mean is (9.526,12.378). This confidence interval contains 10, therefore there is evidence that the population mean is 10. 2.8. A computer program has produced the following output for the hypothesis testing problem: Difference in sample means: 2.35 Degrees of freedom: 18 Standard error of the difference in the sample means: ? Test statistic: to = 2.01 P-Value = 0.0298 Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-4 (a) What is the missing value for the standard error? 1 2 0 1 2 2.35 2.01 1 1 2.35 / 2.01 1.169 p y y t StdError S n n StdError        (b) Is this a two-sided or one-sided test? One-sided test for a t0 = 2.01 is a P-value of 0.0298. (c) If =0.05, what are your conclusions? Reject the null hypothesis and conclude that there is a difference in the two samples. (d) Find a 90% two-sided CI on the difference in the means.     1 2 1 2 1 2 2, 2 1 1 1 2 2, 2 1 2 1 2 1 2 0.05,18 1 1 1 2 0.05,18 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 1 2.35 1.734 1.169 2.35 1.734 1.169 0.323 4.377 n n p n n p p p y y t S y y t S n n n n y y t S y y t S n n n n                                         2.9S. A computer program has produced the following output for the hypothesis testing problem: Difference in sample means: 11.5 Degrees of freedom: 24 Standard error of the difference in the sample means: ? Test statistic: to = -1.88 P-Value = 0.0723 (a) What is the missing value for the standard error? 1 2 0 1 2 11.5 1.88 1 1 11.5 / 1.88 6.12 p y y t StdError S n n StdError            (b) Is this a two-sided or one-sided test? Two-sided test for a t0 = -1.88 is a P-value of 0.0723. (c) If =0.05, what are your conclusions? Accept the null hypothesis, there is no difference in the means. Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-5 (d) Find a 90% two-sided CI on the difference in the means.     1 2 1 2 1 2 2, 2 1 1 1 2 2, 2 1 2 1 2 1 2 0.05,24 1 1 1 2 0.05,24 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 1 11.5 1.711 6.12 11.5 1.711 6.12 21.97 1.03 n n p n n p p p y y t S y y t S n n n n y y t S y y t S n n n n                                             2.10. A two-sample t-test has been conducted and the sample sizes are n1 = n2 = 10. The computed value of the test statistic is t0 = 2.15. If the null hypothesis is two-sided, an upper bound on the P-value is (a) 0.10 (b) 0.05 (c) 0.025 (d) 0.01 (e) None of the above. 2.11. A two-sample t-test has been conducted and the sample sizes are n1 = n2 = 12. The computed value of the test statistic is t0 = 2.27. If the null hypothesis is two-sided, an upper bound on the P-value is (a) 0.10 (b) 0.05 (c) 0.025 (d) 0.01 (e) None of the above. 2.12S. Suppose that we are testing H0: µ = µ0 versus H1: µ > µ0 with a sample size of n = 15. Calculate bounds on the P-value for the following observed values of the test statistic: (a) t0 = 2.35 Table P-value = 0.01, 0.025 Computer P-value = 0.01698 (b) t0 = 3.55 Table P-value = 0.001, 0.0025 Computer P-value = 0.00160 (c) t0 = 2.00 Table P-value = 0.025, 0.005 Computer P-value = 0.03264 Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-6 (d) t0 = 1.55 Table P-value = 0.05, 0.10 Computer P-value = 0.07172 The degrees of freedom are 15 – 1 = 14. This is a one-sided test. 2.13. Suppose that we are testing H0: µ = µ0 versus H1: µ ≠ µ0 with a sample size of n = 10. Calculate bounds on the P-value for the following observed values of the test statistic: (a) t0 = 2.48 Table P-value = 0.02, 0.05 Computer P-value = 0.03499 (b) t0 = -3.95 Table P-value = 0.002, 0.005 Computer P-value = 0.00335 (c) t0 = 2.69 Table P-value = 0.02, 0.05 Computer P-value = 0.02480 (d) t0 = 1.88 Table P-value = 0.05, 0.10 Computer P-value = 0.09281 (e) t0 = -1.25 Table P-value = 0.20, 0.50 Computer P-value = 0.24282 2.14. Consider the computer output shown below. One-Sample T: Y Test of mu = 25 vs > 25 Variable N Mean Std. Dev. SE Mean 95% Lower Bound T P Y 12 25.6818 ? 0.3360 ? ? 0.034 (a) How many degrees of freedom are there on the t-test statistic? (N-1) = (12 – 1) = 11 (b) Fill in the missing information. Std. Dev. = 1.1639 95% Lower Bound = 2.0292 2.15. Consider the computer output shown below. Two-Sample T-Test and CI: Y1, Y2 Two-sample T for Y1 vs Y2 N Mean Std. Dev. SE Mean Y1 20 50.19 1.71 0.38 Y2 20 52.52 2.48 0.55 Difference = mu (X1) – mu (X2) Estimate for difference: -2.33341 95% CI for difference: (-3.69547, -0.97135) T-Test of difference = 0 (vs not = ) : T-Value = -3.47 P-Value = 0.01 DF = 38 Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-7 Both use Pooled Std. Dev. = 2.1277 (a) Can the null hypothesis be rejected at the 0.05 level? Why? Yes, the P-Value of 0.001 is much less than 0.05. (b) Is this a one-sided or two-sided test? Two-sided. (c) If the hypothesis had been H0: µ1 - µ2 = 2 versus H1: µ1 - µ2 ≠ 2 would you reject the null hypothesis at the 0.05 level? Yes. (d) If the hypothesis had been H0: µ1 - µ2 = 2 versus H1: µ1 - µ2 < 2 would you reject the null hypothesis at the 0.05 level? Can you answer this question without doing any additional calculations? Why? Yes, no additional calculations are required because the test is naturally becoming more significant with the change from -2.33341 to -4.33341. (e) Use the output and the t table to find a 95 percent upper confidence bound on the difference in means? 95% upper confidence bound = -1.21. (f) What is the P-value if the alternative hypotheses are H0: µ1 - µ2 = 2 versus H1: µ1 - µ2 ≠ 2? P-value = 1.4E-07. 2.16. The breaking strength of a fiber is required to be at least 150 psi. Past experience has indicated that the standard deviation of breaking strength is  = 3 psi. A random sample of four specimens is tested. The results are y1 =145, y2 =153, y3 =150 and y4 =147. (a) State the hypotheses that you think should be tested in this experiment. H0 :  = 150 H1 :  > 150 (b) Test these hypotheses using  = 0.05. What are your conclusions? n = 4,  = 3, y = 1/4 (145 + 153 + 150 + 147) = 148.75           148.75 150 1.25 0.8333 3 3 4 2 o o y z n Since z 0.05 = 1.645, do not reject. (c) Find the P-value for the test in part (b). Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-8 From the z-table:             P 1 0.7967 2 3 0.7995 0.7967 0.2014 (d) Construct a 95 percent confidence interval on the mean breaking strength. The 95% confidence interval is                     2 2 148.75 1.96 3 2 148.75 1.96 3 2 y z y z n n 145.81 151.69    2.17. The viscosity of a liquid detergent is supposed to average 800 centistokes at 25C. A random sample of 16 batches of detergent is collected, and the average viscosity is 812. Suppose we know that the standard deviation of viscosity is  = 25 centistokes. (a) State the hypotheses that should be tested. H0 :  = 800 H1 :   800 (b) Test these hypotheses using  = 0.05. What are your conclusions?         .92 25 25 16 4 o o y z n Since z/2 = z 0.025 = 1.96, do not reject. (c) What is the P-value for the test? (d) Find a 95 percent confidence interval on the mean. The 95% confidence interval is          2 2 y z y z n n                    812 1..96 25 4 812 12.25 812 12.25 799.75 824.25 2.18. A normally distributed random variable has an unknown mean  and a known variance  2 = 9. Find the sample size required to construct a 95 percent confidence interval on the mean that has total length of 1.0. Since y  N(,9), a 95% two-sided confidence interval on  is If the total interval is to have width 1.0, then the half-interval is 0.5. Since z/2 = z 0.025 = 1.96, Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-9                           2 3 1.96 0.5 1.96 3 1.96 11.76 0.5 11.76 138.30 139 n n n 2.19S. The shelf life of a carbonated beverage is of interest. Ten bottles are randomly selected and tested, and the following results are obtained: Days 108 138 124 163 124 159 106 134 115 139 (a) We would like to demonstrate that the mean shelf life exceeds 120 days. Set up appropriate hypotheses for investigating this claim. H0 :  = 120 H1 :  > 120 (b) Test these hypotheses using  = 0.01. What are your conclusions? y = 131 S 2 = 3438 / 9 = 382 S   382 19.54       0 0 131 120 1.78 19.54 10 y t S n since t 0.01,9 = 2.821; do not reject H0 Minitab Output T-Test of the Mean Test of mu = 120.00 vs mu > 120.00 Variable N Mean StDev SE Mean T P Shelf Life 10 131.00 19.54 6.18 1.78 0.054 T Confidence Intervals Variable N Mean StDev SE Mean 99.0 % CI Shelf Life 10 131.00 19.54 6.18 ( 110.91, 151.09) (c) Find the P-value for the test in part (b). P=0.054 (d) Construct a 99 percent confidence interval on the mean shelf life. The 99% confidence interval is       , 1 , 1    2 2 n n S S y t y t n n with  = 0.01.                      19.54 19.54 131 3.250 131 3.250 10 10 Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-10 110.91 151.08    2.20. Consider the shelf life data in Problem 2.19S. Can shelf life be described or modeled adequately by a normal distribution? What effect would violation of this assumption have on the test procedure you used in solving Problem 2.22? A normal probability plot, obtained from Minitab, is shown. There is no reason to doubt the adequacy of the normality assumption. If shelf life is not normally distributed, then the impact of this on the t-test in problem 2.19 is not too serious unless the departure from normality is severe. 2.21. The time to repair an electronic instrument is a normally distributed random variable measured in hours. The repair time for 16 such instruments chosen at random are as follows: Hours (a) You wish to know if the mean repair time exceeds 225 hours. Set up appropriate hypotheses for investigating this issue. H0 :  = 225 H1 :  > 225 (b) Test the hypotheses you formulated in part (a). What are your conclusions? Use  = 0.05. y = 241.50 S 2 =146202 / (16 - 1) = 9746.80 S   9746.8 98.73 P-Value: 0.606 A-Squared: 0.266 Anderson-Darling Normality Test N: 10 StDev: 19.5448 Average: 131 155 165 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Shelf Life Normal Probability Plot Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-11       241.50 225 0.67 98.73 16 o o y t S n since t 0.05,15 = 1.753; do not reject H0 Minitab Output T-Test of the Mean Test of mu = 225.0 vs mu > 225.0 Variable N Mean StDev SE Mean T P Hours 16 241.5 98.7 24.7 0.67 0.26 T Confidence Intervals Variable N Mean StDev SE Mean 95.0 % CI Hours 16 241.5 98.7 24.7 ( 188.9, 294.1) (c) Find the P-value for this test. P=0.26 (d) Construct a 95 percent confidence interval on mean repair time. The 95% confidence interval is       , 1 , 1    2 2 n n S S y t y t n n                      98.73 98.73 241.50 2.131 241.50 2.131 16 16 188.9 294.1    2.22. Reconsider the repair time data in Problem 2.21. Can repair time, in your opinion, be adequately modeled by a normal distribution? The normal probability plot below does not reveal any serious problem with the normality assumption. P-Value: 0.163 A-Squared: 0.514 Anderson-Darling Normality Test N: 16 StDev: 98.7259 Average: 241.5 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Hours Normal Probability Plot Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-12 2.23. Two machines are used for filling plastic bottles with a net volume of 16.0 ounces. The filling processes can be assumed to be normal, with standard deviation of 1 = 0.015 and 2 = 0.018. The quality engineering department suspects that both machines fill to the same net volume, whether or not this volume is 16.0 ounces. An experiment is performed by taking a random sample from the output of each machine. Machine 1 Machine 2 16.03 16.01 16.02 16.03 16.04 15.96 15.97 16.04 16.05 15.98 15.96 16.02 16.05 16.02 16.01 16.01 16.02 15.99 15.99 16.00 (a) State the hypotheses that should be tested in this experiment. H0 : 1 = 2 H1 : 1  2 (b) Test these hypotheses using =0.05. What are your conclusions?     1 1 1 16.015 0.015 10 y n     2 2 2 16.005 0.018 10 y n 1 2 2 2 2 2 1 2 1 2 16.015 16.018 1.35 0.015 0.018 10 10 o y y z n n          z 0.025 = 1.96; do not reject (c) What is the P-value for the test? P = 0.1770 (d) Find a 95 percent confidence interval on the difference in the mean fill volume for the two machines. The 95% confidence interval is 2 2 2 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 y y z y y z n n n n                  2 2 2 2 1 2 0.015 0.018 0.015 0.018 (16.015 16.005) (1.96) (16.015 16.005) (1.96)                0.0045 0.0245   1 2 2.24. Two types of plastic are suitable for use by an electronic calculator manufacturer. The breaking strength of this plastic is important. It is known that 1 = 2 = 1.0 psi. From random samples of n1 = 10 and n2 = 12 we obtain y 1 = 162.5 and y 2 = 155.0. The company will not adopt plastic 1 unless its breaking strength exceeds that of plastic 2 by at least 10 psi. Based on the sample information, should they use plastic 1? In answering this questions, set up and test appropriate hypotheses using  = 0.01. Construct a 99 percent confidence interval on the true mean difference in breaking strength. H0 : 1 - 2 =10 H1 : 1 - 2 >10 Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-13 1 1 1 162.5 1 10 y n     2 2 2 155.0 1 10 y n                 1 2 2 2 2 2 1 2 1 2 10 162.5 155.0 10 5.84 1 1 10 12 o y y z n n z 0.01 = 2.325; do not reject The 99 percent confidence interval is 2 2 2 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 y y z y y z n n n n                  2 2 2 2 1 2 1 1 1 1 (162.5 155.0) (2.575) (162.5 155.0) (2.575)            1 2 6.40 8.60      2.25. The following are the burning times (in minutes) of chemical flares of two different formulations. The design engineers are interested in both the means and variance of the burning times. Type 1 Type 2 (a) Test the hypotheses that the two variances are equal. Use  = 0.05. 2 2 0 1 2 2 2 1 1 2 : : H H       Do not reject. (b) Using the results of (a), test the hypotheses that the mean burning times are equal. Use  = 0.05. What is the P-value for this test? Do not reject. From the computer output, t=0.05; do not reject. Also from the computer output P=0.96 Minitab Output Two Sample T-Test and Confidence Interval Two sample T for Type 1 vs Type 2 N Mean StDev SE Mean Type 1 10 70.40 9.26 2.9 Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-14 Type 2 10 70.20 9.37 3.0 95% CI for mu Type 1 - mu Type 2: ( -8.6, 9.0) T-Test mu Type 1 = mu Type 2 (vs not =): T = 0.05 P = 0.96 DF = 18 Both use Pooled StDev = 9.32 (c) Discuss the role of the normality assumption in this problem. Check the assumption of normality for both types of flares. The assumption of normality is required in the theoretical development of the t-test. However, moderate departure from normality has little impact on the performance of the t-test. The normality assumption is more important for the test on the equality of the two variances. An indication of nonnormality would be of concern here. The normal probability plots shown below indicate that burning time for both formulations follow the normal distribution. 2.26S. An article in Solid State Technology, "Orthogonal Design of Process Optimization and Its Application to Plasma Etching" by G.Z. Yin and D.W. Jillie (May, 1987) describes an experiment to determine the effect of C2F6 flow rate on the uniformity of the etch on a silicon wafer used in integrated circuit manufacturing. Data for two flow rates are as follows: P-Value: 0.409 A-Squared: 0.344 Anderson-Darling Normality Test N: 10 StDev: 9.26403 Average: 70.4 60 70 80 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Type 1 Normal Probability Plot P-Value: 0.876 A-Squared: 0.186 Anderson-Darling Normality Test N: 10 StDev: 9.36661 Average: 70.2 60 70 80 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Type 2 Normal Probability Plot Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-15 C2F6 Uniformity Observation (SCCM) 1 2 3 4 5 6 125 2.7 4.6 2.6 3.0 3.2 3.8 200 4.6 3.4 2.9 3.5 4.1 5.1 (a) Does the C2F6 flow rate affect average etch uniformity? Use  = 0.05. No, C2F6 flow rate does not affect average etch uniformity. Minitab Output Two Sample T-Test and Confidence Interval Two sample T for Uniformity Flow Rat N Mean StDev SE Mean 125 6 3.317 0.760 0.31 200 6 3.933 0.821 0.34 95% CI for mu (125) - mu (200): ( -1.63, 0.40) T-Test mu (125) = mu (200) (vs not =): T = -1.35 P = 0.21 DF = 10 Both use Pooled StDev = 0.791 (b) What is the P-value for the test in part (a)? From the Minitab output, P=0.21 (c) Does the C2F6 flow rate affect the wafer-to-wafer variability in etch uniformity? Use  = 0.05. 2 2 0 1 2 2 2 1 1 2 0.025,5,5 0.975,5,5 0 : : 7.15 0.14 0.5776 0.86 0.6724 H H F F F           Do not reject; C2F6 flow rate does not affect wafer-to-wafer variability. (d) Draw box plots to assist in the interpretation of the data from this experiment. The box plots shown below indicate that there is little difference in uniformity at the two gas flow rates. Any observed difference is not statistically significant. See the t-test in part (a). 125 200 5 4 3 Flow Rate Uniformity Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-16 2.27. Photoresist is a light-sensitive material applied to semiconductor wafers so that the circuit pattern can be imaged on to the wafer. After application, the coated wafers are baked to remove the solvent in the photoresist mixture and to harden the resist. Here are measurements of photoresist thickness (in kÅ) for eight wafers baked at two different temperatures. Assume that all of the runs were made in random order. 95 ºC 100 ºC 11.176 5.623 7.089 6.748 8.097 7.461 11.739 7.015 11.291 8.133 10.759 7.418 6.467 3.772 8.315 8.963 (a) Is there evidence to support the claim that the higher baking temperature results in wafers with a lower mean photoresist thickness? Use  = 0.05.                             0 1 2 1 1 2 2 2 2 1 1 2 2 1 2 1 2 0 1 2 0.05,14 : : ( 1) ( 1) (8 1)(4.41) (8 1)(2.54) 3.48 2 8 8 2 1.86 9.37 6.89 2.65 1 1 1 1 1.86 8 8 1.761 p p p H H n S n S S n n S y y t S n n t Since t0.05,14 = 1.761, reject H0. There appears to be a lower mean thickness at the higher temperature. This is also seen in the computer output. Minitab Output Two-Sample T-Test and CI: Thickness, Temp Two-sample T for Thick@95 vs Thick@100 N Mean StDev SE Mean Thick@95 8 9.37 2.10 0.74 Thick@10 8 6.89 1.60 0.56 Difference = mu Thick@95 - mu Thick@100 Estimate for difference: 2.475 95% lower bound for difference: 0.833 T-Test of difference = 0 (vs >): T-Value = 2.65 P-Value = 0.009 DF = 14 Both use Pooled StDev = 1.86 (b) What is the P-value for the test conducted in part (a)? P = 0.009 (c) Find a 95% confidence interval on the difference in means. Provide a practical interpretation of this interval. Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-17 From the computer output the 95% lower confidence bound is     1 2 0.833 . This lower confidence bound is greater than 0; therefore, there is a difference in the two temperatures on the thickness of the photoresist. (d) Draw dot diagrams to assist in interpreting the results from this experiment. Thickness 3.6 4.8 6.0 7.2 8.4 9.6 10.8 12.0 Temp 95 100 Dotplot of Thickness vs Temp (e) Check the assumption of normality of the photoresist thickness. P-Value: 0.161 A-Squared: 0.483 Anderson-Darling Normality Test N: 8 StDev: 2.09956 Average: 9.36662 7 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Thick@95 Normal Probability Plot Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-18 There are no significant deviations from the normality assumptions. (f) Find the power of this test for detecting an actual difference in means of 2.5 kÅ. Minitab Output Power and Sample Size 2-Sample t Test Testing mean 1 = mean 2 (versus not =) Calculating power for mean 1 = mean 2 + difference Alpha = 0.05 Sigma = 1.86 Sample Difference Size Power 2.5 8 0.7056 (g) What sample size would be necessary to detect an actual difference in means of 1.5 kÅ with a power of at least 0.9?. Minitab Output Power and Sample Size 2-Sample t Test Testing mean 1 = mean 2 (versus not =) Calculating power for mean 1 = mean 2 + difference Alpha = 0.05 Sigma = 1.86 Sample Target Actual Difference Size Power Power 1.5 34 0.9000 0.9060 This result makes intuitive sense. More samples are needed to detect a smaller difference. P-Value: 0.457 A-Squared: 0.316 Anderson-Darling Normality Test N: 8 StDev: 1.59509 Average: 6.89163 4 5 6 7 8 9 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Thick@100 Normal Probability Plot Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-19 2.28S. Front housings for cell phones are manufactured in an injection molding process. The time the part is allowed to cool in the mold before removal is thought to influence the occurrence of a particularly troublesome cosmetic defect, flow lines, in the finished housing. After manufacturing, the housings are inspected visually and assigned a score between 1 and 10 based on their appearance, with 10 corresponding to a perfect part and 1 corresponding to a completely defective part. An experiment was conducted using two cool-down times, 10 seconds and 20 seconds, and 20 housings were evaluated at each level of cooldown time. All 40 observations in this experiment were run in random order. The data are shown below. 10 Seconds 20 Seconds 1 3 7 6 2 6 8 9 1 5 5 5 3 3 9 7 5 2 5 4 1 1 8 6 5 6 6 8 2 8 4 5 3 2 6 8 5 3 7 7 (a) Is there evidence to support the claim that the longer cool-down time results in fewer appearance defects? Use  = 0.05. From the analysis shown below, there is evidence that the longer cool-down time results in fewer appearance defects. Minitab Output Two-Sample T-Test and CI: 10 seconds, 20 seconds Two-sample T for 10 seconds vs 20 seconds N Mean StDev SE Mean 10 secon 20 3.35 2.01 0.45 20 secon 20 6.50 1.54 0.34 Difference = mu 10 seconds - mu 20 seconds Estimate for difference: -3.150 95% upper bound for difference: -2.196 T-Test of difference = 0 (vs <): T-Value = -5.57 P-Value = 0.000 DF = 38 Both use Pooled StDev = 1.79 (b) What is the P-value for the test conducted in part (a)? From the Minitab output, P = 0.000 (c) Find a 95% confidence interval on the difference in means. Provide a practical interpretation of this interval. From the Minitab output,   1 2   2.196 . This lower confidence bound is less than 0. The two samples are different. The 20 second cooling time gives a cosmetically better housing. Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-20 (d) Draw dot diagrams to assist in interpreting the results from this experiment. Ranking 2 4 6 8 C4 10 sec 20 sec Dotplot of Ranking vs C4 (e) Check the assumption of normality for the data from this experiment. P-Value: 0.043 A-Squared: 0.748 Anderson-Darling Normality Test N: 20 StDev: 2.00722 Average: 3.35 1 2 3 4 5 6 7 8 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability 10 seconds Normal Probability Plot Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-21 There are no significant departures from normality. 2.29. The diameter of a ball bearing was measured by 12 inspectors, each using two different kinds of calipers. The results were: Inspector Caliper 1 Caliper 2 Difference Difference^2 1 0.265 0.264 0.001 0.000001 2 0.265 0.265 0 0 3 0.266 0.264 0.002 0.000004 4 0.267 0.266 0.001 0.000001 5 0.267 0.267 0 0 6 0.265 0.268 -0.003 0.000009 7 0.267 0.264 0.003 0.000009 8 0.267 0.265 0.002 0.000004 9 0.265 0.265 0 0 10 0.268 0.267 0.001 0.000001 11 0.268 0.268 0 0 12 0.265 0.269 -0.004 0.000016  0.003  0.000045 P-Value: 0.239 A-Squared: 0.457 Anderson-Darling Normality Test N: 20 StDev: 1.53897 Average: 6.5 4 5 6 7 8 9 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability 20 seconds Normal Probability Plot Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-22 (a) Is there a significant difference between the means of the population of measurements represented by the two samples? Use  = 0.05. 0 1 2 1 1 2 : : H H       or equivalently 0 1 : 0 : 0 d d H H     Minitab Output Paired T-Test and Confidence Interval Paired T for Caliper 1 - Caliper 2 N Mean StDev SE Mean Caliper 12 0.266250 0.001215 0.000351 Caliper 12 0.266000 0.001758 0.000508 Difference 12 0.000250 0.002006 0.000579 95% CI for mean difference: (-0.001024, 0.001524) T-Test of mean difference = 0 (vs not = 0): T-Value = 0.43 P-Value = 0.674 (b) Find the P-value for the test in part (a). P=0.674 (c) Construct a 95 percent confidence interval on the difference in the mean diameter measurements for the two types of calipers.  1 2  , 1 , 1 2 2 0.002 0.002 0.00025 2.201 0.00025 2.201 12 12 0.00102 0.00152 d d n n D d d S S d t d t n n                       2.30. An article in the journal of Neurology (1998, Vol. 50, pp.) observed that the monozygotic twins share numerous physical, psychological and pathological traits. The investigators measured an intelligence score of 10 pairs of twins. The data are obtained as follows: Pair Birth Order: 1 Birth Order: 2 1 6.08 5.73 2 6.22 5.80 3 7.99 8.42 4 7.44 6.84 5 6.48 6.43 6 7.99 8.76 7 6.32 6.32 8 7.60 7.62 9 6.03 6.59 10 7.52 7.67 Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-23 (a) Is the assumption that the difference in score is normally distributed reasonable? Minitab Output -0.75 -0.50 -0.25 0.00 0.25 0.50 Median Mean -0.50 -0.25 0.00 0.25 0.50 1st Q uartile -0.462500 Median -0.010000 3rd Q uartile 0.367500 Maximum 0.600000 -0.366415 0.264415 -0.474505 0.373964 0.303280 0.804947 A -Squared 0.19 P-V alue 0.860 Mean -0.051000 StDev 0.440919 V ariance 0.194410 Skewness -0.182965 Kurtosis -0.817391 N 10 Minimum -0.770000 A nderson-Darling Normality Test 95% C onfidence Interv al for Mean 95% C onfidence Interv al for Median 95% C onfidence Interv al for StDev 95% Confidence Intervals Summary for Difference By plotting the differences, the output shows that the Anderson-Darling Normality Test shows a P-Value of 0.860. The data is assumed to be normal. (b) Find a 95% confidence interval on the difference in the mean score. Is there any evidence that mean score depends on birth order? The 95% confidence interval on the difference in mean score is (-0.366415, 0.264415) contains the value of zero. There is no difference in birth order. (c) Test an appropriate set of hypothesis indicating that the mean score does not depend on birth order. 0 1 2 1 1 2 : : H H       or equivalently 0 1 : 0 : 0 d d H H     Minitab Output Paired T for Birth Order: 1 - Birth Order: 2 N Mean StDev SE Mean Birth Order: 1 10 6.967 0.810 0.256 Birth Order: 2 10 7.018 1.053 0.333 Difference 10 -0.051 0.441 0.139 95% CI for mean difference: (-0.366, 0.264) T-Test of mean difference = 0 (vs not = 0): T-Value = -0.37 P-Value = 0.723 Do not reject. The P-value is 0.723. Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-24 2.31S. An article in the Journal of Strain Analysis (vol.18, no. 2, 1983) compares several procedures for predicting the shear strength for steel plate girders. Data for nine girders in the form of the ratio of predicted to observed load for two of these procedures, the Karlsruhe and Lehigh methods, are as follows: Girder Karlsruhe Method Lehigh Method Difference Difference^2 S1/1 1.186 1.061 0.125 0.015625 S2/1 1.151 0.992 0.159 0.025281 S3/1 1.322 1.063 0.259 0.067081 S4/1 1.339 1.062 0.277 0.076729 S5/1 1.200 1.065 0.135 0.018225 S2/1 1.402 1.178 0.224 0.050176 S2/2 1.365 1.037 0.328 0.107584 S2/3 1.537 1.086 0.451 0.203401 S2/4 1.559 1.052 0.507 0.257049 Sum = 2.465 0.821151 Average = 0.274 (a) Is there any evidence to support a claim that there is a difference in mean performance between the two methods? Use  = 0.05. or equivalently   1 1 1 2.465 0.274 9 n i i d d n      1 2 2 1 2 2 2 1 1 1 1 0.821151 (2.465) 9 0.135 1 9 1 n n i i i i d d d n s n                                          0 0.274 6.08 0.135 9 d d t S n    2 , 1 0.025,8 2.306 n t t     , reject the null hypothesis. Minitab Output Paired T-Test and Confidence Interval Paired T for Karlsruhe - Lehigh N Mean StDev SE Mean Karlsruh 9 1.3401 0.1460 0.0487 Lehigh 9 1.0662 0.0494 0.0165 Difference 9 0.2739 0.1351 0.0450 95% CI for mean difference: (0.1700, 0.3777) T-Test of mean difference = 0 (vs not = 0): T-Value = 6.08 P-Value = 0.000 1 1 2 0 1 2 : :       H H 0 0 1 0   d d H : H :   Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-25 (b) What is the P-value for the test in part (a)? P=0.0002 (c) Construct a 95 percent confidence interval for the difference in mean predicted to observed load. , 1 , 1 2 2 0.135 0.135 0.274 2.306 0.274 2.306 9 9 0.17023 0.37777 d d d n n d d S S d t d t n n                  (d) Investigate the normality assumption for both samples. The normal probability plots of the observations for each method follow. There are no serious concerns with the normality assumption, but there is an indication of a possible outlier (1.178) in the Lehigh method data. P-Value: 0.537 A-Squared: 0.286 Anderson-Darling Normality Test N: 9 StDev : 0.146031 Av erage: 1.34011 1.15 1.25 1.35 1.45 1.55 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Karlsruhe Normal Probability Plot P-Value: 0.028 A-Squared: 0.772 Anderson-Darling Normality Test N: 9 StDev : 0.0493806 Av erage: 1.06622 1.00 1.05 1.10 1.15 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Lehigh Normal Probability Plot Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-26 (a) Investigate the normality assumption for the difference in ratios for the two methods. There is no issue with normality in the difference of ratios of the two methods. (b) Discuss the role of the normality assumption in the paired t-test. As in any t-test, the assumption of normality is of only moderate importance. In the paired t-test, the assumption of normality applies to the distribution of the differences. That is, the individual sample measurements do not have to be normally distributed, only their difference. 2.32. The deflection temperature under load for two different formulations of ABS plastic pipe is being studied. Two samples of 12 observations each are prepared using each formulation, and the deflection temperatures (in F) are reported below: Formulation 1 Formulation 2 198 188 189 203 (a) Construct normal probability plots for both samples. Do these plots support assumptions of normality and equal variance for both samples? P-Value: 0.464 A-Squared: 0.318 Anderson-Darling Normality Test N: 9 StDev : 0.135099 Av erage: 0.273889 0.12 0.22 0.32 0.42 0.52 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Difference Normal Probability Plot Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-27 (b) Do the data support the claim that the mean deflection temperature under load for formulation 1 exceeds that of formulation 2? Use  = 0.05. No, formulation 1 does not exceed formulation 2 per the Minitab output below. Minitab Output Two Sample T-Test and Confidence Interval N Mean StDev SE Mean Form 1 12 194.5 10.2 2.9 Form 2 12 193.08 9.95 2.9 Difference = mu Form 1 - mu Form 2 Estimate for difference: 1.42 95% lower bound for difference: -5.64 T-Test of difference = 0 (vs >): T-Value = 0.34 P-Value = 0.367 DF = 22 Both use Pooled StDev = 10.1 P-Value: 0.227 A-Squared: 0.450 Anderson-Darling Normality Test N: 12 StDev: 10.1757 Average: 194.5 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Form 1 Normal Probability Plot P-Value: 0.236 A-Squared: 0.443 Anderson-Darling Normality Test N: 12 StDev : 9.94949 Av erage: 193.083 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Form 2 Normal Probability Plot Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-28 (c) What is the P-value for the test in part (a)? P = 0.367 2.33. Refer to the data in problem 2.32. Do the data support a claim that the mean deflection temperature under load for formulation 1 exceeds that of formulation 2 by at least 3 F? No, formulation 1 does not exceed formulation 2 by at least 3 F. Minitab Output Two-Sample T-Test and CI: Form1, Form2 Two-sample T for Form 1 vs Form 2 N Mean StDev SE Mean Form 1 12 194.5 10.2 2.9 Form 2 12 193.08 9.95 2.9 Difference = mu Form 1 - mu Form 2 Estimate for difference: 1.42 95% lower bound for difference: -5.64 T-Test of difference = 3 (vs >): T-Value = -0.39 P-Value = 0.648 DF = 22 Both use Pooled StDev = 10.1 2.34S. In semiconductor manufacturing, wet chemical etching is often used to remove silicon from the backs of wafers prior to metalization. The etch rate is an important characteristic of this process. Two different etching solutions are being evaluated. Eight randomly selected wafers have been etched in each solution and the observed etch rates (in mils/min) are shown below: Solution 1 Solution 2 9.9 10.6 10.2 10.6 9.4 10.3 10.0 10.2 10.0 9.3 10.7 10.4 10.3 9.8 10.5 10.3 (a) Do the data indicate that the claim that both solutions have the same mean etch rate is valid? Use  = 0.05 and assume equal variances. No, the solutions do not have the same mean etch rate. See the Minitab output below. Minitab Output Two Sample T-Test and Confidence Interval Two-sample T for Solution 1 vs Solution 2 N Mean StDev SE Mean Solution 8 9.950 0.450 0.16 Solution 8 10.363 0.233 0.082 Difference = mu Solution 1 - mu Solution 2 Estimate for difference: -0.413 95% CI for difference: (-0.797, -0.028) T-Test of difference = 0 (vs not =): T-Value = -2.30 P-Value = 0.037 DF = 14 Both use Pooled StDev = 0.358 Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-29 (b) Find a 95% confidence interval on the difference in mean etch rate. From the Minitab output, -0.797 to –0.028. (c) Use normal probability plots to investigate the adequacy of the assumptions of normality and equal variances. Both the normality and equality of variance assumptions are valid. 2.35. Two popular pain medications are being compared on the basis of the speed of absorption by the body. Specifically, tablet 1 is claimed to be absorbed twice as fast as tablet 2. Assume that 2 1 and 2  2 are known. Develop a test statistic for H0 : 21 = 2 H1 : 21  2 P-Value: 0.764 A-Squared: 0.216 Anderson-Darling Normality Test N: 8 StDev: 0.450397 Average: 9.95 9.5 10.0 10.5 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Solution 1 Normal Probability Plot P-Value: 0.919 A-Squared: 0.158 Anderson-Darling Normality Test N: 8 StDev: 0.232609 Average: 10.3625 10.0 10.1 10.2 10.3 10.4 10.5 10.6 10.7 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Solution 2 Normal Probability Plot Solutions from Montgomery, D. C. (2019) Design and Analysis of Experiments, Wiley, NY 2-30 2 2 1 2 1 2 1 2 1 2 4 2 ~ 2 , y y N n n              , assuming that the data is normally distributed. The test statistic is: 1 2 2 2 1 2 1 2 2 4 o y y z n n      , reject if 2 o z z   2.36. Continuation of Problem 2.35. An article in Nature (1972, pp.225-226) reported on the levels of monoamine oxidase in blood platelets for a sample of 43 schizophrenic patients resulting in