Solutions Manual for Design and Analysis of Experiments Part 1

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 Y 9 19.96 ? 3.12 ? SE Mean = 1.04 Minimum Maximum 15.94 27.16 Variance = 9.73 2.2. Computer output for a random sample of data is shown below. Some of the quantities are missing. Compute the values of the missing quantities. Mean = 24.991 Variable N Mean Y 16 ? SE Mean 0.159 Std. Dev. ? Sum 399.851 Std. Dev. = 0.636 2.3. Suppose that we are testing H 0 : µ = µ 0 versus H 1 : µ ≠ µ 0 . Calculate the P-value for the following observed values of the test statistic: (a) Z 0 = 2.25 P-value = 0.02445 (b) Z 0 = 1.55 P-value = 0.12114 (c) Z 0 = 2.10 P-value = 0.03573 (d) Z 0 = 1.95 P-value = 0.05118 (e) Z 0 = -0.10 P-value = 0.92034 2.4. Suppose that we are testing H 0 : µ = µ 0 versus H 1 : µ > µ 0 . Calculate the P-value for the following observed values of the test statistic: (a) Z 0 = 2.45 P-value = 0.00714 (b) Z 0 = -1.53 P-value = 0.93699 (c) Z 0 = 2.15 P-value = 0.01578 (d) Z 0 = 1.95 P-value = 0.02559 (e) Z 0 = -0.25 P-value = 0.59871 2-1 2.5. Consider the computer output shown below. One-Sample Z Test of mu = 30 vs not = 30 The assumed standard deviation = 1.2 (a) Mean 31.2000 SE Mean 0.3000 95% CI Z (30.6120, 31.7880) ? P ? Fill in the missing values in the output. What conclusion would you draw? Z=4 (b) N 16 P = 0.00006; therefore, the mean is not equal to 30. Is this a one-sided or two-sided test? Two-sided. (c) Use the output and the normal table to find a 99 percent CI on the mean. CI = 30.42725, 31.97275 (d) What is the P-value if the alternative hypothesis is H 1 : µ > 30 P-value = 0.00003 2.6. Suppose that we are testing H 0 : µ 1 = µ 2 versus H 1 : µ 1 = µ 2 with a sample size of n 1 = n 2 = 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) t 0 = 2.30 Table P-value = 0.02, 0.05 Computer P-value = 0.0313 (b) t 0 = 3.41 Table P-value = 0.002, 0.005 Computer P-value = 0.0025 (c) t 0 = 1.95 Table P-value = 0.1, 0.05 Computer P-value = 0.0640 (d) t 0 = -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 2.7. Suppose that we are testing H 0 : µ 1 = µ 2 versus H 1 : µ 1 > µ 2 with a sample size of n 1 = n 2 = 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) t 0 = 2.31 Table P-value = 0.01, 0.025 Computer P-value = 0.01648 (b) t 0 = 3.60 Table P-value = 0.001, 0.0005 Computer P-value = 0.00102 (c) t 0 = 1.95 Table P-value = 0.05, 0.025 Computer P-value = 0.03346 2-2 (d) t 0 = 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. 2.8. 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 Summary for Sample Data A nderson-Darling Normality Test A -Squared P -V alue M ean StD ev V ariance Skewness Kurtosis N 8 9 10 11 12 M inimum 1st Q uartile M edian 3rd Q uartile M aximum 13 0.33 0.435 10.952 1.993 3.974 -0.45131 -1.06746 10 7.750 9.080 11.345 13.067 13.210 95% C onfidence Interv al for M ean 9.526 12.378 95% C onfidence Interv al for M edian 8.973 13.078 95% C onfidence Interv al for S tDev 9 5 % Conf ide nce I nt e r v a ls 1.371 3.639 Mean Median 9 10 11 12 13 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. 2.9. 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: t o = 2.01 P-Value = 0.0298 (a) What is the missing value for the standard error? 2-3 2.35 y1 − y2 = = 2.01 StdError 1 1 + Sp n1 n2 StdError = 2.35 / 2.01 = 1.169 t0 = (b) Is this a two-sided or one-sided test? One-sided test for a t 0 = 2.01 is a P-value of 0.0298. (c) If α =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. y1 − y2 − tα 2,n1 +n2 −2 Sp y1 − y2 − t0.05,18 Sp 1 n1 1 n1 + + 1 n2 1 n2 ≤ µ1− µ1≤ y1− y2+ tα 2,n1+ n2− 2 ≤ µ1− µ1 ≤ y1 − y2 + t0.05,18 Sp 1 Sp 1 n1 n1 + + 1 n2 1 n2 2.35 −1.734 (1.169) ≤ µ1 − µ1 ≤ 2.35+ 1.734( 1.169) 0.323 ≤ µ1 − µ1 ≤ 4.377 2.10. 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: t o = -1.88 P-Value = 0.0723 (a) What is the missing value for the standard error? − 11 .5 y1 − y2 = = −1.88 1 1 StdError + Sp n1 n2 StdError = −11 .5 /− 1.88 = 6.12 t0 = (b) Is this a two-sided or one-sided test? Two-sided test for a t 0 = -1.88 is a P-value of 0.0723. (c) If α =0.05, what are your conclusions? Accept the null hypothesis, there is no difference in the means. (d) Find a 90% two-sided CI on the difference in the means. 2-4 y1 − y2 − tα 2,n1 +n2 −2 Sp y1 − y2 − t0.05,24 Sp 1 n1 + 1 n2 ≤ µ1− µ1≤ y1− y2+ tα 2,n1+ n2− 2 Sp 1 n1 + 1 n2 1 1 1 1 + ≤ µ1− µ1≤ y1− y2+ t0.05,24 Sp + n1 n 2 n1 n 2 −11.5 −1.711 (6.12) ≤ µ1− µ1 ≤ − 11.5+ 1.711( 6.12) −21.97 ≤ µ1− µ≤1 − 1.03 2.11. Suppose that we are testing H 0 : µ = µ 0 versus H 1 : µ > µ 0 with a sample size of n = 15. Calculate bounds on the P-value for the following observed values of the test statistic: (a) t 0 = 2.35 Table P-value = 0.01, 0.025 Computer P-value = 0.01698 (b) t 0 = 3.55 Table P-value = 0.001, 0.0025 Computer P-value = 0.00160 (c) t 0 = 2.00 Table P-value = 0.025, 0.005 Computer P-value = 0.03264 (d) t 0 = 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. 2.12. Suppose that we are testing H 0 : µ = µ 0 versus H 1 : µ ≠ µ 0 with a sample size of n = 10. Calculate bounds on the P-value for the following observed values of the test statistic: (a) t 0 = 2.48 Table P-value = 0.02, 0.05 Computer P-value = 0.03499 (b) t 0 = -3.95 Table P-value = 0.002, 0.005 Computer P-value = 0.00335 (c) t 0 = 2.69 Table P-value = 0.02, 0.05 Computer P-value = 0.02480 (d) t 0 = 1.88 Table P-value = 0.05, 0.10 Computer P-value = 0.09281 (e) t 0 = -1.25 Table P-value = 0.20, 0.50 Computer P-value = 0.24282 2.13. Consider the computer output shown below. One-Sample T: Y Test of mu = 91 vs. not = 91 Variable Y (a) N Mean 25 92.5805 Std. Dev. ? SE Mean 0.4675 95% CI (91.6160, ? ) T P 3.38 0.002 Fill in the missing values in the output. Can the null hypothesis be rejected at the 0.05 level? Why? Std. Dev. = 2.3365 UCI = 93.5450 Yes, the null hypothesis can be rejected at the 0.05 level because the P-value is much lower at 0.002. (b) Is this a one-sided or two-sided test? 2-5 Two-sided. (c) If the hypothesis had been H 0 : µ = 90 versus H 1 : µ ≠ 90 would you reject the null hypothesis at the 0.05 level? Yes. (d) Use the output and the t table to find a 99 percent two-sided CI on the mean. CI = 91.2735, 93.8875 (e) What is the P-value if the alternative hypothesis is H 1 : µ > 91? P-value = 0.001. 2.14. Consider the computer output shown below. One-Sample T: Y Test of mu = 25 vs > 25 Variable Y (a) N Mean 12 25.6818 Std. Dev. ? SE Mean 95% Lower Bound 0.3360 ? How many degrees of freedom are there on the t-test statistic? 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 Both use Pooled Std. Dev. = 2.1277 (a) Can the null hypothesis be rejected at the 0.05 level? Why? 2-6 P ? (N-1) = (12 – 1) = 11 (b) T 0.034 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 H 0 : µ 1 - µ 2 = 2 versus H 1 : µ 1 - µ 2 ≠ 2 would you reject the null hypothesis at the 0.05 level? Yes. (d) If the hypothesis had been H 0 : µ 1 - µ 2 = 2 versus H 1 : µ 1 - µ 2 < 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 H 0 : µ 1 - µ 2 = 2 versus H 1 : µ 1 - µ 2 ≠ 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 σ = 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: µ = 150 H1: µ > 150 (b) Test these hypotheses using α = 0.05. What are your conclusions? n = 4, σ = 3, y = 1/4 (145 + 153 + 150 + 147) = 148.75 y − µ 148.75 −150 =− 1.25 =− 0.8333 zo = σ o = 3 3 2 4 n Since z0.05 = 1.645, do not reject. (c) Find the P-value for the test in part (b). ) ] = 0.2014 From the z-table: P ≅1 − [0.7967 + (2 3)( 0.7995 − 0.7967 (d) Construct a 95 percent confidence interval on the mean breaking strength. 2-7 The 95% confidence interval is σ σ ≤ µ ≤ y + zα 2 n n )( 3)2 148.75 − (1.96)( 3 2) ≤ µ ≤ 148.75 +( 1.96 y − zα 2 145. 81 ≤ µ≤ 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 σ = 25 centistokes. (a) State the hypotheses that should be tested. H0: µ = 800 H1: µ ≠ 800 (b) Test these hypotheses using α = 0.05. What are your conclusions? 812 −800 12 y −µ = = 1.92 zo = σ o = 25 25 4 16 n Since zα /2 = z0.025 = 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 y − zα 2 σ n ≤ µ ≤ y + zα2 σ n )( 25 4) ≤ µ ≤ 812 +( 1 .96 ) (25 4) 812− (1. 96 812 − 12 .25 ≤ µ ≤ 812 + 12 .25 799 .75 ≤ µ ≤ 824 .25 2.18. The diameters of steel shafts produced by a certain manufacturing process should have a mean diameter of 0.255 inches. The diameter is known to have a standard deviation of σ = 0.0001 inch. A random sample of 10 shafts has an average diameter of 0.2545 inches. (a) Set up the appropriate hypotheses on the mean µ. H0: µ = 0.255 H1: µ ≠ 0.255 (b) Test these hypotheses using α = 0.05. What are your conclusions? n = 10, σ = 0.0001, y = 0.2545 2-8 y − µ 0.2545 − 0.255 =− 15.81 zo = σ o = 0.0001 10 n Since z0.025 = 1.96, reject H0. (c) Find the P-value for this test. P = 2.6547x10-56 (d) Construct a 95 percent confidence interval on the mean shaft diameter. The 95% confidence interval is y − zα2 σ n ≤ µ ≤ y + zα2 σ n 0.0001  0.0001 )  0.2545 − (1.96)  ≤ µ≤ 0.2545+ ( 1.96  10   10 0. 254438 ≤ µ≤ 0. 254562 2.19. A normally distributed random variable has an unknown mean µ and a known variance σ 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(µ,9), a 95% two-sided confidence interval on µ is If the total interval is to have width 1.0, then the half-interval is 0.5. Since zα /2 = z0.025 = 1.96, (1. 96) (3 n) = 0. 5 n = (1 .96)( 3 0 .5) = 11 .76 2 n = (11. 76) =138. 30 ≅ 139 2.20. The shelf life of a carbonated beverage is of interest. Ten bottles are randomly selected and tested, and the following results are obtained: 108 124 124 106 115 Days 138 163 159 134 139 (a) We would like to demonstrate that the mean shelf life exceeds 120 days. Set up appropriate hypotheses for investigating this claim. H0: µ = 120 H1: µ > 120 (b) Test these hypotheses using α = 0.01. What are your conclusions? 2-9 y = 131 S2 = 3438 / 9 = 382 S = 382 = 19.54 y − µ0 = 131 − 120 = 1.78 t0 = S n 19.54 10 since t0.01,9 = 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. S ≤ µ≤ + S The 99% confidence interval is y − t α 2,n−1 with α = 0.01. y tα 2,n− 1 n n 19.54  19.54 )  131 − (3.250)  ≤ µ ≤ 131+ ( 3.250  10   10 110.91 ≤ µ≤ 151.08 2.21. Consider the shelf life data in Problem 2.20. 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 2.20? 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 2.20 is not too serious unless the departure from normality is severe. 2-10 Normal Probability Plot .999 Probability .99 .95 .80 .50 .20 .05 .01 .001 105 115 125 135 145 155 165 Shelf Life Average: 131 StDev: 19.5448 N: 10 Anderson-Darling Normality Test A-Squared: 0.266 P-Value: 0.606 2.22. 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 280 101 379 179 362 168 260 485 159 224 222 149 212 264 250 170 (a) You wish to know if the mean repair time exceeds 225 hours. Set up appropriate hypotheses for investigating this issue. H0: µ = 225 H1: µ > 225 (b) Test the hypotheses you formulated in part (a). What are your conclusions? Use α = 0.05. y = 241.50 S2 =146202 / (16 - 1) = 9746.80 S = 9746.8 = 98.73 to = y − µo 241.50 − 225 = = 0.67 98.73 S 16 n since t0.05,15 = 1.753; do not reject H0 Minitab Output T-Test of the Mean Test of mu = 225.0 vs mu > 225.0 Variable N Hours 16 Mean StDev SE Mean 241.5 98.7 24.7 0.67 T P 0.26 2-11 T Confidence Intervals Variable N Hours 16 Mean StDev SE Mean 95.0 % CI 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 y − tα 2,n −1 S ≤µ ≤ + y tα n S 2,n− 1 n 98.73  98.73 241.50 − (2.131)  )  ≤ µ≤ 241.50+ ( 2.131  16   16 188.9 ≤ µ ≤ 294.1 2.23. Reconsider the repair time data in Problem 2.22. 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. Normal Probability Plot .999 .99 Probability .95 .80 .50 .20 .05 .01 .001 100 200 300 400 500 Hours Average: 241.5 StDev: 98.7259 N: 16 Anderson-Darling Normality Test A-Squared: 0.514 P-Value: 0.163 2.24. 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 σ 1 = 0.015 and σ 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 16.03 16.01 16.04 15.96 16.05 15.98 16.05 16.02 16.02 15.99 Machine 2 16.02 16.03 15.97 16.04 15.96 16.02 16.01 16.01 15.99 16.00 2-12 (a) State the hypotheses that should be tested in this experiment. H 0: µ 1 = µ 2 H1: µ1 ≠ µ2 (b) Test these hypotheses using α =0.05. What are your conclusions? y1 = 16. 015 y 2 = 16. 005 σ 1 = 0. 015 σ 2 = 0 .018 n1 = 10 n 2 = 10 y1 − y2 zo = σ 21 σ n1 + 16 .015 −16 .018 = 2 2 0. 0152 0. 0182 + 10 10 n2 = 1. 35 z0.025 = 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 y1 − y2 − zα 2 σ 21 n1 + σ 22 ≤ µ1 − µ2 ≤ y1 − y2 + z α2 n2 σ 2 1 n1 + σ 22 n2 0.015 2 0.018 2 0.015 2 0.018 2 (16.015− 16.005) − (1.96) + ≤ µ1 − µ 2 ≤ (16.015− 16.005)+ (1.96) + 10 10 10 10 − 0.0045 ≤ µ1 − µ2 ≤ 0.0245 2.25. 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 σ 1 = σ 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 α = 0.01. Construct a 99 percent confidence interval on the true mean difference in breaking strength. H0: µ1 - µ2 =10 H1: µ1 - µ2 >10 y1 = 162 .5 y 2 = 155. 0 σ1 = 1 σ2 = 1 n1 = 10 n 2 = 10 zo = y1 − y2− 10 σ12 σ n1 + 2 2 = 162.5− 155.0− 10 n2 2 2 1 1 + 10 12 z0.01 = 2.325; do not reject 2-13 =−5.84 The 99 percent confidence interval is y1 − y2 − zα 2 σ 12 n1 + σ 22 n2 ≤ µ1 − µ 2 ≤ y1 − y2 + zα 2 σ 21 n1 + σ 22 n2 1 2 12 1 2 12 + ≤ µ1 − µ2 ≤ (162.5 − 155.0) + ( 2.575) + (162.5 − 155.0) − ( 2.575) 10 12 10 12 6.40 ≤ µ 1 − µ 2 ≤ 8.60 2.26. 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 65 81 57 66 82 Type 2 64 56 71 69 83 74 59 82 65 79 82 67 59 75 70 (a) Test the hypotheses that the two variances are equal. Use α = 0.05. 2 H0 : σ 1 =σ 2 1 H1 : σ ≠σ 2 2 2 2 Do not reject. (b) Using the results of (a), test the hypotheses that the mean burning times are equal. Use α = 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 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 2-14 (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. Normal Probability Plot .999 Probability .99 .95 .80 .50 .20 .05 .01 .001 60 70 80 Type 1 Average: 70.4 StDev: 9.26403 N: 10 Anderson-Darling Normality Test A-Squared: 0.344 P-Value: 0.409 Normal Probability Plot .999 Probability .99 .95 .80 .50 .20 .05 .01 .001 60 70 80 Type 2 Average: 70.2 StDev: 9.36661 N: 10 Anderson-Darling Normality Test A-Squared: 0.186 P-Value: 0.876 2.27. 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 C 2 F 6 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: C2F6 (SCCM) 125 200 1 2.7 4.6 2 4.6 3.4 Uniformity Observation 3 4 2.6 3.0 2.9 3.5 2-15 5 3.2 4.1 6 3.8 5.1 (a) Does the C2 F 6 flow rate affect average etch uniformity? Use α = 0.05. No, C 2 F 6 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 C2 F 6 flow rate affect the wafer-to-wafer variability in etch uniformity? Use α = 0.05. H 0 : σ 12 =σ 2 2 2 2 H1 : σ 1 ≠σ 2 F0.025,5,5 = 7.15 F0.975,5,5 = 0.14 0.5776 = 0.86 F0 = 0.6724 Do not reject; C 2 F 6 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). Uniformity 5 4 3 125 200 Flow Rate 2-16