(Solution Manual) Montgomery, D. C. Design and Analysis of Experiments, Wiley, NY

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Solutions from Montgomery, D. C. Design and Analysis of Experiments , Wiley, NY 2-1 Chapter 2 Simple Comparative Experiments Solutions 2-1 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 z = y − o = 148.75 −150 = −1.25 = −0.8333 o  3 3 4 2 Since z0.05 = 1.645, do not reject. (c) Find the P-value for the test in part (b). 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  y − z 2     y + z 2 148.75 − (1.96)(3 2)    148.75 + (1.96)(3 2) 145.81    151.69 2-2 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? Solutions from Montgomery, D. C. Design and Analysis of Experiments , Wiley, NY 2-2 n n 10 10     z = y − o = 812 − 800 = 12 = 1.92 Since z/2 = z0.025 = 1.96, do not reject. o  25 25 16 4 (c) What is the P-value for the test? P = 2(0.0274) = 0.0549 (d) Find a 95 percent confidence interval on the mean. The 95% confidence interval is  y − z 2    y + z 2 812 − (1.96)(25 4)    812 + (1.96)(25 4) 812 −12.25    812 + 12.25 799.75    824.25 2-3 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  = 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 . H0:  = 0.255 H1:   0.255 (b) Test these hypotheses using  = 0.05. What are your conclusions? n = 10,  = 0.0001, y = 0.2545 z = y − o = 0.2545 − 0.255 = −15.81 o  Since z0.025 = 1.96, reject H0. 0.0001 10 (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 − z 2    y + z 2 0.2545 − (1.96) 0.0001     0.2545 + (1.96) 0.0001      0.254438    0.254562 2-4 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 width of 1.0. n n n   Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-3 n n n n 382 T-Test of the Mean Test of mu = 120.00 vs mu > 120.00 Since y  N(,9), a 95% two-sided confidence interval on  is y − z     y + z  2 2 y − (1.96) 3    y + (1.96) 3 If the total interval is to have width 1.0, then the half-interval is 0.5. Since z /2 = z0.025 = 1.96, (1.96)(3 n )= 0.5 = (1.96)(3 0.5) = 11.76 n = (11.76)2 = 138.30  139 2-5 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 s2 = [ (108 - 131)2 + (124 - 131)2 + (124 - 131)2 + (106 - 131)2 + (115 - 131)2 + (138 - 131)2 + (163 - 131)2 + (159 - 131)2 + (134 - 131)2 + ( 139 - 131)2 ] / (10 - 1) s2 = 3438 / 9 = 382 s = = 19.54 t = y − o = o s = 1.78 since t0.01,9 = 2.821; do not reject H0 Minitab Output Variable N Mean StDev SE Mean T P Shelf Life 10 T Confidence Intervals 131.00 19.54 6.18 1.78 0.054 n n 131− 120 19.54 10 Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-4     (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 95% confidence interval is y − t  2 ,n −1    y + t  2 ,n −1 131− (3.250) 1954     131+ (3.250) 1954      110.91    151.09 2-6 Consider the shelf life data in Problem 2 -5. 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 2-5? 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 2-5 is not too serious unless the departure from normality is severe. Normal Probability Plot for Shelf Life ML Estimates 99 ML Estimates Mean 95 StDev 90 131 18.5418 80 Goodness of Fit 70 AD* 60 50 40 30 20 10 5 1 1.292 86 96 106 116 126 136 146 156 166 176 Data 2-7 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 159 280 101 212 224 379 179 264 222 362 168 250 Variable N Shelf Life 10 Mean StDev SE Mean 99.0 % CI 19.54 6.18 ( 110.91, 151.09) s n s n Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-5 149 260 485 170 Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-6 9746.8 T-Test of the Mean Test of mu = 225.0 vs mu > 225.0 16 16     (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 = 247.50 s2 =146202 / (16 - 1) = 9746.80 s = = 98.73 t = y − o = 241.50 − 225 = 0.67 o 98.73 16 since t0.05,15 = 1.753; do not reject H0 Minitab Output 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 y − t  2 ,n−1    y + t  2 ,n −1 241.50 − (2.131) 98.73     241.50 + (2.131) 98.73      188.9    294.1 2-8 Reconsider the repair time data in Problem 2-7. 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. s n s n Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-7 y1 − y2 2 n1 n2 1 + 2 2 Normal Probability Plot for Hours ML Estimates 99 ML Estimates Mean 95 StDev 90 241.5 95.5909 80 Goodness of Fit 70 AD* 60 50 40 30 20 10 5 1 1.185 50 150 250 Data 350 450 2-9 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? y1 = 16.015 1 = 0.015 n1 = 10 y2 = 16.005 2 = 0.018 n2 = 10 zo = = = 1.35 z0.025 = 1.96; do not reject (c) What is the P-value for the test? P=0.1770 16.015 − 16.018 0.0152 + 0.0182 10 10 Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-8 (d) Find a 95 percent confidence interval on the difference in the mean fill volume for the two machines. Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-9 2  2 n1 n2 1 + 2 12 + 12 10 12  2  2 o The 95% confidence interval is  2  2  2  2 y1 − y2 − z 2 1 + 2 n1 n2  1 −  2  y1 − y2 + z 2 1 + 2 n1 n2 (16.015 −16.005) − (19.6)  1 − 2  (16.015 −16.005) + (19.6) − 0.0045  1 −  2  0.0245 2-10 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 y1 = 162.5 1 = 1 n1 = 10 y2 = 155.0  2 = 1 n2 = 10 z = y1 − y2 − 10 = 162.5 − 155.0 − 10 = −5.85 z0.01 = 2.225; do not reject The 99 percent confidence interval is y1 − y2 − z  1 −  2  y1 − y2 + z (162.5 − 155.0) − (2.575)  1 − 2  (162.5 − 155.0) + (2.575) 6.40  1 −  2  8.60 2-11 The following are the burning times 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 65 82 64 56 81 67 71 69 57 59 83 74 66 75 59 82 82 70 65 79 0.0152 + 0.0182 10 10 0.0152 + 0.0182 10 10  2  2 12 + 12 12 + 12 Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-10 (a) Test the hypotheses that the two variances are equal. Use  = 0.05. Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-11 Two Sample T-Test and Confidence Interval Two sample T for Type 1 vs Type 2 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 1 2 1 2 H0 : 2 =  2 S = 9.264 H1: 2   2 S2 = 9.367 1 2 S 2 85.82 F0 = 1 = = 0.98 F = 4.03 F 2 87.73 = 1 = 1 = 0.248 Do not reject. 0.025,9,9 0.975,9,9 F0.025,9,9 4.03 (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? 2 (n1 − 1)S 2 + (n2 − 1)S 2 1561.95 S p = S p = 9.32 1 n1 + n2 − 2 2 = = 86.775 18 t = y1 − y2 = 70.4 − 70.2 = 0.048 0 1 1 1 1 S p + n n 9.32 + 1 2 10 10 t0.025,18 = 2.101 Do not reject. From the computer output, t=0.05; do not reject. Also from the computer output P=0.96 Minitab Output N Mean StDev SE Mean Type 1 10 70.40 9.26 2.9 Type 2 10 70.20 9.37 3.0 (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. S Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-12 Normal Probability Plot for Type 1 ML Estimates 99 ML Estimates Mean 95 StDev 90 70.4 8.78863 80 Goodness of Fit 70 AD* 60 50 40 30 20 10 5 1 1.387 50 60 70 80 90 Data Normal Probability Plot for Type 2 ML Estimates 99 ML Estimates Mean 95 StDev 90 70.2 8.88594 80 Goodness of Fit 70 AD* 60 50 40 30 20 10 5 1 1.227 50 60 70 80 90 Data 2-12 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 2F6 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 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. Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-13 Minitab Output Two Sample T-Test and Confidence Interval Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-14 Two sample T for Uniformity 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 1 2 Flow Rat N Mean StDev SE Mean 125 6 3.317 0.760 0.31 200 6 3.933 0.821 0.34 (b) What is the P-value for the test in part (a)? From the computer printout, P=0.21 (c) Does the C2F6 flow rate affect the wafer -to-wafer variability in etch uniformity? Use  = 0.05. H0: 2 =  2 H1: 2   2 1 2 F0.05,5,5 = 5.05 F = 0.5776 = 0.86 0 0.6724 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). 5 4 3 125 Flow Rate 200 2-13 A new filtering device is installed in a chemical unit. Before its installation, a random sample yielded the following information about the percentage of impurity: y = 12.5, S 2 =101.17, and n = 8. After installation, a random sample yielded y = 10.2, S 2 = 94.73, n 1 1 1 = 9. 2 2 2 (a) Can you concluded that the two variances are equal? Use  = 0.05. Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-15 S n n 1 + 1 1 2 S 2 ( )( )  1 2 1 2 H0 :  2 =  2 H1 :  2   2 F0.025 ,7 ,8 = 4.53 S 2 F0 = 1 = 2 101.17 94.73 = 1.07 Do Not Reject. Assume that the variances are equal. (b) Has the filtering device reduced the percentage of impurity significantly? Use  = 0.05. H 0 : 1 = 2 H1 : 1  2 2 ( n1 − 1)S 2 + ( n − 1)S 2 ( 8 − 1)(101.17 ) + ( 9 −1)( 94.73 ) S p = 1 2 2 = = 97.74 S p = 9.89 n1 + n2 − 2 8 + 9 − 2 t0 = y1 − y2 = = 0.479 t0.05,15 = 1.753 Do not reject. There is no evidence to indicate that the new filtering device has affected the mean 2-14 Twenty observations on etch uniformity on silicon wafers are taken during a qualification experiment for a plasma etcher. The data are as follows: 5.34 6.65 4.76 5.98 7.25 6.00 7.55 5.54 5.62 6.21 5.97 7.35 5.44 4.39 4.98 5.25 6.35 4.61 6.00 5.32 (a) Construct a 95 percent confidence interval estimate of 2. (n −1) S 2  2   2  (n −1) S 2  2  2,n−1 (1− 2 ),n−1 20 −1 0.88907 2   2  (20 −1)(0.88907)2 32.852 8.907 0.457   2  1.686 (b) Test the hypothesis that 2 = 1.0. Use  = 0.05. What are your conclusions? H0 : 2 = 1 H1: 2  1  2 = SS = 15.019 0 2 12.5 − 10.2 9.89 1 + 1 8 9 Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-16 0 2 0.025,19 = 32.852 2 0.975,19 = 8.907   Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-17 1 Do not reject. There is no evidence to indicate that  2  1 (c) Discuss the normality assumption and its role in this problem. The normality assumption is much more important when analyzing variances then when analyzing means. A moderate departure from normality could cause problems with both statistical tests and confidence intervals. Specifically, it will cause the reported significance levels to be incorrect. (d) Check normality by constructing a normal probability plot. What are your conclusions? The normal probability plot indicates that there is not any serious problem with the normality assumption. Normal Probability Plot for Uniformity ML Estimates 99 ML Estimates Mean 95 StDev 90 5.828 0.866560 80 Goodness of Fit 70 AD* 60 50 40 30 20 10 5 1 0.835 3.8 4.8 5.8 Data 6.8 7.8 2-15 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 .001 .000001 2 0.265 0.265 .000 0 3 0.266 0.264 .002 .000004 4 0.267 0.266 .001 .000001 5 0.267 0.267 .000 0 6 0.265 0.268 -.003 .000009 7 0.267 0.264 .003 .000009 8 0.267 0.265 .002 .000004 9 0.265 0.265 .000 0 10 0.268 0.267 .001 .000001 11 0.268 0.268 .000 0 12 0.265 0.269 -.004 = 0.003 .000016 = 0.000045 (a) Is there a significant difference between the means of the population of measurements represented by the two samples? Use  = 0.05. Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-18 Paired T-Test and Confidence Interval Paired T for Caliper 1 - Caliper 2 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  H0 : 1 = 2 H1: 1  2 or equivalently H 0 : d = 0 H1 : d  0 Minitab Output 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 (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. d − t Sd 2,n−1  D (= 1 − 2 )  d + t Sd 2,n−1 0.00025 − 2.2010.002   12 d  0.00025 + 2.2010.002 −0.00102  d  0.00152 2-16 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. H0 : 1 = 2 H1: 1  2 or equivalently H 0 : d = 0 H1 : d  0 d = 1 n d n i=1 i Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-19 = 1 (2.465) = 0.274 9 Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-20 n 9 1 2  n n 2  2   d 2 − 1   d    0.821151 − 1 (2.465)2  i n  i    s =  i=1  i=1   =  9  = 0.135 d  n −1   9 −1        t = d = 0.274 = 6.08 0 Sd 0.135 9 Minitab Output t ,n−1 = t0.025,9 = 2.306 , reject the null hypothesis. (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. d − t  2 ,n −1 Sd    d + t  2 ,n−1 Sd 0.274 − 2.306 0.135   9  0.274 + 2.306 0.135 0.17023  d  0.37777 (d) Investigate the normality assumption for both samples. Normal Probability Plot .999 .99 .95 .80 .50 .20 .05 .01 .001 1.15 1.25 1.35 Karlsruhe 1.45 1.55 Av erage: 1.34011 StDev : 0.146031 N: 9 Anderson -Darling Normality Test A- Squared: 0.286 P-Value: 0.537 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 2 n n d d Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-21 .999 .99 .95 .80 .50 .20 .05 .01 .001 Normal Probability Plot Av erage: 1.06622 StDev : 0.0493806 N: 9 1.00 1.05 1.10 Lehigh 1.15 Anders on -Darling Normality Test A-Squared: 0.772 P-Value: 0.028 (e) Investigate the normality assumption for the difference in ratios for the two methods. .999 .99 .95 .80 .50 .20 .05 .01 .001 Normal Probability Plot 0.12 0.22 0.32 Difference 0.42 0.52 Av erage: 0.273889 StDev : 0.135099 N: 9 Anderson -Darling Normality Test A- Squared: 0.318 P-Value: 0.464 (f) 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. 2-17 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 2 212 199 198 177 176 198 194 213 216 197 185 188 Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-22 211 191 200 206 200 189 193 195 184 201 197 203 Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-23 Two Sample T-Test and Confidence Interval Two sample T for Form 1 vs Form 2 95% CI for mu Form 1 - mu Form 2: ( -1.1, 15.9) T-Test mu Form 1 = mu Form 2 (vs >): T = 1.81 P = 0.042 DF = 22 Both use Pooled StDev = 10.1 (a) Construct normal probability plots for both samples. Do these plots support assumptions of normality and equal variance for both samples? .999 .99 .95 .80 .50 .20 .05 .01 .001 Normal Probability Plot 185 195 Form 1 205 215 Av erage: 200.5 StDev : 10.1757 N: 12 .999 .99 .95 .80 .50 .20 .05 .01 .001 Anderson -Darling Normality Test A-Squared: 0.450 P-Value: 0.227 Normal Probability Plot Av erage: 193.083 StDev : 9.94949 N: 12 175 185 Form 2 195 205 Anderson -Darling Normality Test A- Squared: 0.443 P-Value: 0.236 (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. Minitab Output N Mean StDev SE Mean Form 1 12 200.5 10.2 2.9 Form 2 12 193.08 9.95 2.9 (c) What is the P-value for the test in part (a)? P = 0.042 Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-24 Two-Sample T-Test and CI: Form1, Form2 Two-sample T for Form1 vs Form2 Difference = mu Form1 - mu Form2 Estimate for difference: 7.42 95% lower bound for difference: 0.36 T-Test of difference = 3 (vs >): T-Value = 1.08 P-Value = 0.147 DF = 22 Both use Pooled StDev = 10.1 Two Sample T-Test and Confidence Interval Two sample T for Solution 1 vs Solution 2 95% CI for mu Solution - mu Solution: ( -0.83, -0.043) T-Test mu Solution = mu Solution (vs not =):T = -2.38 P = 0.032 DF = 14 Both use Pooled StDev = 0.368 2-18 Refer to the data in problem 2-17. 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? Yes, formulation 1 exceeds formulation 2 by at least 3 F. Minitab Output N Mean StDev SE Mean Form1 12 200.5 10.2 2.9 Form2 12 193.08 9.95 2.9r 2-19 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 solutionsare being evaluated. Eight randomly selected wa fers 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. See the Minitab output below. Minitab Output N Mean StDev SE Mean Solution 8 9.925 0.465 0.16 Solution 8 10.362 0.233 0.082 (b) Find a 95% confidence interval on the difference in mean etch rate. From the Minitab output, -0.83 to –0.043. (c) Use normal probability plots to investigate the adequacy of the assumptions of normality and equal variances. Solutions from Montgomery, D. C. (2001) Design and Analysis of Experiments , Wiley, NY 2-25 2 y1 − y2 4 2 n1 n2 1 + 2 2 1 2 .999 .99 .95 .80 .50 .20 .05 .01 .001 Normal Probability Plot Av erage: 9.925 StDev : 0.465219 N: 8 9.5 10.0 Solution 1 10.5 Anders on -Darling Normality Test A-Squared: 0.222 P-Value: 0.743 .999 .99 .95 .80 .50 .20 .05 .01 .001 Normal Probability Plot 10.0 10.1 10.2 10.3 10.4 10.5 10.6 10.7 Av erage: 10.3625 StDev : 0.232609 N: 8 Solution 2 Anderson -Darling Normality Test A- Squared: 0.158 P-Value: 0.919 Both the normality and equality of variance assumptions are valid. 2-20 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 and  2 are known. Develop a test statistic for H0: 21 = 2 H1: 21  2   4 2  2  2 y1 − y2 ~ N  21 − 2 , 1 + 2  , assuming that the data is normally distributed.  n1 The test statistic is: zo = n2   , reject if zo  z 2