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Solutions. Design and Analysis of Experiments. Montgomery

• 1. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 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 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) = 148.75 148.75 150 1.25 0.8333 3 3 24 o o y z n µ σ − − − = = = = − Since z0.05 = 1.645, do not reject. (c) Find the P-value for the test in part (b). From the z-table: ( )( )[ ] 20140796707995032796701 ....P =−+−≅ (d) Construct a 95 percent confidence interval on the mean breaking strength. The 95% confidence interval is ( )( ) ( )( )2396.175.1482396.175.148 22 +≤≤− +≤≤− µ σ µ σ αα n zy n zy 145 81 151 69. .≤ ≤µ 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 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? 2-1
• 2. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 812 800 12 1.92 25 25 416 o o y z n µ σ − − = = = = Since zα/2 = z0.025 = 1.96, do not reject. (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 n zy n zy σ µ σ αα 22 +≤≤− ( )( ) ( )( ) 2582475799 25128122512812 425961812425961812 .. .. .. ≤≤ +≤≤− +≤≤− µ µ µ 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 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 0.2545 0.255 15.81 0.0001 10 o o y z n µ σ − − = = = − 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 n zy n zy σ µ σ αα 22 +≤≤− ( ) ( ) 0.0001 0.0001 0.2545 1.96 0.2545 1.96 10 10 µ ⎛ ⎞ ⎛ ⎞ − ≤ ≤ +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 0 254438 0 254562. .≤ ≤µ 2-4 A normally distributed random variable has an unknown mean µ and a known variance σ2 = 9. Find the sample size required to construct a 95 percent confidence interval on the mean, that has total length of 1.0. 2-2
• 3. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Since y ∼ N(µ,9), a 95% two-sided confidence interval on µ is y z n y z n − ≤ ≤ +α α σ µ σ 2 2 y n y n − ≤ ≤ +( . ) ( . )196 3 196 3 µ If the total interval is to have width 1.0, then the half-interval is 0.5. Since zα/2 = z0.025 = 1.96, ( )( ) ( )( ) ( ) 139301387611 7611503961 503961 2 ≅== == = ..n ...n .n. 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: µ = 120 H1: µ > 120 (b) Test these hypotheses using α = 0.01. What are your conclusions? y = 131 S2 = 3438 / 9 = 382 382 19.54S = = 131 120 1.78 19.54 10 o o y t S n µ− − = = = 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) 2-3
• 4. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY (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 n S y t y t n n α µ− − ≤ ≤ + 2 n S α − with α = 0.01. ( ) ( ) 1954 1954 131 3.250 131 3.250 10 10 µ ⎛ ⎞ ⎛ ⎞ − ≤ ≤ +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 110 91 15109. .≤ ≤µ 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. 1761661561461361261161069686 99 95 90 80 70 60 50 40 30 20 10 5 1 Data Percent 1.292AD* Goodness of Fit Normal Probability Plot for Shelf Life ML Estimates Mean StDev 131 18.5418 ML Estimates 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 159 280 101 212 224 379 179 264 222 362 168 250 149 260 485 170 (a) You wish to know if the mean repair time exceeds 225 hours. Set up appropriate hypotheses for investigating this issue. 2-4
• 5. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY H0: µ = 225 H1: µ > 225 (b) Test the hypotheses you formulated in part (a). What are your conclusions? Use α = 0.05. y = 247.50 S2 =146202 / (16 - 1) = 9746.80 9746.8 98.73S = = 241.50 225 0.67 98.73 16 o o y t S n µ− − = = = 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 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 α αµ− − − ≤ ≤ + ( ) ( ) 98.73 98.73 241.50 2.131 241.50 2.131 16 16 µ ⎛ ⎞ ⎛ ⎞ − ≤ ≤ +⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ 12949188 .. ≤≤ µ 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. 2-5
• 6. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 45035025015050 99 95 90 80 70 60 50 40 30 20 10 5 1 Data Percent 1.185AD* Goodness of Fit Normal Probability Plot for Hours ML Estimates Mean StDev 241.5 95.5909 ML Estimates 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 σ1 = 0.015 and σ2 = 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: µ1 = µ2 H1: µ1 ≠ µ2 (b) Test these hypotheses using α=0.05. What are your conclusions? y n 1 1 1 16 015 0 015 10 = = = . .σ y n 2 2 2 16 005 0 018 10 = = = . .σ z y y n n o = − + = − + =1 2 1 2 1 2 2 2 2 2 16 015 16 018 0 015 10 0 018 10 1 35 σ σ . . . . . 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. 2-6
• 7. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY The 95% confidence interval is 2 2 2 1 2 1 2121 2 2 2 1 2 1 21 22 nn zyy nn zyy σσ µµ σσ αα ++−≤−≤+−− 10 018.0 10 015.0 )6.19()005.16015.16( 10 018.0 10 015.0 )6.19()005.16015.16( 22 21 22 ++−≤−≤+−− µµ 0245.00045.0 21 ≤−≤− µµ 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 σ1 = σ2 = 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 10 1 5162 1 1 1 = = = n .y σ 10 1 0155 2 2 2 = = = n .y σ z y y n n o = − − + = − − + = −1 2 1 2 1 2 2 2 2 2 10 162 5 155 0 10 1 10 1 12 5 85 σ σ . . . z0.01 = 2.225; do not reject The 99 percent confidence interval is 2 2 2 1 2 1 2121 2 2 2 1 2 1 21 22 nn zyy nn zyy σσ µµ σσ αα ++−≤−≤+−− 12 1 10 1 )575.2()0.1555.162( 12 1 10 1 )575.2()0.1555.162( 22 21 22 ++−≤−≤+−− µµ 60.840.6 21 ≤−≤ µµ 2-11 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 65 82 64 56 81 67 71 69 57 59 83 74 66 75 59 82 82 70 65 79 (a) Test the hypotheses that the two variances are equal. Use α = 0.05. 2-7
• 8. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 2 2 0 1 2 2 1 1 2 : : H H 2 σ σ σ σ = ≠ S S 1 2 9 264 9 367 = = . . F S S 0 1 2 2 2 8582 87 73 0 98= = = . . . F0 025 9 9 4 03. , , .= F F 0 975 9 9 0 025 9 9 1 1 4 03 0 248. , , . , , . .= = = 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? S n S n S n n S t y y S n n p p p 2 1 1 2 2 2 2 1 2 0 1 2 1 2 1 1 2 156195 18 86 775 9 32 1 1 70 4 70 2 9 32 1 10 1 10 0 048 = − + − + − = = = = − + = − + = ( ) ( ) . . . . . . . t0 025 18 2101. , .= 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 (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. 2-8
• 9. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 9080706050 99 95 90 80 70 60 50 40 30 20 10 5 1 Data Percent 1.387AD* Goodness of Fit Normal Probability Plot for Type 1 ML Estimates Mean StDev 70.4 8.78863 ML Estimates 9080706050 99 95 90 80 70 60 50 40 30 20 10 5 1 Data Percent 1.227AD* Goodness of Fit Normal Probability Plot for Type 2 ML Estimates Mean StDev 70.2 8.88594 ML Estimates 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 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: 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 α = 0.05. No, C2F6 flow rate does not affect average etch uniformity. 2-9
• 10. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 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 computer printout, P=0.21 (c) Does the C2F6 flow rate affect the wafer-to-wafer variability in etch uniformity? Use α = 0.05. 2 2 0 1 2 2 2 1 1 2 0.05,5,5 0 : : 5.05 0.5776 0.86 0.6724 H H F F σ σ σ σ = ≠ = = = 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). 200125 5 4 3 Flow Rate Uniformity 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 1 = 12.5, S =101.17, and n1 2 1 = 8. After installation, a random sample yielded y 2 = 10.2, S = 94.73, n2 2 2 = 9. (a) Can you concluded that the two variances are equal? Use α = 0.05. 2-10
• 11. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 071 7394 17101 534 2 2 2 1 0 870250 2 2 2 11 2 2 2 10 . . . S S F .F :H :H ,,. === = ≠ = σσ σσ Do Not Reject. Assume that the variances are equal. (b) Has the filtering device reduced the percentage of impurity significantly? Use α = 0.05. 7531 4790 9 1 8 1 899 210512 11 899 7497 298 7394191710118 2 11 15050 21 21 0 21 2 22 2 112 211 210 .t . . .. nn S yy t .S . ).)(().)(( nn S)n(S)n( S :H :H ,. p p p = = + − = + − = = = −+ −+− = −+ −+− = ≠ = µµ µµ Do not reject. There is no evidence to indicate that the new filtering device has affected the mean 2-14 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 5.263 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 α = 0.05. 2-11
• 12. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 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 µ µ µ µ = ≠ − + − − + − = = + − + − = − − = = = + + = = 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 Thickness Temp N Mean StDev SE Mean 95 8 9.37 2.10 0.74 100 8 6.89 1.60 0.56 Difference = mu ( 95) - mu (100) Estimate for difference: 2.475 95% CI for difference: (0.476, 4.474) T-Test of difference = 0 (vs not =): T-Value = 2.65 P-Value = 0.019 DF = 14 Both use Pooled StDev = 1.86 (b) What is the P-value for the test conducted in part (a)? P = 0.019 (c) Find a 95% confidence interval on the difference in means. Provide a practical interpretation of this interval. From the computer output the 95% confidence interval is . This confidence interval doesnot include 0 in it, there for there is a difference in the two temperatures on the thickness of the photo resist. 1 20.476 4.474µ µ≤ − ≤ (d) Draw dot diagrams to assist in interpreting the results from this experiment. 2-12
• 13. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Thickness 12.010.89.68.47.26.04.83.6 Temp 95 100 Dotplot of Thickness vs Temp (e) Check the assumption of normality of the photoresist thickness. 5 10 15 1 5 10 20 30 40 50 60 70 80 90 95 99 Data Percent AD* 1.767 Goodness of Fit Normal Probability Plot for Thick@95 ML Estimates - 95% CI Mean StDev 9.36663 1.96396 ML Estimates 2-13
• 14. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 2 7 12 1 5 10 20 30 40 50 60 70 80 90 95 99 Data Percent AD* 1.567 Goodness of Fit Normal Probability Plot for Thick@100 ML Estimates - 95% CI Mean StDev 6.89163 1.49207 ML Estimates 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 2-14
• 15. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY This result makes intuitive sense. More samples are needed to detect a smaller difference. 2-15 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 cool- down time. 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 α = 0.05. 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% CI for difference: (-4.295, -2.005) T-Test of difference = 0 (vs not =): 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 computer output, . This interval does not contain 0. The two samples are different. The 20 second cooling time gives a cosmetically better housing. 1 24.295 2.005µ µ− ≤ − ≤ − (d) Draw dot diagrams to assist in interpreting the results from this experiment. 2-15
• 16. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Ranking 8642 C4 10 sec 20 sec Dotplot of Ranking vs C4 (e) Check the assumption of normality for the data from this experiment. 0 4 8 1 5 10 20 30 40 50 60 70 80 90 95 99 Data Percent AD* 1.252 Goodness of Fit Normal Probability Plot for 10 seconds ML Estimates - 95% CI Mean StDev 3.35 1.95640 ML Estimates 2-16
• 17. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 2 3 4 5 6 7 8 9 10 11 1 5 10 20 30 40 50 60 70 80 90 95 99 Data Percent AD* 0.988 Goodness of Fit Normal Probability Plot for 20 seconds ML Estimates - 95% CI Mean StDev 6.5 1.50000 ML Estimates There are no significant departures from normality. 2-16 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 σ2. ( ) ( ) ( )( ) ( )( ) 2 2 2 2 2 , 1 (1 ), 1 2 2 2 2 2 2 1 1 20 1 0.88907 20 1 0.88907 32.852 8.907 0.457 1.686 n n n S n S α α σ χ χ σ σ − − − − − ≤ ≤ − − ≤ ≤ ≤ ≤ (b) Test the hypothesis that σ2 = 1.0. Use α = 0.05. What are your conclusions? H H 0 2 1 2 1 1 : : σ σ = ≠ χ σ 0 2 0 2 15019= = SS . 2-17
• 18. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY χ0 025 19 2 32852. , .= χ0 975 19 2 8 907. , .= Do not reject. There is no evidence to indicate that 2 1σ ≠ (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. 7.86.85.84.83.8 99 95 90 80 70 60 50 40 30 20 10 5 1 Data Percent 0.835AD* Goodness of Fit Normal Probability Plot for Uniformity ML Estimates Mean StDev 5.828 0.866560 ML Estimates 2-17 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 .000016 =∑ 0 003. =∑ 0 000045. 2-18
• 19. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY (a) Is there a significant difference between the means of the population of measurements represented by the two samples? Use α = 0.05. H H 0 1 2 1 1 2 : : µ µ µ µ = ≠ or equivalently 0 0 1 0 ≠ = d d :H :H µ µ 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 Dn n d d S S d t d t n n α αµ µ µ µ µ − − − ≤ = − ≤ + − ≤ ≤ + − ≤ ≤ 2-18 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 α = 0.05. H H 0 1 2 1 1 2 : : µ µ µ µ = ≠ or equivalently 0 0 1 0 ≠ = d d :H :H µ µ 2-19
• 20. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY ( ) 1 1 1 2.465 0.274 9 n i i d d n = = = =∑ 1 22 1 22 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 = = = 30629025012 .tt ,.n, ==−α , 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 (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. 377770170230 9 1350 30622740 9 1350 30622740 1 2 1 2 .. . .. . .. n S td n S td d d d n,d d n, ≤≤ +≤≤− +≤≤− −− µ µ µ αα (d) Investigate the normality assumption for both samples. P-Value: 0.537 A-Squared: 0.286 Anderson-Darling Normality Test N: 9 StDev : 0.146031 Av erage: 1.34011 1.551.451.351.251.15 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Karlsruhe Normal Probability Plot 2-20
• 21. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY P-Value: 0.028 A-Squared: 0.772 Anderson-Darling Normality Test N: 9 StDev : 0.0493806 Av erage: 1.06622 1.151.101.051.00 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Lehigh Normal Probability Plot (e) Investigate the normality assumption for the difference in ratios for the two methods. P-Value: 0.464 A-Squared: 0.318 Anderson-Darling Normality Test N: 9 StDev : 0.135099 Av erage: 0.273889 0.520.420.320.220.12 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Difference Normal Probability Plot (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. 2-19 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 2 212 199 198 177 176 198 194 213 216 197 185 188 211 191 200 206 200 189 193 195 184 201 197 203 2-21
• 22. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY (a) Construct normal probability plots for both samples. Do these plots support assumptions of normality and equal variance for both samples? P-Value: 0.227 A-Squared: 0.450 Anderson-Darling Normality Test N: 12 StDev : 10.1757 Av erage: 200.5 215205195185 .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 205195185175 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Form 2 Normal Probability Plot (b) Do the data support the claim that the mean deflection temperature under load for formulation 1 exceeds that of formulation 2? Use α = 0.05. Minitab Output Two Sample T-Test and Confidence Interval Two sample T for Form 1 vs Form 2 N Mean StDev SE Mean Form 1 12 200.5 10.2 2.9 Form 2 12 193.08 9.95 2.9 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 (c) What is the P-value for the test in part (a)? P = 0.042 2-22
• 23. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 2-20 Refer to the data in problem 2-19. 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 Two-Sample T-Test and CI: Form1, Form2 Two-sample T for Form1 vs Form2 N Mean StDev SE Mean Form1 12 200.5 10.2 2.9 Form2 12 193.08 9.95 2.9r 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 2-21 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 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 α = 0.05 and assume equal variances. 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 9.925 0.465 0.16 Solution 8 10.362 0.233 0.082 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 (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. 2-23
• 24. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY P-Value: 0.743 A-Squared: 0.222 Anderson-Darling Normality Test N: 8 StDev : 0.465219 Av erage: 9.925 10.510.09.5 .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 Av erage: 10.3625 10.710.610.510.410.310.210.110.0 .999 .99 .95 .80 .50 .20 .05 .01 .001 Probability Solution 2 Normal Probability Plot Both the normality and equality of variance assumptions are valid. 2-22 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 and are known. Develop a test statistic for 2 1σ 2 2σ H0: 2µ1 = µ2 H1: 2µ1 ≠ µ2 2 2 1 2 1 2 1 2 1 2 4 2 ~ 2 ,y y N n n σ σ µ µ ⎛ − − +⎜ ⎝ ⎠ ⎞ ⎟ , assuming that the data is normally distributed. The test statistic is: z y y n n o = − + 2 4 1 2 1 2 1 2 2 2 σ σ , reject if z zo > α 2 2-23 Suppose we are testing H0: µ1 = µ2 H1: µ1 ≠ µ2 2-24
• 25. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY where and are known. Our sampling resources are constrained such that n1 + n2 = N. How should we allocate the N observations between the two populations to obtain the most powerful test? 2 1σ 2 2σ The most powerful test is attained by the n1 and n2 that maximize zo for given y y1 2− . Thus, we chose n1 and n2 to max z y y n n o = − + 1 2 1 2 1 2 2 2 σ σ , subject to n1 + n2 = N. This is equivalent to min L n n n N n = + = + − σ σ σ σ1 2 1 2 2 2 1 2 1 2 2 1 , subject to n1 + n2 = N. Now ( ) 2 2 1 2 22 1 1 1 0 dL dn n N n σ σ− = + = − , implies that n1 / n2 = σ1 / σ2. Thus n1 and n2 are assigned proportionally to the ratio of the standard deviations. This has intuitive appeal, as it allocates more observations to the population with the greatest variability. 2-24 Develop Equation 2-46 for a 100(1 - α) percent confidence interval for the variance of a normal distribution. 2 12 ~ n SS χ σ − . Thus, 2 2 1 , 1 , 1 2 2 2 1n n SS P α αχ χ α σ− − − ⎧ ≤ ≤ = −⎨ ⎩ ⎭ ⎫ ⎬ . Therefore, 2 2 , 1 1 , 1 2 2 2 1 n n SS SS P α α σ α χ χ− − − ⎧ ⎫ ⎪ ⎪ ≤ ≤ = −⎨ ⎬ ⎪ ⎪⎩ ⎭ , so 2 2 , 1 1 , 1 2 2 , n n SS SS α αχ χ− − − ⎡ ⎤ ⎢ ⎢ ⎥ ⎣ ⎦ ⎥ is the 100(1 - α)% confidence interval on σ2. 2-25 Develop Equation 2-50 for a 100(1 - α) percent confidence interval for the ratio / , where and are the variances of two normal distributions. σ1 2 2 2σ 2 1σ 2 2σ 2 1 2 2 2 2 1, 12 2 1 1 ~ n n S F S σ σ − − 2 12 2 2 2 2 2 1 , 1, 1 2 2 , 1, 12 11 1 1n n n n S P F F S α α σ α σ− − − − − ⎧ ⎫ ≤ ≤ =⎨ ⎬ ⎩ ⎭ − or 2 12 2 2 2 2 1 1 1 1 , 1, 12 2 2 , 1, 12 1 2 2 2 1n n n n S S P F F S S α α σ α σ− − − − − ⎧ ⎫ ≤ ≤ = −⎨ ⎬ ⎩ ⎭ 2-26 Develop an equation for finding a 100(1 - α) percent confidence interval on the difference in the means of two normal distributions where ≠ . Apply your equation to the portland cement experiment data, and find a 95% confidence interval. 2 1σ 2 2σ 2-25
• 26. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY ( ) ( ) 2 1 2 1 2 ,2 2 1 2 1 2 ~ y y t S S n n α υ µ µ− − − + ( ) ( )2 2 2 2 2 2 1 2 1 2 1 2 1 2, , 1 2 1 2 S S S S t y y t n n n n α αυ υµ µ+ ≤ − − − ≤ + ( ) ( ) ( )2 2 2 2 2 1 2 1 2 1 2 1 2 1 2, , 1 2 1 2 S S S S y y t y y t n n n n α αυ υµ µ− − + ≤ − ≤ − + + 2 where 2 2 2 1 2 1 2 2 2 2 2 1 2 1 2 1 21 1 S S n n S S n n n n υ ⎛ ⎞ +⎜ ⎟ ⎝ ⎠= ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠+ − − Using the data from Table 2-1 n y S 1 1 1 2 10 16 764 0100138 = = = . . n y S 2 2 2 2 10 17 343 0 0614622 = = = . . ( ) ( )1 2 0.100138 0.0614622 16.764 17.343 2.110 10 10 µ µ− − + ≤ − ≤ ( ) 0.100138 0.0614622 16.764 17.343 2.110 10 10 − + + where 1702417 110 10 06146220 110 10 1001380 10 06146220 10 1001380 22 2 ≅= − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = . .. .. υ ( )1 21.426 0.889µ µ− ≤ − ≤ − This agrees with the result in Table 2-2. 2-27 Construct a data set for which the paired t-test statistic is very large, but for which the usual two- sample or pooled t-test statistic is small. In general, describe how you created the data. Does this give you any insight regarding how the paired t-test works? A B delta 7.1662 8.2416 1.07541 2.3590 2.4555 0.09650 19.9977 21.1018 1.10412 0.9077 2.3401 1.43239 -15.9034 -15.0013 0.90204 -6.0722 -5.5941 0.47808 2-26
• 27. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 9.9501 10.6910 0.74085 -1.0944 -0.1358 0.95854 -4.6907 -3.3446 1.34615 -6.6929 -5.9303 0.76256 Minitab Output Paired T-Test and Confidence Interval Paired T for A - B N Mean StDev SE Mean A 10 0.59 10.06 3.18 B 10 1.48 10.11 3.20 Difference 10 -0.890 0.398 0.126 95% CI for mean difference: (-1.174, -0.605) T-Test of mean difference = 0 (vs not = 0): T-Value = -7.07 P-Value = 0.000 Two Sample T-Test and Confidence Interval Two sample T for A vs B N Mean StDev SE Mean A 10 0.6 10.1 3.2 B 10 1.5 10.1 3.2 95% CI for mu A - mu B: ( -10.4, 8.6) T-Test mu A = mu B (vs not =): T = -0.20 P = 0.85 DF = 18 Both use Pooled StDev = 10.1 These two sets of data were created by making the observation for A and B moderately different within each pair (or block), but making the observations between pairs very different. The fact that the difference between pairs is large makes the pooled estimate of the standard deviation large and the two-sample t-test statistic small. Therefore the fairly small difference between the means of the two treatments that is present when they are applied to the same experimental unit cannot be detected. Generally, if the blocks are very different, then this will occur. Blocking eliminates the variabiliy associated with the nuisance variable that they represent. 2-28 Consider the experiment described in problem 2-11. If the mean burning times of the two flames differ by as much as 2 minutes, find the power of the test. What sample size would be required to detect an actual difference in mean burning time of 1 minute with a power of at least 0.90? 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 = 9.32 Sample Target Actual Difference Size Power Power 2 458 0.9000 0.9004 2-29 Reconsider the bottle filling experiment described in Problem 2-9. Rework this problem assuming that the two population variances are unknown but equal. 2-27
• 28. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Minitab Output Two-Sample T-Test and CI: Machine 1, Machine 2 Two-sample T for Machine 1 vs Machine 2 N Mean StDev SE Mean Machine 10 16.0150 0.0303 0.0096 Machine 10 16.0050 0.0255 0.0081 Difference = mu Machine 1 - mu Machine 2 Estimate for difference: 0.0100 95% CI for difference: (-0.0163, 0.0363) T-Test of difference = 0 (vs not =): T-Value = 0.80 P-Value = 0.435 DF = 18 Both use Pooled StDev = 0.0280 The hypothesis test is the same: H0: µ1 = µ2 H1: µ1 ≠ µ2 The conclusions are the same as Problem 2-9, do not reject H0. There is no difference in the machines. The P-value for this anlysis is 0.435. The confidence interval is (-0.0163, 0.0363). This interval contains 0. There is no difference in machines. 2-29 Consider the data from problem 2-9. If the mean fill volume of the two machines differ by as much as 0.25 ounces, what is the power of the test used in problem 2-9? What sample size could result in a power of at least 0.9 if the actual difference in mean fill volume is 0.25 ounces? 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 = 0.028 Sample Difference Size Power 0.25 10 1.0000 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 = 0.028 Sample Target Actual Difference Size Power Power 0.25 2 0.9000 0.9805 2-28
• 29. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Chapter 3 Experiments with a Single Factor: The Analysis of Variance Solutions 3-1 The tensile strength of portland cement is being studied. Four different mixing techniques can be used economically. The following data have been collected: Mixing Technique Tensile Strength (lb/in2 ) 1 3129 3000 2865 2890 2 3200 3300 2975 3150 3 2800 2900 2985 3050 4 2600 2700 2600 2765 (a) Test the hypothesis that mixing techniques affect the strength of the cement. Use α = 0.05. Design Expert Output Response: Tensile Strengthin lb/in^2 ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 4.897E+005 3 1.632E+005 12.73 0.0005 significant A 4.897E+005 3 1.632E+005 12.73 0.0005 Residual 1.539E+005 12 12825.69 Lack of Fit 0.000 0 Pure Error 1.539E+005 12 12825.69 Cor Total 6.436E+005 15 The Model F-value of 12.73 implies the model is significant. There is only a 0.05% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error 1-1 2971.00 56.63 2-2 3156.25 56.63 3-3 2933.75 56.63 4-4 2666.25 56.63 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > |t| 1 vs 2 -185.25 1 80.08 -2.31 0.0392 1 vs 3 37.25 1 80.08 0.47 0.6501 1 vs 4 304.75 1 80.08 3.81 0.0025 2 vs 3 222.50 1 80.08 2.78 0.0167 2 vs 4 490.00 1 80.08 6.12 < 0.0001 3 vs 4 267.50 1 80.08 3.34 0.0059 The F-value is 12.73 with a corresponding P-value of .0005. Mixing technique has an effect. (b) Construct a graphical display as described in Section 3-5.3 to compare the mean tensile strengths for the four mixing techniques. What are your conclusions? 62556 4 712825 . . n MS S E y .i === 3-1
• 30. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 2 7 0 0 2 8 0 0 2 9 0 0 3 0 0 0 3 1 0 0 T e n s ile S tre n g th S c a le d t D is trib u tio n (3 ) (2 )(1 )(4 ) Based on examination of the plot, we would conclude that 1µ and 3µ are the same; that 4µ differs from 1µ and 3µ , that 2µ differs from 1µ and 3µ , and that 2µ and 4µ are different. (c) Use the Fisher LSD method with α=0.05 to make comparisons between pairs of means. 4951748564121792 4 7128252 2 4160250 2 ...LSD ).( tLSD n MS tLSD ,. aN, E == = = − −α Treatment 2 vs. Treatment 4 = 3156.250 - 2666.250 = 490.000 > 174.495 Treatment 2 vs. Treatment 3 = 3156.250 - 2933.750 = 222.500 > 174.495 Treatment 2 vs. Treatment 1 = 3156.250 - 2971.000 = 185.250 > 174.495 Treatment 1 vs. Treatment 4 = 2971.000 - 2666.250 = 304.750 > 174.495 Treatment 1 vs. Treatment 3 = 2971.000 - 2933.750 = 37.250 < 174.495 Treatment 3 vs. Treatment 4 = 2933.750 - 2666.250 = 267.500 > 174.495 The Fisher LSD method is also presented in the Design-Expert computer output above. The results agree with the graphical method for this experiment. (d) Construct a normal probability plot of the residuals. What conclusion would you draw about the validity of the normality assumption? There is nothing unusual about the normal probability plot of residuals. 3-2
• 31. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Residual Normal%probability Normal plot of residuals -181.25 -96.4375 -11.625 73.1875 158 1 5 10 20 30 50 70 80 90 95 99 (e) Plot the residuals versus the predicted tensile strength. Comment on the plot. There is nothing unusual about this plot. 22 Predicted Residuals Residuals vs. Predicted -181.25 -96.4375 -11.625 73.1875 158 2666.25 2788.75 2911.25 3033.75 3156.25 (f) Prepare a scatter plot of the results to aid the interpretation of the results of this experiment. Design-Expert automatically generates the scatter plot. The plot below also shows the sample average for each treatment and the 95 percent confidence interval on the treatment mean. 3-3
• 32. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Technique TensileStrength One Factor Plot 1 2 3 4 2579.01 2759.26 2939.51 3119.75 3300 22 3-2 (a) Rework part (b) of Problem 3-1 using Tukey’s test with α = 0.05. Do you get the same conclusions from Tukey’s test that you did from the graphical procedure and/or the Fisher LSD method? Minitab Output Tukey's pairwise comparisons Family error rate = 0.0500 Individual error rate = 0.0117 Critical value = 4.20 Intervals for (column level mean) - (row level mean) 1 2 3 2 -423 53 3 -201 -15 275 460 4 67 252 30 543 728 505 No, the conclusions are not the same. The mean of Treatment 4 is different than the means of Treatments 1, 2, and 3. However, the mean of Treatment 2 is not different from the means of Treatments 1 and 3 according to the Tukey method, they were found to be different using the graphical method and the Fisher LSD method. (b) Explain the difference between the Tukey and Fisher procedures. Both Tukey and Fisher utilize a single critical value; however, Tukey’s is based on the studentized range statistic while Fisher’s is based on t distribution. 3-3 Reconsider the experiment in Problem 3-1. Find a 95 percent confidence interval on the mean tensile strength of the portland cement produced by each of the four mixing techniques. Also find a 95 3-4
• 33. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY percent confidence interval on the difference in means for techniques 1 and 3. Does this aid in interpreting the results of the experiment? n MS ty n MS ty E aN,.ii E aN,.i −− +≤≤− 22 αα µ Treatment 1: 4 1282837 17922971 .± 3871232971 .± 38730946132847 1 .. ≤≤ µ Treatment 2: 3156.25±123.387 63732798633032 2 .. ≤≤ µ Treatment 3: 2933.75±123.387 13730573632810 3 .. ≤≤ µ Treatment 4: 2666.25±123.387 63727898632542 4 .. ≤≤ µ Treatment 1 - Treatment 3: n MS tyy n MS tyy E aN,.j.iji E aN,.j.i 22 22 −− +−≤−≤−− αα µµ ( ) 4 7128252 1792752933002971 . ... ±− 745211245137 31 .. ≤−≤− µµ 3-4 A product developer is investigating the tensile strength of a new synthetic fiber that will be used to make cloth for men’s shirts. Strength is usually affected by the percentage of cotton used in the blend of materials for the fiber. The engineer conducts an experiment with five levels of cotton content and replicated the experiment five times. The data are shown in the following table. Cotton Weight Percentage Observations 15 7 7 15 11 9 20 12 17 12 18 18 25 14 19 19 18 18 30 19 25 22 19 23 35 7 10 11 15 11 (a) Is there evidence to support the claim that cotton content affects the mean tensile strength? Use α = 0.05. Minitab Output One-way ANOVA: Tensile Strength versus Cotton Percentage Analysis of Variance for Tensile Source DF SS MS F P Cotton P 4 475.76 118.94 14.76 0.000 Error 20 161.20 8.06 Total 24 636.96 Yes, the F-value is 14.76 with a corresponding P-value of 0.000. The percentage of cotton in the fiber appears to have an affect on the tensile strength. 3-5
• 34. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY (b) Use the Fisher LSD method to make comparisons between the pairs of means. What conclusions can you draw? Minitab Output Fisher's pairwise comparisons Family error rate = 0.264 Individual error rate = 0.0500 Critical value = 2.086 Intervals for (column level mean) - (row level mean) 15 20 25 30 20 -9.346 -1.854 25 -11.546 -5.946 -4.054 1.546 30 -15.546 -9.946 -7.746 -8.054 -2.454 -0.254 35 -4.746 0.854 3.054 7.054 2.746 8.346 10.546 14.546 In the Minitab output the pairs of treatments that do not contain zero in the pair of numbers indicates that there is a difference in the pairs of the treatments. 15% cotton is different than 20%, 25% and 30%. 20% cotton is different than 30% and 35% cotton. 25% cotton is different than 30% and 35% cotton. 30% cotton is different than 35%. (c) Analyze the residuals from this experiment and comment on model adequacy. -4 -3 -2 -1 0 1 2 3 4 5 6 -2 -1 0 1 2 NormalScore Residual Normal Probability Plot of the Residuals (response is Tensile) 3-6
• 35. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 10 15 20 -4 -3 -2 -1 0 1 2 3 4 5 6 Fitted Value Residual Residuals Versus the Fitted Values (response is Tensile) The residuals show nothing unusual. 3-5 Reconsider the experiment described in Problem 3-4. Suppose that 30 percent cotton content is a control. Use Dunnett’s test with α = 0.05 to compare all of the other means with the control. For this problem: a = 5, a-1 = 4, f=20, n=5 and α = 0.05 0.05 2 2(8.06) (4,20) 2.65 4.76EMS d n n = = 1. 4. 2. 4. 3. 4. 5. 4. 1 vs. 4 : 9.8 21.6 11.8* 2 vs. 4 : 15.4 21.6 6.2* 3 vs. 4 : 17.6 21.6 4.0 5 vs. 4 : 10.8 21.6 10.6* y y y y y y y y − = − = − − = − = − − = − = − − = − = − The control treatment, treatment 4, differs from treatments 1,2 and 5. 3-6 A pharmaceutical manufacturer wants to investigate the bioactivity of a new drug. A completely randomized single-factor experiment was conducted with three dosage levels, and the following results were obtained. Dosage Observations 20g 24 28 37 30 30g 37 44 31 35 3-7
• 36. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 40g 42 47 52 38 (a) Is there evidence to indicate that dosage level affects bioactivity? Use α = 0.05. Minitab Output One-way ANOVA: Activity versus Dosage Analysis of Variance for Activity Source DF SS MS F P Dosage 2 450.7 225.3 7.04 0.014 Error 9 288.3 32.0 Total 11 738.9 There appears to be a different in the dosages. (b) If it is appropriate to do so, make comparisons between the pairs of means. What conclusions can you draw? Because there appears to be a difference in the dosages, the comparison of means is appropriate. Minitab Output Tukey's pairwise comparisons Family error rate = 0.0500 Individual error rate = 0.0209 Critical value = 3.95 Intervals for (column level mean) - (row level mean) 20g 30g 30g -18.177 4.177 40g -26.177 -19.177 -3.823 3.177 The Tukey comparison shows a difference in the means between the 20g and the 40g dosages. (c) Analyze the residuals from this experiment and comment on the model adequacy. 3-8
• 37. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 50-5 1 0 -1 -2 NormalScore Residual Normal Probability Plot of the Residuals (response is Activity) 30 35 40 45 -5 0 5 Fitted Value Residual Residuals Versus the Fitted Values (response is Activity) There is nothing too unusual about the residuals. 3-7 A rental car company wants to investigate whether the type of car rented affects the length of the rental period. An experiment is run for one week at a particular location, and 10 rental contracts are selected at random for each car type. The results are shown in the following table. 3-9
• 38. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Type of Car Observations Sub-compact 3 5 3 7 6 5 3 2 1 6 Compact 1 3 4 7 5 6 3 2 1 7 Midsize 4 1 3 5 7 1 2 4 2 7 Full Size 3 5 7 5 10 3 4 7 2 7 (a) Is there evidence to support a claim that the type of car rented affects the length of the rental contract? Use α = 0.05. If so, which types of cars are responsible for the difference? Minitab Output One-way ANOVA: Days versus Car Type Analysis of Variance for Days Source DF SS MS F P Car Type 3 16.68 5.56 1.11 0.358 Error 36 180.30 5.01 Total 39 196.98 There is no difference. (b) Analyze the residuals from this experiment and comment on the model adequacy. -4 -3 -2 -1 0 1 2 3 4 5 -2 -1 0 1 2 NormalScore Residual Normal Probability Plot of the Residuals (response is Days) 3-10
• 39. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 3.5 4.5 5.5 -4 -3 -2 -1 0 1 2 3 4 5 Fitted Value Residual Residuals Versus the Fitted Values (response is Days) There is nothing unusual about the residuals. (c) Notice that the response variable in this experiment is a count. Should the cause any potential concerns about the validity of the analysis of variance? Because the data is count data, a square root transformation could be applied. The analysis is shown below. It does not change the interpretation of the data. Minitab Output One-way ANOVA: Sqrt Days versus Car Type Analysis of Variance for Sqrt Day Source DF SS MS F P Car Type 3 1.087 0.362 1.10 0.360 Error 36 11.807 0.328 Total 39 12.893 3-8 I belong to a golf club in my neighborhood. I divide the year into three golf seasons: summer (June- September), winter (November-March) and shoulder (October, April and May). I believe that I play my best golf during the summer (because I have more time and the course isn’t crowded) and shoulder (because the course isn’t crowded) seasons, and my worst golf during the winter (because all of the part- year residents show up, and the course is crowded, play is slow, and I get frustrated). Data from the last year are shown in the following table. Season Observations Summer 83 85 85 87 90 88 88 84 91 90 Shoulde r 91 87 84 87 85 86 83 Winter 94 91 87 85 87 91 92 86 3-11
• 40. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY (a) Do the data indicate that my opinion is correct? Use α = 0.05. Minitab Output One-way ANOVA: Score versus Season Analysis of Variance for Score Source DF SS MS F P Season 2 35.61 17.80 2.12 0.144 Error 22 184.63 8.39 Total 24 220.24 The data do not support the author’s opinion. (b) Analyze the residuals from this experiment and comment on model adequacy. -4 -3 -2 -1 0 1 2 3 4 5 -2 -1 0 1 2 NormalScore Residual Normal Probability Plot of the Residuals (response is Score) 3-12
• 41. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 89888786 5 4 3 2 1 0 -1 -2 -3 -4 Fitted Value Residual Residuals Versus the Fitted Values (response is Score) There is nothing unusual about the residuals. 3-9 A regional opera company has tried three approaches to solicit donations from 24 potential sponsors. The 24 potential sponsors were randomly divided into three groups of eight, and one approach was used for each group. The dollar amounts of the resulting contributions are shown in the following table. Approac h Contributions (in \$) 1 1000 1500 1200 1800 1600 1100 1000 1250 2 1500 1800 2000 1200 2000 1700 1800 1900 3 900 1000 1200 1500 1200 1550 1000 1100 (a) Do the data indicate that there is a difference in results obtained from the three different approaches? Use α = 0.05. Minitab Output One-way ANOVA: Contribution versus Approach Analysis of Variance for Contribution Source DF SS MS F P Approach 2 1362708 681354 9.41 0.001 Error 21 1520625 72411 Total 23 2883333 There is a difference between the approaches. The Tukey test will indicate which are different. Approach 2 is different than approach 3. Minitab Output Tukey's pairwise comparisons Family error rate = 0.0500 Individual error rate = 0.0200 Critical value = 3.56 3-13
• 42. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Intervals for (column level mean) - (row level mean) 1 2 2 -770 -93 3 -214 218 464 895 (b) Analyze the residuals from this experiment and comment on the model adequacy. -500 0 500 -2 -1 0 1 2 NormalScore Residual Normal Probability Plot of the Residuals (response is Contribu) 3-14
• 43. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 1150 1250 1350 1450 1550 1650 1750 -500 0 500 Fitted Value Residual Residuals Versus the Fitted Values (response is Contribu) There is nothing unusual about the residuals. 3-10 An experiment was run to determine whether four specific firing temperatures affect the density of a certain type of brick. The experiment led to the following data: Temperature Density 100 21.8 21.9 21.7 21.6 21.7 125 21.7 21.4 21.5 21.4 150 21.9 21.8 21.8 21.6 21.5 175 21.9 21.7 21.8 21.4 (a) Does the firing temperature affect the density of the bricks? Use α = 0.05. No, firing temperature does not affect the density of the bricks. Refer to the Design-Expert output below. Design Expert Output Response: Density ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 0.16 3 0.052 2.02 0.1569 not significant A 0.16 3 0.052 2.02 0.1569 Residual 0.36 14 0.026 Lack of Fit 0.000 0 Pure Error 0.36 14 0.026 Cor Total 0.52 17 The "Model F-value" of 2.02 implies the model is not significant relative to the noise. There is a 15.69 % chance that a "Model F-value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) 3-15
• 44. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Estimated Standard Mean Error 1-100 21.74 0.072 2-125 21.50 0.080 3-150 21.72 0.072 4-175 21.70 0.080 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > |t| 1 vs 2 0.24 1 0.11 2.23 0.0425 1 vs 3 0.020 1 0.10 0.20 0.8465 1 vs 4 0.040 1 0.11 0.37 0.7156 2 vs 3 -0.22 1 0.11 -2.05 0.0601 2 vs 4 -0.20 1 0.11 -1.76 0.0996 3 vs 4 0.020 1 0.11 0.19 0.8552 (b) Is it appropriate to compare the means using the Fisher LSD method in this experiment? The analysis of variance tells us that there is no difference in the treatments. There is no need to proceed with Fisher’s LSD method to decide which mean is difference. (c) Analyze the residuals from this experiment. Are the analysis of variance assumptions satisfied? There is nothing unusual about the residual plots. Residual Normal%probability Normal plot of residuals -0.3 -0.175 -0.05 0.075 0.2 1 5 10 20 30 50 70 80 90 95 99 22 22 22 Predicted Residuals Residuals vs. Predicted -0.3 -0.175 -0.05 0.075 0.2 21.50 21.56 21.62 21.68 21.74 (d) Construct a graphical display of the treatments as described in Section 3-5.3. Does this graph adequately summarize the results of the analysis of variance in part (b). Yes. 3-16
• 45. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 2 1 .2 2 1 .3 2 1 .4 2 1 .5 2 1 .6 2 1 .7 2 1 .8 M e a n D e n s ity S c a le d t D is trib u tio n (12 5 ) (1 7 5,15 0 ,1 0 0) 3-11 Rework Part (d) of Problem 3-10 using the Tukey method. What conclusions can you draw? Explain carefully how you modified the procedure to account for unequal sample sizes. When sample sizes are unequal, the appropriate formula for the Tukey method is ( , ) 1 1 2 E i j q a f T MS n n α α ⎛ ⎞ = +⎜ ⎟⎜ ⎟ ⎝ ⎠ Treatment 1 vs. Treatment 2= 21.74 – 21.50 = 0.24 < 0.994 Treatment 1 vs. Treatment 3= 21.74 – 21.72 = 0.02 < 0.937 Treatment 1 vs. Treatment 4= 21.74 – 21.70 = 0.04 < 0.994 Treatment 3 vs. Treatment 2= 21.72 – 21.50 = 0.22 < 1.048 Treatment 4 vs. Treatment 2= 21.70 – 21.50 = 0.20 < 1.048 Treatment 3 vs. Treatment 4= 21.72 – 21.70 = 0.02 < 0.994 All pairwise comparisons do not identify differences. Notice that there are different critical values for the comparisons depending on the sample sizes of the two groups being compared. Because we could not reject the hypothesis of equal means using the analysis of variance, we should never have performed the Tukey test (or any other multiple comparison procedure, for that matter). If you ignore the analysis of variance results and run multiple comparisons, you will likely make type I errors. 3-12 A manufacturer of television sets is interested in the effect of tube conductivity of four different types of coating for color picture tubes. The following conductivity data are obtained: Coating Type Conductivity 1 143 141 150 146 2 152 149 137 143 3 134 136 132 127 4 129 127 132 129 (a) Is there a difference in conductivity due to coating type? Use α = 0.05. Yes, there is a difference in means. Refer to the Design-Expert output below.. 3-17
• 46. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Design Expert Output ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 844.69 3 281.56 14.30 0.0003 significant A 844.69 3 281.56 14.30 0.0003 Residual 236.25 12 19.69 Lack of Fit 0.000 0 Pure Error 236.25 12 19.69 Cor Total 1080.94 15 The Model F-value of 14.30 implies the model is significant. There is only a 0.03% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error 1-1 145.00 2.22 2-2 145.25 2.22 3-3 132.25 2.22 4-4 129.25 2.22 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > |t| 1 vs 2 -0.25 1 3.14 -0.080 0.9378 1 vs 3 12.75 1 3.14 4.06 0.0016 1 vs 4 15.75 1 3.14 5.02 0.0003 2 vs 3 13.00 1 3.14 4.14 0.0014 2 vs 4 16.00 1 3.14 5.10 0.0003 3 vs 4 3.00 1 3.14 0.96 0.3578 (b) Estimate the overall mean and the treatment effects. 68758937513725129 68755937513725132 31257937513725145 06257937513700145 9375137162207 44 33 22 11 ...yyˆ ...yyˆ ...yyˆ ...yyˆ ./ˆ ... ... ... ... −=−=−= −=−=−= =−=−= =−=−= == τ τ τ τ µ (c) Compute a 95 percent interval estimate of the mean of coating type 4. Compute a 99 percent interval estimate of the mean difference between coating types 1 and 4. Treatment 4: 4 6919 179225129 . .. ± 08451344155124 4 .. ≤≤ µ Treatment 1 - Treatment 4: ( ) ( ) 4 69192 055325129145 . .. ±− 336251646 41 .. ≤−≤ µµ (d) Test all pairs of means using the Fisher LSD method with α=0.05. Refer to the Design-Expert output above. The Fisher LSD procedure is automatically included in the output. 3-18
• 47. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY The means of Coating Type 2 and Coating Type 1 are not different. The means of Coating Type 3 and Coating Type 4 are not different. However, Coating Types 1 and 2 produce higher mean conductivity that does Coating Types 3 and 4. (e) Use the graphical method discussed in Section 3-5.3 to compare the means. Which coating produces the highest conductivity? 2192 4 9616 . . n MS S E y .i === Coating types 1 and 2 produce the highest conductivity. 1 3 0 1 3 5 1 4 0 1 4 5 1 5 0 C o n d u c tiv ity S c a le d t D is trib u tio n (3 ) (2 )(1 )(4 ) (f) Assuming that coating type 4 is currently in use, what are your recommendations to the manufacturer? We wish to minimize conductivity. Since coatings 3 and 4 do not differ, and as they both produce the lowest mean values of conductivity, use either coating 3 or 4. As type 4 is currently being used, there is probably no need to change. 3-13 Reconsider the experiment in Problem 3-12. Analyze the residuals and draw conclusions about model adequacy. There is nothing unusual in the normal probability plot. A funnel shape is seen in the plot of residuals versus predicted conductivity indicating a possible non-constant variance. 3-19
• 48. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Residual Normal%probability Normal plot of residuals -8.25 -4.5 -0.75 3 6.75 1 5 10 20 30 50 70 80 90 95 99 22 Predicted Residuals Residuals vs. Predicted -8.25 -4.5 -0.75 3 6.75 129.25 133.25 137.25 141.25 145.25 22 Coating Type Residuals Residuals vs. Coating Type -8.25 -4.5 -0.75 3 6.75 1 2 3 4 3-14 An article in the ACI Materials Journal (Vol. 84, 1987. pp. 213-216) describes several experiments investigating the rodding of concrete to remove entrapped air. A 3” x 6” cylinder was used, and the number of times this rod was used is the design variable. The resulting compressive strength of the concrete specimen is the response. The data are shown in the following table. Rodding Level Compressive Strength 10 1530 1530 1440 15 1610 1650 1500 20 1560 1730 1530 25 1500 1490 1510 (a) Is there any difference in compressive strength due to the rodding level? Use α = 0.05. There are no differences. Design Expert Output 3-20
• 49. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 28633.33 3 9544.44 1.87 0.2138 not significant A 28633.33 3 9544.44 1.87 0.2138 Residual 40933.33 8 5116.67 Lack of Fit 0.000 0 Pure Error 40933.33 8 5116.67 Cor Total 69566.67 11 The "Model F-value" of 1.87 implies the model is not significant relative to the noise. There is a 21.38 % chance that a "Model F-value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error 1-10 1500.00 41.30 2-15 1586.67 41.30 3-20 1606.67 41.30 4-25 1500.00 41.30 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > |t| 1 vs 2 -86.67 1 58.40 -1.48 0.1761 1 vs 3 -106.67 1 58.40 -1.83 0.1052 1 vs 4 0.000 1 58.40 0.000 1.0000 2 vs 3 -20.00 1 58.40 -0.34 0.7408 2 vs 4 86.67 1 58.40 1.48 0.1761 3 vs 4 106.67 1 58.40 1.83 0.1052 (b) Find the P-value for the F statistic in part (a). From computer output, P=0.2138. (c) Analyze the residuals from this experiment. What conclusions can you draw about the underlying model assumptions? There is nothing unusual about the residual plots. Residual Normal%probability Normal plot of residuals -86.6667 -34.1667 18.3333 70.8333 123.333 1 5 10 20 30 50 70 80 90 95 99 22 Predicted Residuals Residuals vs. Predicted -86.6667 -34.1667 18.3333 70.8333 123.333 1500.00 1526.67 1553.33 1580.00 1606.67 3-21
• 50. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 22 Rodding Level Residuals Residuals vs. Rodding Level -86.6667 -34.1667 18.3333 70.8333 123.333 1 2 3 4 (d) Construct a graphical display to compare the treatment means as describe in Section 3-5.3. 1 4 1 8 1 4 5 9 1 5 0 0 1 5 4 1 1 5 8 2 1 6 2 3 1 6 6 4 M e a n C o m p re s s ive S tre n g th S c a le d t D is trib u tio n (1 0 , 2 5 ) (1 5 ) (2 0 ) 3-15 An article in Environment International (Vol. 18, No. 4, 1992) describes an experiment in which the amount of radon released in showers was investigated. Radon enriched water was used in the experiment and six different orifice diameters were tested in shower heads. The data from the experiment are shown in the following table. Orifice Dia. Radon Released (%) 0.37 80 83 83 85 0.51 75 75 79 79 0.71 74 73 76 77 1.02 67 72 74 74 1.40 62 62 67 69 1.99 60 61 64 66 (a) Does the size of the orifice affect the mean percentage of radon released? Use α = 0.05. 3-22
• 51. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Yes. There is at least one treatment mean that is different. Design Expert Output Response: Radon Released in % ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 1133.38 5 226.68 30.85 < 0.0001 significant A 1133.38 5 226.68 30.85 < 0.0001 Residual 132.25 18 7.35 Lack of Fit 0.000 0 Pure Error 132.25 18 7.35 Cor Total 1265.63 23 The Model F-value of 30.85 implies the model is significant. There is only a 0.01% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) EstimatedStandard Mean Error 1-0.37 82.75 1.36 2-0.51 77.00 1.36 3-0.71 75.00 1.36 4-1.02 71.75 1.36 5-1.40 65.00 1.36 6-1.99 62.75 1.36 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > |t| 1 vs 2 5.75 1 1.92 3.00 0.0077 1 vs 3 7.75 1 1.92 4.04 0.0008 1 vs 4 11.00 1 1.92 5.74 < 0.0001 1 vs 5 17.75 1 1.92 9.26 < 0.0001 1 vs 6 20.00 1 1.92 10.43 < 0.0001 2 vs 3 2.00 1 1.92 1.04 0.3105 2 vs 4 5.25 1 1.92 2.74 0.0135 2 vs 5 12.00 1 1.92 6.26 < 0.0001 2 vs 6 14.25 1 1.92 7.43 < 0.0001 3 vs 4 3.25 1 1.92 1.70 0.1072 3 vs 5 10.00 1 1.92 5.22 < 0.0001 3 vs 6 12.25 1 1.92 6.39 < 0.0001 4 vs 5 6.75 1 1.92 3.52 0.0024 4 vs 6 9.00 1 1.92 4.70 0.0002 5 vs 6 2.25 1 1.92 1.17 0.2557 (b) Find the P-value for the F statistic in part (a). P=3.161 x 10-8 (c) Analyze the residuals from this experiment. There is nothing unusual about the residuals. 3-23
• 52. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Residual Normal%probability Normal plot of residuals -4.75 -2.5625 -0.375 1.8125 4 1 5 10 20 30 50 70 80 90 95 99 22 22 22 22 22 Predicted Residuals Residuals vs. Predicted -4.75 -2.5625 -0.375 1.8125 4 62.75 67.75 72.75 77.75 82.75 22 22 22 22 22 Orifice Diameter Residuals Residuals vs. Orifice Diameter -4.75 -2.5625 -0.375 1.8125 4 1 2 3 4 5 6 (d) Find a 95 percent confidence interval on the mean percent radon released when the orifice diameter is 1.40. Treatment 5 (Orifice =1.40): 4 357 10126 . .± 8486715262 .. ≤≤ µ (e) Construct a graphical display to compare the treatment means as describe in Section 3-5.3. What conclusions can you draw? 3-24
• 53. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 6 0 6 5 7 0 7 5 8 0 C o n d u c tivity S c a le d t D is trib u tio n (6 ) (5 ) (3 ) (2 ) (1 )(4 ) Treatments 5 and 6 as a group differ from the other means; 2, 3, and 4 as a group differ from the other means, 1 differs from the others. 3-16 The response time in milliseconds was determined for three different types of circuits that could be used in an automatic valve shutoff mechanism. The results are shown in the following table. Circuit Type Response Time 1 9 12 10 8 15 2 20 21 23 17 30 3 6 5 8 16 7 (a) Test the hypothesis that the three circuit types have the same response time. Use α = 0.01. From the computer printout, F=16.08, so there is at least one circuit type that is different. Design Expert Output Response: Response Time in ms ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 543.60 2 271.80 16.08 0.0004 significant A 543.60 2 271.80 16.08 0.0004 Residual 202.80 12 16.90 Lack of Fit 0.000 0 Pure Error 202.80 12 16.90 Cor Total 746.40 14 The Model F-value of 16.08 implies the model is significant. There is only a 0.04% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error 1-1 10.80 1.84 2-2 22.20 1.84 3-3 8.40 1.84 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > |t| 3-25
• 54. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 1 vs 2 -11.40 1 2.60 -4.38 0.0009 1 vs 3 2.40 1 2.60 0.92 0.3742 2 vs 3 13.80 1 2.60 5.31 0.0002 (b) Use Tukey’s test to compare pairs of treatment means. Use α = 0.01. 83851 5 1690 . n MS S E y .i === ( ) 045123010 .q ,,. = ( ) 2669045838510 ...t == 1 e computer output) gives the same results. (c) eans. What conclusions can you draw? How do they compare with the conclusions from part (a). case, the large difference between the mean of treatment 2 vs. 2: ⏐10.8-22.2⏐=11.4 > 9.266 1 vs. 3: ⏐10.8-8.4⏐=2.4 < 9.266 2 vs. 3: ⏐22.2-8.4⏐=13.8 > 9.266 1 and 2 are different. 2 and 3 are different. Notice that the results indicate that the mean of treatment 2 differs from the means of both treatments 1 and 3, and that the means for treatments 1 and 3 are the same. Notice also that the Fisher LSD procedure (see th Use the graphical procedure in Section 3-5.3 to compare the treatment m The scaled-t plot agrees with part (b). In this and the other two treatments is very obvious. 5 1 0 1 5 2 0 2 5 T e n s ile S tre n g th S c a le d t D is trib u tio n (3 ) (2 )(1 ) Construct a set of orthogonal contrasts, assuming that at the outset(d) of the experiment you suspected the response time of circuit type 2 to be different from the other two. ( ) 0 1 2 3 1 1 2 3 1 1. 2. 3. 1 2 0 2 0 2 54 2 111 42 126 H H C y y y C µ µ µ µ µ µ = − + = = − + ≠ = − + = − + = − 3-26
• 55. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY ( ) ( ) 2 1 1 126 529.2 5 6 529.2 31.31 16.9 C C SS F − = = = = Type 2 differs from the average of type 1 and type 3. (e) If you were a design engineer and you wished to minimize the response time, which circuit type would you select? Either type 1 or type 3 as they are not different from each other and have the lowest response time. (f) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied? The normal probability plot has some points that do not lie along the line in the upper region. This may indicate potential outliers in the data. Residual Normal%probability Normal plot of residuals -5.2 -1.95 1.3 4.55 7.8 1 5 10 20 30 50 70 80 90 95 99 Predicted Residuals Residuals vs. Predicted -5.2 -1.95 1.3 4.55 7.8 8.40 11.85 15.30 18.75 22.20 Circuit Type Residuals Residuals vs. Circuit Type -5.2 -1.95 1.3 4.55 7.8 1 2 3 3-27
• 56. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 3-17 The effective life of insulating fluids at an accelerated load of 35 kV is being studied. Test data have been obtained for four types of fluids. The results were as follows: Fluid Type Life (in h) at 35 kV Load 1 17.6 18.9 16.3 17.4 20.1 21.6 2 16.9 15.3 18.6 17.1 19.5 20.3 3 21.4 23.6 19.4 18.5 20.5 22.3 4 19.3 21.1 16.9 17.5 18.3 19.8 (a) Is there any indication that the fluids differ? Use α = 0.05. At α = 0.05 there are no difference, but at since the P-value is just slightly above 0.05, there is probably a difference in means. Design Expert Output Response: Life in in h ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 30.17 3 10.06 3.05 0.0525 not significant A 30.16 3 10.05 3.05 0.0525 Residual 65.99 20 3.30 Lack of Fit 0.000 0 Pure Error 65.99 20 3.30 Cor Total 96.16 23 The Model F-value of 3.05 implies there is a 5.25% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error 1-1 18.65 0.74 2-2 17.95 0.74 3-3 20.95 0.74 4-4 18.82 0.74 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > |t| 1 vs 2 0.70 1 1.05 0.67 0.5121 1 vs 3 -2.30 1 1.05 -2.19 0.0403 1 vs 4 -0.17 1 1.05 -0.16 0.8753 2 vs 3 -3.00 1 1.05 -2.86 0.0097 2 vs 4 -0.87 1 1.05 -0.83 0.4183 3 vs 4 2.13 1 1.05 2.03 0.0554 (b) Which fluid would you select, given that the objective is long life? Treatment 3. The Fisher LSD procedure in the computer output indicates that the fluid 3 is different from the others, and it’s average life also exceeds the average lives of the other three fluids. (c) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied? There is nothing unusual in the residual plots. 3-28
• 57. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Residual Normal%probability Normal plot of residuals -2.65 -1.25 0.15 1.55 2.95 1 5 10 20 30 50 70 80 90 95 99 Predicted Residuals Residuals vs. Predicted -2.65 -1.25 0.15 1.55 2.95 17.95 18.70 19.45 20.20 20.95 Fluid Type Residuals Residuals vs. Fluid Type -2.65 -1.25 0.15 1.55 2.95 1 2 3 4 3-18 Four different designs for a digital computer circuit are being studied in order to compare the amount of noise present. The following data have been obtained: Circuit Design Noise Observed 1 19 20 19 30 8 2 80 61 73 56 80 3 47 26 25 35 50 4 95 46 83 78 97 (a) Is the amount of noise present the same for all four designs? Use α = 0.05. No, at least one treatment mean is different. Design Expert Output Response: Noise ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] 3-29
• 58. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Sum of Mean F Source Squares DF Square Value Prob > F Model 12042.00 3 4014.00 21.78 < 0.0001 significant A 12042.00 3 4014.00 21.78 < 0.0001 Residual 2948.80 16 184.30 Lack of Fit 0.000 0 Pure Error 2948.80 16 184.30 Cor Total 14990.80 19 The Model F-value of 21.78 implies the model is significant. There is only a 0.01% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error 1-1 19.20 6.07 2-2 70.00 6.07 3-3 36.60 6.07 4-4 79.80 6.07 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > |t| 1 vs 2 -50.80 1 8.59 -5.92 < 0.0001 1 vs 3 -17.40 1 8.59 -2.03 0.0597 1 vs 4 -60.60 1 8.59 -7.06 < 0.0001 2 vs 3 33.40 1 8.59 3.89 0.0013 2 vs 4 -9.80 1 8.59 -1.14 0.2705 3 vs 4 -43.20 1 8.59 -5.03 0.0001 (b) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied? There is nothing too unusual about the residual plots, although there is a mild outlier present. Residual Normal%probability Normal plot of residuals -33.8 -21.05 -8.3 4.45 17.2 1 5 10 20 30 50 70 80 90 95 99 22 22 Predicted Residuals Residuals vs. Predicted -33.8 -21.05 -8.3 4.45 17.2 19.20 34.35 49.50 64.65 79.80 3-30
• 59. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 22 22 Circuit Design Residuals Residuals vs. Circuit Design -33.8 -21.05 -8.3 4.45 17.2 1 2 3 4 (c) Which circuit design would you select for use? Low noise is best. From the Design Expert Output, the Fisher LSD procedure comparing the difference in means identifies Type 1 as having lower noise than Types 2 and 4. Although the LSD procedure comparing Types 1 and 3 has a P-value greater than 0.05, it is less than 0.10. Unless there are other reasons for choosing Type 3, Type 1 would be selected. 3-19 Four chemists are asked to determine the percentage of methyl alcohol in a certain chemical compound. Each chemist makes three determinations, and the results are the following: Chemist Percentage of Methyl Alcohol 1 84.99 84.04 84.38 2 85.15 85.13 84.88 3 84.72 84.48 85.16 4 84.20 84.10 84.55 (a) Do chemists differ significantly? Use α = 0.05. There is no significant difference at the 5% level, but chemists differ significantly at the 10% level. Design Expert Output Response: Methyl Alcohol in % ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 1.04 3 0.35 3.25 0.0813 not significant A 1.04 3 0.35 3.25 0.0813 Residual 0.86 8 0.11 Lack of Fit 0.000 0 Pure Error 0.86 8 0.11 Cor Total 1.90 11 The Model F-value of 3.25 implies there is a 8.13% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard 3-31
• 60. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Mean Error 1-1 84.47 0.19 2-2 85.05 0.19 3-3 84.79 0.19 4-4 84.28 0.19 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > |t| 1 vs 2 -0.58 1 0.27 -2.18 0.0607 1 vs 3 -0.32 1 0.27 -1.18 0.2703 1 vs 4 0.19 1 0.27 0.70 0.5049 2 vs 3 0.27 1 0.27 1.00 0.3479 2 vs 4 0.77 1 0.27 2.88 0.0205 3 vs 4 0.50 1 0.27 1.88 0.0966 (b) Analyze the residuals from this experiment. There is nothing unusual about the residual plots. Residual Normal%probability Normal plot of residuals -0.43 -0.1925 0.045 0.2825 0.52 1 5 10 20 30 50 70 80 90 95 99 Predicted Residuals Residuals vs. Predicted -0.43 -0.1925 0.045 0.2825 0.52 84.28 84.48 84.67 84.86 85.05 Chemist Residuals Residuals vs. Chemist -0.43 -0.1925 0.045 0.2825 0.52 1 2 3 4 3-32
• 61. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY (c) If chemist 2 is a new employee, construct a meaningful set of orthogonal contrasts that might have been useful at the start of the experiment. Chemists Total C1 C2 C3 1 253.41 1 -2 0 2 255.16 -3 0 0 3 254.36 1 1 -1 4 252.85 1 1 1 Contrast Totals: -4.86 0.39 -1.51 ( ) ( ) ( ) ( ) ( ) ( ) 543 107270 3800 3800 23 511 0750 107270 0080 0080 63 390 1156 107270 6560 6560 123 864 3 2 3 2 2 2 1 2 1 . . . F. . SS . . . F. . SS *. . . F. . SS CC CC CC === − = ==== === − = Only contrast 1 is significant at 5%. 3-20 Three brands of batteries are under study. It is s suspected that the lives (in weeks) of the three brands are different. Five batteries of each brand are tested with the following results: Weeks of Life Brand 1 Brand 2 Brand 3 100 76 108 96 80 100 92 75 96 96 84 98 92 82 100 (a) Are the lives of these brands of batteries different? Yes, at least one of the brands is different. Design Expert Output Response: Life in Weeks ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 1196.13 2 598.07 38.34 < 0.0001 significant A 1196.13 2 598.07 38.34 < 0.0001 Residual 187.20 12 15.60 Lack of Fit 0.000 0 Pure Error 187.20 12 15.60 Cor Total 1383.33 14 The Model F-value of 38.34 implies the model is significant. There is only a 0.01% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error 1-1 95.20 1.77 2-2 79.40 1.77 3-3 100.40 1.77 3-33
• 62. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > |t| 1 vs 2 15.80 1 2.50 6.33 < 0.0001 1 vs 3 -5.20 1 2.50 -2.08 0.0594 2 vs 3 -21.00 1 2.50 -8.41 < 0.0001 (b) Analyze the residuals from this experiment. There is nothing unusual about the residuals. Residual Normal%probability Normal plot of residuals -4.4 -1.4 1.6 4.6 7.6 1 5 10 20 30 50 70 80 90 95 99 2 2 2 2 22 Predicted Residuals Residuals vs. Predicted -4.4 -1.4 1.6 4.6 7.6 79.40 84.65 89.90 95.15 100.40 2 2 2 2 22 Brand Residuals Residuals vs. Brand -4.4 -1.4 1.6 4.6 7.6 1 2 3 (c) Construct a 95 percent interval estimate on the mean life of battery brand 2. Construct a 99 percent interval estimate on the mean difference between the lives of battery brands 2 and 3. n MS ty E aN,.i − ± 2 α 3-34
• 63. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Brand 2: 5 6015 1792479 . .. ± 84934079 .. ± 2498355175 2 .. ≤≤ µ Brand 2 - Brand 3: n MS tyy E aN,.j.i 2 2 − ±− α ( ) 5 60152 05534100479 . ... ±− 3691363128 32 .. −≤−≤− µµ (d) Which brand would you select for use? If the manufacturer will replace without charge any battery that fails in less than 85 weeks, what percentage would the company expect to replace? Chose brand 3 for longest life. Mean life of this brand in 100.4 weeks, and the variance of life is estimated by 15.60 (MSE). Assuming normality, then the probability of failure before 85 weeks is: ( ) 000050903 6015 410085 .. . . =−=⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ΦΦ That is, about 5 out of 100,000 batteries will fail before 85 week. 3-21 Four catalysts that may affect the concentration of one component in a three component liquid mixture are being investigated. The following concentrations are obtained: Catalyst 1 2 3 4 58.2 56.3 50.1 52.9 57.2 54.5 54.2 49.9 58.4 57.0 55.4 50.0 55.8 55.3 51.7 54.9 (a) Do the four catalysts have the same effect on concentration? No, their means are different. Design Expert Output Response: Concentration ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 85.68 3 28.56 9.92 0.0014 significant A 85.68 3 28.56 9.92 0.0014 Residual 34.56 12 2.88 Lack of Fit 0.000 0 Pure Error 34.56 12 2.88 Cor Total 120.24 15 The Model F-value of 9.92 implies the model is significant. There is only a 0.14% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard 3-35
• 64. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Mean Error 1-1 56.90 0.76 2-2 55.77 0.85 3-3 53.23 0.98 4-4 51.13 0.85 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > |t| 1 vs 2 1.13 1 1.14 0.99 0.3426 1 vs 3 3.67 1 1.24 2.96 0.0120 1 vs 4 5.77 1 1.14 5.07 0.0003 2 vs 3 2.54 1 1.30 1.96 0.0735 2 vs 4 4.65 1 1.20 3.87 0.0022 3 vs 4 2.11 1 1.30 1.63 0.1298 (b) Analyze the residuals from this experiment. There is nothing unusual about the residual plots. Residual Normal%probability Normal plot of residuals -3.13333 -1.80833 -0.483333 0.841667 2.16667 1 5 10 20 30 50 70 80 90 95 99 Predicted Residuals Residuals vs. Predicted -3.13333 -1.80833 -0.483333 0.841667 2.16667 51.13 52.57 54.01 55.46 56.90 Catalyst Residuals Residuals vs. Catalyst -3.13333 -1.80833 -0.483333 0.841667 2.16667 1 2 3 4 (c) Construct a 99 percent confidence interval estimate of the mean response for catalyst 1. 3-36
• 65. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY n MS ty E aN,.i − ± 2 α Catalyst 1: 5 882 0553956 . .. ± 31862956 .. ± 218659581454 1 .. ≤≤ µ 3-22 An experiment was performed to investigate the effectiveness of five insulating materials. Four samples of each material were tested at an elevated voltage level to accelerate the time to failure. The failure times (in minutes) is shown below. Material Failure Time (minutes) 1 110 157 194 178 2 1 2 4 18 3 880 1256 5276 4355 4 495 7040 5307 10050 5 7 5 29 2 (a) Do all five materials have the same effect on mean failure time? No, at least one material is different. Design Expert Output Response: Failure Timein Minutes ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 1.032E+008 4 2.580E+007 6.19 0.0038 significant A 1.032E+008 4 2.580E+007 6.19 0.0038 Residual 6.251E+00715 4.167E+006 Lack of Fit 0.000 0 Pure Error 6.251E+00715 4.167E+006 Cor Total 1.657E+00819 The Model F-value of 6.19 implies the model is significant. There is only a 0.38% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error 1-1 159.75 1020.67 2-2 6.25 1020.67 3-3 2941.75 1020.67 4-4 5723.00 1020.67 5-5 10.75 1020.67 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > |t| 1 vs 2 153.50 1 1443.44 0.11 0.9167 1 vs 3 -2782.00 1 1443.44 -1.93 0.0731 1 vs 4 -5563.25 1 1443.44 -3.85 0.0016 1 vs 5 149.00 1 1443.44 0.10 0.9192 2 vs 3 -2935.50 1 1443.44 -2.03 0.0601 2 vs 4 -5716.75 1 1443.44 -3.96 0.0013 2 vs 5 -4.50 1 1443.44 -3.118E-003 0.9976 3 vs 4 -2781.25 1 1443.44 -1.93 0.0732 3 vs 5 2931.00 1 1443.44 2.03 0.0604 3-37
• 66. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 4 vs 5 5712.25 1 1443.44 3.96 0.0013 (b) Plot the residuals versus the predicted response. Construct a normal probability plot of the residuals. What information do these plots convey? Predicted Residuals Residuals vs. Predicted -5228 -2839.25 -450.5 1938.25 4327 6.25 1435.44 2864.62 4293.81 5723.00 Residual Normal%probability Normal plot of residuals -5228 -2839.25 -450.5 1938.25 4327 1 5 10 20 30 50 70 80 90 95 99 The plot of residuals versus predicted has a strong outward-opening funnel shape, which indicates the variance of the original observations is not constant. The normal probability plot also indicates that the normality assumption is not valid. A data transformation is recommended. (c) Based on your answer to part (b) conduct another analysis of the failure time data and draw appropriate conclusions. A natural log transformation was applied to the failure time data. The analysis in the log scale identifies that there exists at least one difference in treatment means. Design Expert Output Response: Failure Timein Minutes Transform: Natural log Constant: 0.000 ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 165.06 4 41.26 37.66 < 0.0001 significant A 165.06 4 41.26 37.66 < 0.0001 Residual 16.44 15 1.10 Lack of Fit 0.000 0 Pure Error 16.44 15 1.10 Cor Total 181.49 19 The Model F-value of 37.66 implies the model is significant. There is only a 0.01% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error 1-1 5.05 0.52 2-2 1.24 0.52 3-3 7.72 0.52 4-4 8.21 0.52 5-5 1.90 0.52 3-38
• 67. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > |t| 1 vs 2 3.81 1 0.74 5.15 0.0001 1 vs 3 -2.66 1 0.74 -3.60 0.0026 1 vs 4 -3.16 1 0.74 -4.27 0.0007 1 vs 5 3.15 1 0.74 4.25 0.0007 2 vs 3 -6.47 1 0.74 -8.75 < 0.0001 2 vs 4 -6.97 1 0.74 -9.42 < 0.0001 2 vs 5 -0.66 1 0.74 -0.89 0.3856 3 vs 4 -0.50 1 0.74 -0.67 0.5116 3 vs 5 5.81 1 0.74 7.85 < 0.0001 4 vs 5 6.31 1 0.74 8.52 < 0.0001 There is nothing unusual about the residual plots when the natural log transformation is applied. Residual Normal%probability Normal plot of residuals -2.00945 -1.09511 -0.180766 0.733576 1.64792 1 5 10 20 30 50 70 80 90 95 99 Predicted Residuals Residuals vs. Predicted -2.00945 -1.09511 -0.180766 0.733576 1.64792 1.24 2.99 4.73 6.47 8.21 Material Residuals Residuals vs. Material -2.00945 -1.09511 -0.180766 0.733576 1.64792 1 2 3 4 5 3-23 A semiconductor manufacturer has developed three different methods for reducing particle counts on wafers. All three methods are tested on five wafers and the after-treatment particle counts obtained. The data are shown below. 3-39
• 68. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Method Count 1 31 10 21 4 1 2 62 40 24 30 35 3 58 27 120 97 68 (a) Do all methods have the same effect on mean particle count? No, at least one method has a different effect on mean particle count. Design Expert Output Response: Count ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 8963.73 2 4481.87 7.91 0.0064 significant A 8963.73 2 4481.87 7.91 0.0064 Residual 6796.00 12 566.33 Lack of Fit 0.000 0 Pure Error 6796.00 12 566.33 Cor Total 15759.73 14 The Model F-value of 7.91 implies the model is significant. There is only a 0.64% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error 1-1 13.40 10.64 2-2 38.20 10.64 3-3 73.00 10.64 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > |t| 1 vs 2 -24.80 1 15.05 -1.65 0.1253 1 vs 3 -59.60 1 15.05 -3.96 0.0019 2 vs 3 -34.80 1 15.05 -2.31 0.0393 (b) Plot the residuals versus the predicted response. Construct a normal probability plot of the residuals. Are there potential concerns about the validity of the assumptions? The plot of residuals versus predicted appears to be funnel shaped. This indicates the variance of the original observations is not constant. The residuals plotted in the normal probability plot do not fall along a straight line, which suggests that the normality assumption is not valid. A data transformation is recommended. 3-40
• 69. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Predicted Residuals Residuals vs. Predicted -46 -22.75 0.5 23.75 47 13.40 28.30 43.20 58.10 73.00 Residual Normal%probability Normal plot of residuals -46 -22.75 0.5 23.75 47 1 5 10 20 30 50 70 80 90 95 99 (c) Based on your answer to part (b) conduct another analysis of the particle count data and draw appropriate conclusions. For count data, a square root transformation is often very effective in resolving problems with inequality of variance. The analysis of variance for the transformed response is shown below. The difference between methods is much more apparent after applying the square root transformation. Design Expert Output Response: Count Transform: Square root Constant: 0.000 ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 63.90 2 31.95 9.84 0.0030 significant A 63.90 2 31.95 9.84 0.0030 Residual 38.96 12 3.25 Lack of Fit 0.000 0 Pure Error 38.96 12 3.25 Cor Total 102.86 14 The Model F-value of 9.84 implies the model is significant. There is only a 0.30% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error 1-1 3.26 0.81 2-2 6.10 0.81 3-3 8.31 0.81 Mean Standard t for H0 Treatment Difference DF Error Coeff=0 Prob > |t| 1 vs 2 -2.84 1 1.14 -2.49 0.0285 1 vs 3 -5.04 1 1.14 -4.42 0.0008 2 vs 3 -2.21 1 1.14 -1.94 0.0767 3-24 Consider testing the equality of the means of two normal populations, where the variances are unknown but are assumed to be equal. The appropriate test procedure is the pooled t test. Show that the pooled t test is equivalent to the single factor analysis of variance. 3-41
• 70. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY 22 21 0 2 − − = n p .. t~ n S yy t assuming n1 = n2 = n ( ) ( ) ( ) E i n j .ij n j .j n j .j p MS n yy n yyyy S ≡ − − = − −+− = ∑∑∑∑ = === 2222 2 1 1 2 1 1 2 22 1 2 11 for a=2 Furthermore, ( ) n y n yn yy .. i .i .. 22 22 1 2 2 21 −=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ∑= , which is exactly the same as SSTreatments in a one-way classification with a=2. Thus we have shown that E Treatments MS SS t =2 0 . In general, we know that so that . Thus the square of the test statistic from the pooled t-test is the same test statistic that results from a single-factor analysis of variance with a=2. u,u Ft 1 2 = 221 2 0 −n,F~t 3-25 Show that the variance of the linear combination ∑ is . = a i .ii yc 1 ∑= a i ii cn 1 22 σ ( ) ( )∑∑∑ ∑∑∑ === === = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ == ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ii n j .ij a i i a i n j iji a i .ii a i .ii yVcyVcycVycV 11 2 1 1 2 11 , ( ) 2 σ=ijyV ∑= = a i ii nc 1 22 σ 3-26 In a fixed effects experiment, suppose that there are n observations for each of four treatments. Let be single-degree-of-freedom components for the orthogonal contrasts. Prove that . 2 3 2 2 2 1 Q,Q,Q 2 3 2 2 2 1 QQQSSTreatments ++= 2 33433 2 224322 2 1143211 2 3 QSSyyC QSSyyyC QSSyyyyC C.. C... C.... =−= =−−= =−−−= n )yy( Q n )yyy( Q n )yyyy( Q .. ... .... 2 6 2 12 3 2 432 3 2 4322 2 2 43212 1 − = −− = −−− = n yyy QQQ i ji .j.i.i 12 69 4 1 2 2 3 2 2 2 1 ∑ ∑∑= < ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − =++ and since 3-42
• 71. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −= ∑∑∑ =< 4 1 22 2 1 i .i.. ji .j.i yyyy , we have Treatments .. i .ii ...i SS n y n y n yy QQQ =−= − =++ ∑ ∑ = = 412 312 24 1 2 4 1 22 2 3 2 2 2 1 for a=4. 3-27 Use Bartlett's test to determine if the assumption of equal variances is satisfied in Problem 3-14. Use α = 0.05. Did you reach the same conclusion regarding the equality of variance by examining the residual plots? c q .302622 0 =χ , where ( ) ( ) ( ) ( ) ( ) ( ) aN Sn S aNn a c SlognSlogaNq a i ii p a i i a i iip − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −−− − += −−−= ∑ ∑ ∑ = = −− = 1 2 2 1 11 1 2 10 2 10 1 1 13 1 1 1 820 814 211 2 3 2 2 2 1 .S .S .S = = = ( ) ( ) ( ) ( ) ( ) ( ) 615 315 820158141521115 315 820158141521115 2 2 . ... S ... S p p = − −+−+− = − −+−+− = ( ) ( ) ( ) ( ) 13891 12 1 4 3 133 1 1 31515 133 1 1 1 11 .c c a i =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − += ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −−− − += ∑= −− ( ) ( ) ( ) ( 1675015014317514 8208142114615315 1 10101010 1 2 10 2 10 ...q .log.log.log.logq SlognSlogaNq a i iip =−= ++−−= −−−= ∑= ) 33860 13891 16750 30262302622 0 . . . . c q . ===χ 4992 4050 .,. =χ Cannot reject null hypothesis; conclude that the variance are equal. This agrees with the residual plots in Problem 3-16. 3-28 Use the modified Levene test to determine if the assumption of equal variances is satisfied on Problem 3-20. Use α = 0.05. Did you reach the same conclusion regarding the equality of variances by examining the residual plots? The absolute value of Battery Life – brand median is: ij iy y− 3-43
• 72. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY Brand 1 Brand 2 Brand 3 4 4 8 0 0 0 4 5 4 0 4 2 4 2 0 The analysis of variance indicates that there is not a difference between the different brands and therefore the assumption of equal variances is satisfired. Design Expert Output Response: Mod Levine ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Prob > F Model 0.93 2 0.47 0.070 0.9328 A 0.93 2 0.47 0.070 0.9328 Pure Error 80.00 12 6.67 Cor Total 80.93 14 3-29 Refer to Problem 3-16. If we wish to detect a maximum difference in mean response times of 10 milliseconds with a probability of at least 0.90, what sample size should be used? How would you obtain a preliminary estimate of ?2 σ 2 2 2 2 σ Φ a nD = , use MSE from Problem 3-10 to estimate .2 σ ( ) ( )( ) n. . n 9860 91632 10 2 2 ==Φ Letting 050.=α , P(accept) = 0.1 , 211 =−= aυ Trial and Error yields: n 2υ Φ P(accept) 5 12 2.22 0.17 6 15 2.43 0.09 7 18 2.62 0.04 Choose n ≥ 6, therefore N ≥ 18 Notice that we have used an estimate of the variance obtained from the present experiment. This indicates that we probably didn’t use a large enough sample (n was 5 in problem 3-10) to satisfy the criteria specified in this problem. However, the sample size was adequate to detect differences in one of the circuit types. When we have no prior estimate of variability, sometimes we will generate sample sizes for a range of possible variances to see what effect this has on the size of the experiment. Often a knowledgeable expert will be able to bound the variability in the response, by statements such as “the standard deviation is going to be at least…” or “the standard deviation shouldn’t be larger than…”. 3-30 Refer to Problem 3-20. 3-44
• 73. Solutions from Montgomery, D. C. (2004) Design and Analysis of Experiments, Wiley, NY (a) If we wish to detect a maximum difference in mean battery life of 0.5 percent with a probability of at least 0.90, what sample size should be used? Discuss how you would obtain a preliminary estimate of σ2 for answering this question. Use the MSE from Problem 3-14. Φ2 2 2 2 = nD aσ ( ) ( )( ) n. . ..n 0022440 601532 6667910050 2 2 = × =Φ Letting α = 0 05. , P(accept) = 0.1 , 211 =−= aυ Trial and Error yields: n 2υ Φ P(accept) 40 117 1.895 0.18 45 132 2.132 0.10 50 147 2.369 0.05 Choose n ≥ 45, therefore N ≥ 135 See the discussion from the previous problem about the estimate of variance. (b) If the difference between brands is great enough so that the standard deviation of an observation is increased by 25 percent, what sample size should be used if we wish to detect this with a probability of at least 0.90? 211 =−= aυ 123152 =−=−= aNυ 050.=α 10.)accept(P ≤ ( )[ ] ( )( )[ ] n..nP.n 56250112501011101011 22 +=−++=−++=λ Trial and Error yields: n 2υ λ P(accept) 40 117 4.84 0.13 45 132 5.13 0.11 50 147 5.40 0.10 Choose n ≥ 50, therefore N ≥ 150 3-31 Consider the experiment in Problem 3-20. If we wish to construct a 95 percent confidence interval on the difference in two mean battery lives that has an accuracy of ±2 weeks, how many batteries of each brand must be tested? 050.=α 615.MSE = n MS twidth E aN,. 2 0250 −= Trial and Error yields: n υ2 t width 3-45
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