ch3-8-sol
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ch3-8-sol

Course Number: IE 410, Fall 2010

College/University: Shanghai Jiao Tong...

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Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Chapter 3 Experiments with a Single Factor: The Analysis of Variance Solutions 3.1. An experimenter has conducted a single-factor experiment with four levels of the factor, and each factor level has been replicated six times. The computed value of the F-statistic is F 0 = 3.26. Find bounds on the P-value. Table P-value = 0.025,...

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from Solutions Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Chapter 3 Experiments with a Single Factor: The Analysis of Variance Solutions 3.1. An experimenter has conducted a single-factor experiment with four levels of the factor, and each factor level has been replicated six times. The computed value of the F-statistic is F 0 = 3.26. Find bounds on the P-value. Table P-value = 0.025, 0.050 Computer P-value = 0.043 3.2. An experimenter has conducted a single-factor experiment with six levels of the factor, and each factor level has been replicated three times. The computed value of the F-statistic is F 0 = 5.81. Find bounds on the P-value. Table P-value < 0.010 Computer P-value = 0.006 3.3. A computer ANOVA output is shown below. Fill in the blanks. You may give bounds on the Pvalue. One-way ANOVA Source DF SS MS F P Factor 3 36.15 ? ? ? Error ? ? ? Total 19 196.04 Completed table is: One-way ANOVA Source DF SS MS F P Factor 3 36.15 12.05 1.21 0.3395 Error 16 159.89 9.99 Total 19 196.04 3.4. A computer ANOVA output is shown below. Fill in the blanks. You may give bounds on the Pvalue. One-way ANOVA Source DF SS MS F P Factor ? ? 246.93 ? ? Error 25 186.53 ? Total 29 1174.24 3-1 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Completed table is: One-way ANOVA Source DF SS MS F P Factor 4 987.71 246.93 33.09 < 0.0001 Error 25 186.53 7.46 Total 29 1174.24 3.5. An article appeared in The Wall Street Journal on Tuesday, April 27, 2010, with the title “Eating Chocolate Is Linked to Depression.” The article reported on a study funded by the National Heart, Lung and Blood Institute (part of the National Institutes of Health) and conducted by the faculty at the University of California, San Diego, and the University of California, Davis. The research was also published in the Archives of Internal Medicine (2010, pp. 699-703). The study examined 931 adults who were not taking antidepressants and did not have known cardiovascular disease or diabetes. The group was about 70% men and the average age of the group was reported to be about 58. The participants were asked about chocolate consumption and then screened for depression using a questionnaire. People who scored less than 16 on the questionnaire are not considered depressed, while those with scores above 16 and less than or equal to 22 are considered possibly depressed, while those with scores above 22 are considered likely to be depressed. The survey found that people who were not depressed ate an average of 8.4 servings of chocolate per month, while those individuals who scored above 22 were likely to be depressed ate the most chocolate, an average of 11.8 servings per month. No differentiation was made between dark and milk chocolate. Other foods were also examined, but no patterned emerged between other foods and depression. Is this study really a designed experiment? Does it establish a cause-and-effect link between chocolate consumption and depression? How would the study have to be conducted to establish such a link? This is not a designed experiment, and it does not establish a cause-and-effect link between chocolate consumption and depression. An experiment could be run by giving a group of people a defined amount of chocolate servings per month for several months, while not giving another group any chocolate. Ideally it would be good to have the participants not eat any chocolate for a period of time before the experiment, and measure depression for each participant before and after the experiment. 3.6. An article in Bioelectromagnetics (“Electromagnetic Effects on Forearm Disuse Osteopenia: A Randomized, Double-Blind, Sham-Controlled Study,” Vol. 32, 2011, pp. 273 – 282) describes a randomized, double-blind, sham-controlled, feasibility and dosing study to determine if a common pulsing electromagnetic field (PEMF) treatment could moderate the substantial osteopenia that occurs after forearm disuse. Subjects were randomized into four groups after a distal radius fracture, or carpal surgery requiring immobilization in a cast. Active of identical sham PEMF transducers were worn on a distal forearm for 1, 2, or 4h/day for 8 weeks starting after cast removal (“baseline”) when bone density continues to decline. Bone mineral density (BMD) and bone geometry were measured in the distal forearm by dual energy X-ray absorptiometry (DXA) and peripheral quantitative computed tomography (pQCT). The data below are the percent losses in BMD measurements on the radius after 16weeks for patients wearing the active or sham PEMF transducers for 1, 2, or 4h/day (data were constructed to match the means and standard deviations read from a graph in the paper). Sham PEMF 1h/day PEMF 2h/day PEMF 4h/day 4.51 5.32 4.73 7.03 7.95 6.00 5.81 4.65 4.97 5.12 5.69 6.65 3-2 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 3.00 7.08 3.86 5.49 7.97 5.48 4.06 6.98 2.23 6.52 6.56 4.85 3.95 4.09 8.34 7.26 5.64 6.28 3.01 5.92 9.35 7.77 6.71 5.58 6.52 5.68 6.51 7.91 4.96 8.47 1.70 4.90 6.10 4.58 5.89 4.54 7.19 4.11 6.55 8.18 4.03 5.72 5.34 5.42 2.72 5.91 5.88 6.03 9.19 6.89 7.50 7.04 5.17 6.99 3.28 5.17 5.70 4.98 5.38 7.60 5.85 9.94 7.30 7.90 6.45 6.38 5.46 7.91 (a) Is there evidence to support a claim that PEMF usage affects BMD loss? If so, analyze the data to determine which specific treatments produce the differences. The ANOVA from the Minitab output shows that there is no difference between the treatments; P=0.281. Minitab Output One-way ANOVA: Sham, PEMF 1h/day, PEMF 2h/day, PEMF 4h/day Source Factor Error Total DF 3 76 79 S = 1.606 Level Sham PEMF 1h/day PEMF 2h/day PEMF 4h/day SS 10.04 196.03 206.07 MS 3.35 2.58 F 1.30 R-Sq = 4.87% N 20 20 20 20 Mean 5.673 6.165 5.478 6.351 P 0.281 R-Sq(adj) = 1.12% StDev 2.002 1.444 1.645 1.232 Individual 95% CIs For Mean Based on Pooled StDev -+---------+---------+---------+-------(-----------*----------) (-----------*-----------) (-----------*-----------) (-----------*-----------) -+---------+---------+---------+-------4.80 5.40 6.00 6.60 (b) Analyze the residuals from this experiment and comment on the underlying assumptions and model adequacy. The residuals show the model is good. 3-3 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Residual Plots for Sham, PEMF 1h/day, PEMF 2h/day, PEMF 4h/day Normal Probability Plot Versus Fits 99.9 4 90 Residual Percent 99 50 10 1 0.1 2 0 -2 -4 -5.0 -2.5 0.0 Residual 2.5 5.0 5.6 5.8 6.0 Fitted Value 6.2 6.4 Histogram Frequency 16 12 8 4 0 -3.2 -1.6 0.0 Residual 1.6 3.2 3.7. The tensile strength of Portland cement is being studied. Four different mixing techniques can be used economically. A completely randomized experiment was conducted and the following data were collected. Mixing Technique 1 2 3 4 3129 3200 2800 2600 Tensile Strength (lb/in2) 3000 2865 3300 2975 2900 2985 2700 2600 2890 3150 3050 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 Source Squares DF Square Model 4.897E+005 3 1.632E+005 A 4.897E+005 3 1.632E+005 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 F Value 12.73 12.73 Prob > F 0.0005 0.0005 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 Treatment 1 vs 2 1 vs 3 1 vs 4 Mean Difference -185.25 37.25 304.75 DF 1 1 1 Standard Error 80.08 80.08 80.08 t for H0 Coeff=0 -2.31 0.47 3.81 3-4 Prob > |t| 0.0392 0.6501 0.0025 significant Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 2 vs 3 2 vs 4 3 vs 4 222.50 490.00 267.50 1 1 1 80.08 80.08 80.08 2.78 6.12 3.34 0.0167 < 0.0001 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? S yi . = MS E 12825.7 = = 56.625 4 n Scaled t Distribution (4) (3) 2700 2800 2900 (1) 3000 (2) 3100 Tensile Strength Based on examination of the plot, we would conclude that µ1 and µ3 are the same; that µ 4 differs from µ1 and µ3 , that µ 2 differs from µ1 and µ3 , and that µ 2 and µ 4 are different. (c) Use the Fisher LSD method with α=0.05 to make comparisons between pairs of means. LSD = t α 2 ,N − a LSD = t 0.025 ,16 − 4 2 MS E n 2( 12825.7 ) 4 LSD = 2.179 6412.85 = 174.495 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? 3-5 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY There is nothing unusual about the normal probability plot of residuals. Normal plot of residuals 99 Normal % probability 95 90 80 70 50 30 20 10 5 1 -181.25 -96.4375 -11.625 73.1875 158 Residual (e) Plot the residuals versus the predicted tensile strength. Comment on the plot. There is nothing unusual about this plot. Residuals vs. Predicted 158 Res iduals 73.1875 -11.625 2 -96.4375 -181.25 2666.25 2788.75 2911.25 3033.75 3156.25 Predicted (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. 3-6 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY One Factor Plot 3300 Tens ile Strength 3119.75 2939.51 2759.26 2 2579.01 3 2 1 4 Technique 3.8. (a) Rework part (c) of Problem 3.7 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 -201 275 -15 460 4 67 543 252 728 30 505 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 Tukey’s 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. 3.9. Reconsider the experiment in Problem 3.7. 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 percent confidence interval on the difference in means for techniques 1 and 3. Does this aid in interpreting the results of the experiment? 3-7 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY yi . − tα 2 ,N − a MS E MS E ≤ µi ≤ yi . + tα ,N − a n n 2 Treatment 1: 2971 ± 2.179 12825.69 4 2971 ± 123.387 2847.613 ≤ µ1 ≤ 3094.387 Treatment 2: 3156.25±123.387 3032.863 ≤ µ 2 ≤ 3279.637 Treatment 3: 2933.75±123.387 2810.363 ≤ µ3 ≤ 3057.137 Treatment 4: 2666.25±123.387 2542.863 ≤ µ 4 ≤ 2789.637 Treatment 1 - Treatment 3: yi . − y j . − tα 2 ,N − a 2 MS E 2 MS E ≤ µi − µ j ≤ yi . − y j . + tα ,N − a n n 2 2( 12825.7 ) 4 −137.245 ≤ µ1 − µ3 ≤ 211.745 2971.00 − 2933.75 ± 2.179 Because the confidence interval for the difference between means 1 and 3 spans zero, we agree with the statement in Problem 3.5 (b); there is not a statistical difference between these two means. 3.10. 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 a completely randomized experiment with five levels of cotton content and replicated the experiment five times. The data are shown in the following table. Cotton Weight Percentage 15 20 25 30 35 7 12 14 19 7 Observations 15 12 19 22 11 7 17 19 25 10 11 18 18 19 15 9 18 18 23 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 Cotton P 4 475.76 118.94 Error 20 161.20 8.06 Total 24 636.96 F 14.76 P 0.000 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. 3-8 Solutions from Montgomery, D. C. (2012) 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 20 -9.346 -1.854 25 -11.546 -4.054 -5.946 1.546 30 -15.546 -8.054 -9.946 -2.454 -7.746 -0.254 35 -4.746 2.746 0.854 8.346 3.054 10.546 30 7.054 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. The residual plots below show nothing unusual. Normal Probability Plot of the Residuals (response is Tensile Strength) 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 -5.0 -2.5 0.0 Residual 3-9 2.5 5.0 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Residuals Versus the Fitted Values (response is Tensile Strength) 5.0 Residual 2.5 0.0 -2.5 -5.0 10 12 14 16 Fitted Value 18 20 22 3.11. Reconsider the experiment described in Problem 3.10. 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 d 0.05 (4, 20) 2 MS E 2(8.06) = 2.65 = 4.76 n 5 1 vs. 4 : y1. − y4. = 9.8 − 21.6 = −11.8* 2 vs. 4 : y2. − y4. = 15.4 − 21.6 = −6.2 * 3 vs. 4 : y3. − y4. = 17.6 − 21.6 = −4.0 5 vs. 4 : y5. − y4. = 10.8 − 21.6 = −10.8* The control treatment, treatment 4, differs from treatments 1, 2 and 5. 3.12. 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 20g 30g 40g 24 37 42 Observations 28 37 44 31 47 52 30 35 38 (a) Is there evidence to indicate that dosage level affects bioactivity? Use α = 0.05. 3-10 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Minitab Output One-way ANOVA: Activity versus Dosage Analysis of Variance for Activity Source DF SS MS Dosage 2 450.7 225.3 Error 9 288.3 32.0 Total 11 738.9 F 7.04 P 0.014 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 -3.823 -19.177 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. There is nothing too unusual about the residual plots shown below. Normal Probability Plot of the Residuals (response is Activity) 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 -8 -6 -4 -2 0 Residual 3-11 2 4 6 8 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Residuals Versus the Fitted Values (response is Activity) 8 6 Residual 4 2 0 -2 -4 -6 -8 30 32 34 36 38 Fitted Value 40 42 44 46 3.13. 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. Type of Car Sub-compact Compact Midsize Full Size 3 1 4 3 5 3 1 5 3 4 3 7 Observations 7 6 5 7 5 6 5 7 1 5 10 3 3 3 2 4 2 2 4 7 1 1 2 2 6 7 7 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 Car Type 3 16.68 5.56 Error 36 180.30 5.01 Total 39 196.98 F 1.11 P 0.358 There is no difference. (b) Analyze the residuals from this experiment and comment on the model adequacy. 3-12 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Normal Probability Plot of the Residuals (response is Days) 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 -4 -3 -2 -1 0 1 Residual 2 3 4 5 Residuals Versus the Fitted Values (response is Days) 5 4 3 Residual 2 1 0 -1 -2 -3 -4 3.5 4.0 4.5 Fitted Value 5.0 5.5 There is nothing unusual about the residuals. (c) Notice that the response variable in this experiment is a count. Should this 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 Car Type 3 1.087 0.362 Error 36 11.807 0.328 Total 39 12.893 F 1.10 P 0.360 3-13 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 3.14. I belong to a golf club in my neighborhood. I divide the year into three golf seasons: summer (JuneSeptember), 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 partyear 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 Summer Shoulder W inter 83 91 94 85 87 91 85 84 87 87 87 85 Observations 90 88 85 86 87 91 88 83 92 84 91 86 (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 Season 2 35.61 17.80 Error 22 184.63 8.39 Total 24 220.24 F 2.12 P 0.144 The data do not support the author’s opinion. (b) Analyze the residuals from this experiment and comment on model adequacy. Normal Probability Plot of the Residuals (response is Score) 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 -5.0 -2.5 0.0 Residual 3-14 2.5 5.0 90 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Residuals Versus the Fitted Values (response is Score) 5.0 Residual 2.5 0.0 -2.5 -5.0 86.0 86.5 87.0 87.5 88.0 Fitted Value 88.5 89.0 89.5 There is nothing unusual about the residuals. 3.15. 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. Approach 1 2 3 1000 1500 900 1500 1800 1000 Contributions (in $) 1200 1800 1600 1100 2000 1200 2000 1700 1200 1500 1200 1550 1000 1800 1000 1250 1900 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 Approach 2 1362708 681354 9.41 Error 21 1520625 72411 Total 23 2883333 P 0.001 There is a difference between the approaches. The Tukey test will indicate which are different. Approach 2 is different than approach 1 and approach 3. Minitab Output Tukey's pairwise comparisons Family error rate = 0.0500 Individual error rate = 0.0200 Critical value = 3.56 Intervals for (column level mean) - (row level mean) 1 2 3-15 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 2 -770 -93 3 -214 464 218 895 (b) Analyze the residuals from this experiment and comment on the model adequacy. Normal Probability Plot of the Residuals (response is Contribution) 99 95 90 Percent 80 70 60 50 40 30 20 10 5 1 -500 -250 0 Residual 250 500 Residuals Versus the Fitted Values (response is Contribution) 500 Residual 250 0 -250 -500 1200 1300 1400 1500 Fitted Value There is nothing unusual about the residuals. 3-16 1600 1700 1800 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 3.16. An experiment was run to determine whether four specific firing temperatures affect the density of a certain type of brick. A completely randomized experiment led to the following data: Temperature 100 125 150 175 21.8 21.7 21.9 21.9 Density 21.7 21.5 21.8 21.8 21.9 21.4 21.8 21.7 21.6 21.4 21.6 21.4 21.7 21.5 (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 Source Squares DF Square Model 0.16 3 0.052 A 0.16 3 0.052 Residual 0.36 14 0.026 Lack of Fit 0.000 0 Pure Error 0.36 14 0.026 Cor Total 0.52 17 F Value 2.02 2.02 Prob > F 0.1569 0.1569 not significant 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) 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 Treatment 1 vs 2 1 vs 3 1 vs 4 2 vs 3 2 vs 4 3 vs 4 Mean Difference 0.24 0.020 0.040 -0.22 -0.20 0.020 DF 1 1 1 1 1 1 Standard Error 0.11 0.10 0.11 0.11 0.11 0.11 t for H0 Coeff=0 2.23 0.20 0.37 -2.05 -1.76 0.19 Prob > |t| 0.0425 0.8465 0.7156 0.0601 0.0996 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. 3-17 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Normal plot of residuals Residuals vs. Predicted 0.2 99 2 0.075 90 80 70 Res iduals Norm al % probability 95 50 30 20 10 2 -0.05 2 -0.175 5 1 -0.3 -0.3 -0.175 -0.05 0.075 0.2 21.50 21.56 R es idual 21.62 21.68 21.74 Predicted (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. Scaled t Distribution (125) 21.2 21.3 21.4 (175,150,100) 21.5 21.6 21.7 21.8 Mean Density 3.17. Rework Part (d) of Problem 3.16 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 Tα = Treatment 1 Treatment 1 Treatment 1 Treatment 3 Treatment 4 Treatment 3 qα (a, f ) 2 1 1 MS E  +  n n  j i vs. Treatment 2 = 21.74 – 21.50 = 0.24 < 0.314 vs. Treatment 3 = 21.74 – 21.72 = 0.02 < 0.296 vs. Treatment 4 = 21.74 – 21.70 = 0.04 < 0.314 vs. Treatment 2 = 21.72 – 21.50 = 0.22 < 0.314 vs. Treatment 2 = 21.70 – 21.50 = 0.20 < 0.331 vs. Treatment 4 = 21.72 – 21.70 = 0.02 < 0.314 3-18 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 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. 3.18. A manufacturer of television sets is interested in the effect of tube conductivity of four different types of coating for color picture tubes. A completely randomized experiment is conducted and the following conductivity data are obtained: Coating Type 1 2 3 4 143 152 134 129 141 149 136 127 Conductivity 150 137 132 132 146 143 127 129 (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.. Design Expert Output ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean Source Squares DF Square Model 844.69 3 281.56 A 844.69 3 281.56 Residual 236.25 12 19.69 Lack of Fit 0.000 0 Pure Error 236.25 12 19.69 Cor Total 1080.94 15 F Value 14.30 14.30 Prob > F 0.0003 0.0003 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 Treatment 1 vs 2 1 vs 3 1 vs 4 2 vs 3 2 vs 4 3 vs 4 Mean Difference -0.25 12.75 15.75 13.00 16.00 3.00 DF 1 1 1 1 1 1 Standard Error 3.14 3.14 3.14 3.14 3.14 3.14 t for H0 Coeff=0 -0.080 4.06 5.02 4.14 5.10 0.96 (b) Estimate the overall mean and the treatment effects. 3-19 Prob > |t| 0.9378 0.0016 0.0003 0.0014 0.0003 0.3578 significant Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY ö µ = 2207 / 16 = 137.9375 ö1 = y1. − y .. = 145.00 − 137.9375 = 7.0625 τ ö τ 2 = y 2. − y .. = 145.25 − 137.9375 = 7.3125 ö τ 3 = y 3. − y .. = 132.25 − 137.9375 = −5.6875 ö τ 4 = y 4. − y .. = 129.25 − 137.9375 = −8.6875 (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. 19.69 4 124.4155 ≤ µ 4 ≤ 134.0845 Treatment 4: 129.25 ± 2.179 Treatment 1 - Treatment 4: ( − 129.25)± 3.055 145 (2) .69 19 4 6.164 ≤ µ1 − µ 4 ≤ 25.336 (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. 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 than 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? S yi . = MS E 19.96 = = 2.219 Coating types 1 and 2 produce the highest conductivity. n 4 Scaled t Distribution (4) (3) 130 (1) (2) 135 140 145 150 Conductivity (f) Assuming that coating type 4 is currently in use, what are your recommendations to the manufacturer? We wish to minimize conductivity. 3-20 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 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. 3.19. Reconsider the experiment in Problem 3.18. 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. Residuals vs. Predicted Normal plot of residuals 6.75 99 3 90 80 70 Res iduals Norm al % probability 95 50 30 20 10 -0.75 2 -4.5 5 1 -8.25 -8.25 3 -0.75 -4.5 129.25 6.75 133.25 137.25 141.25 145.25 Predicted R es idual Residuals vs. Coating Type 6.75 Res iduals 3 2 -0.75 -4.5 -8.25 1 2 3 4 Coating Type 3.20. 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. 3-21 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Rodding Level 10 15 20 25 Compressive Strength 1530 1530 1610 1650 1560 1730 1500 1490 1440 1500 1530 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 ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean Source Squares DF Square Model 28633.33 3 9544.44 A 28633.33 3 9544.44 Residual 40933.33 8 5116.67 Lack of Fit 0.000 0 Pure Error 40933.33 8 5116.67 Cor Total 69566.67 11 F Value 1.87 1.87 Prob > F 0.2138 0.2138 not significant 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 Treatment 1 vs 2 1 vs 3 1 vs 4 2 vs 3 2 vs 4 3 vs 4 Mean Difference -86.67 -106.67 0.000 -20.00 86.67 106.67 DF 1 1 1 1 1 1 Standard Error 58.40 58.40 58.40 58.40 58.40 58.40 t for H0 Coeff=0 -1.48 -1.83 0.000 -0.34 1.48 1.83 Prob > |t| 0.1761 0.1052 1.0000 0.7408 0.1761 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? Slight inequality of variance can be observed in the residual plots below; however, not enough to be concerned about the assumptions. 3-22 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Residuals vs. Predicted Normal plot of residuals 123.333 99 70.8333 90 80 70 Res iduals Norm al % probability 95 50 2 18.3333 30 20 10 -34.1667 5 1 -86.6667 -86.6667 -34.1667 70.8333 18.3333 1500.00 123.333 1526.67 1553.33 1580.00 Predicted R es idual Residuals vs. Rodding Level 123.333 Res iduals 70.8333 2 18.3333 -34.1667 -86.6667 1 2 3 4 Rodding Level (d) Construct a graphical display to compare the treatment means as describe in Section 3.5.3. 3-23 1606.67 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Scaled t Distribution (10, 25) (15) (20) 1418 1459 1500 1541 1582 1623 1664 Mean Compressive Strength 3.21. 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. 0.37 0.51 0.71 1.02 1.40 1.99 80 75 74 67 62 60 Radon Released (%) 83 83 75 79 73 76 72 74 62 67 61 64 85 79 77 74 69 66 (a) Does the size of the orifice affect the mean percentage of radon released? Use α = 0.05. 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 Source Squares DF Square Model 1133.38 5 226.68 A 1133.38 5 226.68 Residual 132.25 18 7.35 Lack of Fit 0.000 0 Pure Error 132.25 18 7.35 Cor Total 1265.63 23 F Value 30.85 30.85 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 3-24 Prob > F < 0.0001 < 0.0001 significant Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Treatment 1 vs 2 1 vs 3 1 vs 4 1 vs 5 1 vs 6 2 vs 3 2 vs 4 2 vs 5 2 vs 6 3 vs 4 3 vs 5 3 vs 6 4 vs 5 4 vs 6 5 vs 6 Mean Difference 5.75 7.75 11.00 17.75 20.00 2.00 5.25 12.00 14.25 3.25 10.00 12.25 6.75 9.00 2.25 DF 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Standard Error 1.92 1.92 1.92 1.92 1.92 1.92 1.92 1.92 1.92 1.92 1.92 1.92 1.92 1.92 1.92 t for H0 Coeff=0 3.00 4.04 5.74 9.26 10.43 1.04 2.74 6.26 7.43 1.70 5.22 6.39 3.52 4.70 1.17 Prob > |t| 0.0077 0.0008 < 0.0001 < 0.0001 < 0.0001 0.3105 0.0135 < 0.0001 < 0.0001 0.1072 < 0.0001 < 0.0001 0.0024 0.0002 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. Residuals vs. Predicted Normal plot of residuals 4 99 2 2 1.8125 90 80 70 Res iduals Norm al % probability 95 50 30 20 2 -0.375 2 10 -2.5625 5 2 1 -4.75 -4.75 -2.5625 -0.375 1.8125 62.75 4 67.75 72.75 Predicted R es idual 3-25 77.75 82.75 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Residuals vs. Orifice Diameter 4 Res iduals 2 2 1.8125 2 -0.375 2 -2.5625 2 -4.75 1 2 3 6 5 4 Orifice Diam eter (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): 65 ± 2.101 62.152 ≤ µ ≤ 67.848 7.35 4 (e) Construct a graphical display to compare the treatment means as describe in Section 3.5.3. What conclusions can you draw? Scaled t Distribution (6) (5) 60 (5) 65 (4) 70 (3) 75 (2) (1) 80 Conductivity 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. 3-26 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 3.22. 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 1 2 3 9 20 6 Response Time 10 23 8 12 21 5 8 17 16 15 30 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 Source Squares DF Square Model 543.60 2 271.80 A 543.60 2 271.80 Residual 202.80 12 16.90 Lack of Fit 0.000 0 Pure Error 202.80 12 16.90 Cor Total 746.40 14 F Value 16.08 16.08 Prob > F 0.0004 0.0004 significant 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 Treatment 1 vs 2 1 vs 3 2 vs 3 Mean Difference -11.40 2.40 13.80 DF 1 1 1 Standard Error 2.60 2.60 2.60 t for H0 Coeff=0 -4.38 0.92 5.31 Prob > |t| 0.0009 0.3742 0.0002 (b) Use Tukey’s test to compare pairs of treatment means. Use α = 0.01. S yi . = MS E 16.90 = = 1.8385 5 n q0.01,(3,12 ) = 5.04 t0 = 1.8385(5.04) = 9.266 1 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 the computer output) gives the same results. (c) Use the graphical procedure in Section 3.5.3 to compare the treatment means. What conclusions can you draw? How do they compare with the conclusions from part (a). 3-27 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY The scaled-t plot agrees with part (b). In this case, the large difference between the mean of treatment 2 and the other two treatments is very obvious. Scaled t Distribution (3) 5 (1) 10 (2) 15 20 25 Tensile Strength (d) Construct a set of orthogonal contrasts, assuming that at the outset of the experiment you suspected the response time of circuit type 2 to be different from the other two. H 0 = µ1 − 2 µ 2 + µ3 = 0 H1 = µ1 − 2 µ 2 + µ3 ≠ 0 C1 = y1. − 2 y2. + y3. C1 = 54 − 2 (111) + 42 = −126 (−126 ) 5 (6 ) 2 SSC1 = FC1 = = 529.2 529.2 = 31.31 16.9 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. 3-28 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Residuals vs. Predicted Normal plot of residuals 7.8 99 4.55 90 80 70 Res iduals Norm al % probability 95 50 30 20 10 1.3 -1.95 5 1 -5.2 -5.2 -1.95 1.3 8.40 7.8 4.55 11.85 15.30 18.75 22.20 Predicted R es idual Residuals vs. Circuit Type 7.8 Res iduals 4.55 1.3 -1.95 -5.2 1 3 2 Circuit Type 3.23. 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 from a completely randomized experiment were as follows: Fluid Type 1 2 3 4 17.6 16.9 21.4 19.3 18.9 15.3 23.6 21.1 Life (in h) at 35 kV Load 16.3 17.4 18.6 17.1 19.4 18.5 16.9 17.5 20.1 19.5 20.5 18.3 21.6 20.3 22.3 19.8 (a) Is there any indication that the fluids differ? Use α = 0.05. At α = 0.05 there is no difference, but since the P-value is just slightly above 0.05, there is probably a difference in means. 3-29 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Design Expert Output Response: Life in in h ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean Source Squares DF Square Model 30.17 3 10.06 A 30.16 3 10.05 Residual 65.99 20 3.30 Lack of Fit 0.000 0 Pure Error 65.99 20 3.30 Cor Total 96.16 23 F Value 3.05 3.05 Prob > F 0.0525 0.0525 not significant 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 Treatment 1 vs 2 1 vs 3 1 vs 4 2 vs 3 2 vs 4 3 vs 4 Mean Difference 0.70 -2.30 -0.17 -3.00 -0.87 2.13 DF 1 1 1 1 1 1 Standard Error 1.05 1.05 1.05 1.05 1.05 1.05 t for H0 Coeff=0 0.67 -2.19 -0.16 -2.86 -0.83 2.03 Prob > |t| 0.5121 0.0403 0.8753 0.0097 0.4183 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. Residuals vs. Predicted Normal plot of residuals 2.95 99 1.55 90 80 70 Res iduals Norm al % probability 95 50 30 20 10 0.15 -1.25 5 1 -2.65 -2.65 -1.25 0.15 1.55 17.95 2.95 18.70 19.45 Predicted R es idual 3-30 20.20 20.95 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Residuals vs. Fluid Type 2.95 Res iduals 1.55 0.15 -1.25 -2.65 1 4 3 2 Fluid Type 3.24. 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 1 2 3 4 19 80 47 95 20 61 26 46 Noise Observed 19 73 25 83 30 56 35 78 8 80 50 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] Sum of Mean Source Squares DF Square Model 12042.00 3 4014.00 A 12042.00 3 4014.00 Residual 2948.80 16 184.30 Lack of Fit 0.000 0 Pure Error 2948.80 16 184.30 Cor Total 14990.80 19 F Value 21.78 21.78 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 3-31 Prob > F < 0.0001 < 0.0001 significant Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Mean Difference -50.80 -17.40 -60.60 33.40 -9.80 -43.20 Treatment 1 vs 2 1 vs 3 1 vs 4 2 vs 3 2 vs 4 3 vs 4 Standard Error 8.59 8.59 8.59 8.59 8.59 8.59 DF 1 1 1 1 1 1 t for H0 Coeff=0 -5.92 -2.03 -7.06 3.89 -1.14 -5.03 Prob > |t| < 0.0001 0.0597 < 0.0001 0.0013 0.2705 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. Residuals vs. Predicted Normal plot of residuals 17.2 99 2 4.45 90 2 80 70 Res iduals Norm al % probability 95 50 30 20 10 -8.3 -21.05 5 1 -33.8 -33.8 -21.05 4.45 -8.3 19.20 17.2 34.35 49.50 64.65 79.80 Predicted R es idual Residuals vs. Circuit Design 17.2 2 4.45 Res iduals 2 -8.3 -21.05 -33.8 1 2 3 4 Circuit Des ign (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. 3-32 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 3.25. 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 1 2 3 4 Percentage of Methyl Alcohol 84.99 84.04 85.15 85.13 84.72 84.48 84.20 84.10 84.38 84.88 85.16 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 Source Squares DF Square Model 1.04 3 0.35 A 1.04 3 0.35 Residual 0.86 8 0.11 Lack of Fit 0.000 0 Pure Error 0.86 8 0.11 Cor Total 1.90 11 F Value 3.25 3.25 Prob > F 0.0813 0.0813 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 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 Treatment 1 vs 2 1 vs 3 1 vs 4 2 vs 3 2 vs 4 3 vs 4 Mean Difference -0.58 -0.32 0.19 0.27 0.77 0.50 DF 1 1 1 1 1 1 Standard Error 0.27 0.27 0.27 0.27 0.27 0.27 t for H0 Coeff=0 -2.18 -1.18 0.70 1.00 2.88 1.88 (b) Analyze the residuals from this experiment. There is nothing unusual about the residual plots. 3-33 Prob > |t| 0.0607 0.2703 0.5049 0.3479 0.0205 0.0966 not significant Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Residuals vs. Predicted Normal plot of residuals 0.52 99 0.2825 90 80 70 Res iduals Norm al % probability 95 50 30 20 10 0.045 -0.1925 5 1 -0.43 -0.43 -0.1925 0.2825 0.045 84.28 0.52 84.48 84.67 84.86 85.05 Predicted R es idual Residuals vs. Chemist 0.52 Res iduals 0.2825 0.045 -0.1925 -0.43 1 3 2 4 Chem is t (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 1 2 3 4 Total 253.41 255.16 254.36 252.85 Contrast Totals: 3-34 C1 1 -3 1 1 -4.86 C2 -2 0 1 1 0.39 C3 0 0 -1 1 -1.51 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY (− 4.86)2 = 0.656 3( ) 12 (0.39)2 = 0.008 SS C 2 = 3(6 ) (− 1.51)2 = 0.380 SS C 3 = 3(2 ) SS C1 = 0.656 = 6.115* 0.10727 0.008 FC 2 = = 0.075 0.10727 0.380 FC 3 = = 3.54 0.10727 FC1 = Only contrast 1 is significant at 5%. 3.26. Three brands of batteries are under study. It is s suspected that the lives (in weeks) of the three brands are different. Five randomly selected batteries of each brand are tested with the following results: Brand 1 100 96 92 96 92 Weeks of Life Brand 2 Brand 3 76 108 80 100 75 96 84 98 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 Source Squares DF Square Model 1196.13 2 598.07 A 1196.13 2 598.07 Residual 187.20 12 15.60 Lack of Fit 0.000 0 Pure Error 187.20 12 15.60 Cor Total 1383.33 14 F Value 38.34 38.34 Prob > F < 0.0001 < 0.0001 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 Mean Standard Treatment Difference DF Error 1 vs 2 15.80 1 2.50 1 vs 3 -5.20 1 2.50 2 vs 3 -21.00 1 2.50 t for H0 Coeff=0 6.33 -2.08 -8.41 (b) Analyze the residuals from this experiment. There is nothing unusual about the residuals. 3-35 Prob > |t| < 0.0001 0.0594 < from 0.0001 significant Solutions Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Residuals vs. Predicted Normal plot of residuals 7.6 99 4.6 90 80 70 Res iduals Norm al % probability 95 50 30 20 1.6 2 2 10 -1.4 5 1 2 -4.4 -4.4 -1.4 1.6 4.6 79.40 7.6 84.65 89.90 95.15 100.40 Predicted R es idual Residuals vs. Brand 7.6 Res iduals 4.6 1.6 2 2 -1.4 2 -4.4 1 3 2 Brand (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. y i . ± tα 2 ,N − a MS E n Brand 2: 79.4 ± 2.179 15.60 5 79.40 ± 3.849 75.551 ≤ µ 2 ≤ 83.249 Brand 2 - Brand 3: y i . − y j . ± tα 2 ,N − a 2( .60 ) 15 5 −28.631 ≤ µ 2 − µ3 ≤ −13.369 79.4 − 100.4 ± 3.055 3-36 2 MS E n Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY (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:  85 − 100.4   = Φ (− 3.90 ) = 0.00005   15.60  Φ  That is, about 5 out of 100,000 batteries will fail before 85 week. 3.27. Four catalysts that may affect the concentration of one component in a three component liquid mixture are being investigated. The following concentrations are obtained from a completely randomized experiment: Catalyst 2 3 56.3 50.1 54.5 54.2 57.0 55.4 55.3 1 58.2 57.2 58.4 55.8 54.9 4 52.9 49.9 50.0 51.7 (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 Source Squares DF Square Model 85.68 3 28.56 A 85.68 3 28.56 Residual 34.56 12 2.88 Lack of Fit 0.000 0 Pure Error 34.56 12 2.88 Cor Total 120.24 15 F Value 9.92 9.92 Prob > F 0.0014 0.0014 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 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 Treatment 1 vs 2 1 vs 3 1 vs 4 2 vs 3 2 vs 4 3 vs 4 Mean Difference 1.13 3.67 5.77 2.54 4.65 2.11 DF 1 1 1 1 1 1 Standard Error 1.14 1.24 1.14 1.30 1.20 1.30 t for H0 Coeff=0 0.99 2.96 5.07 1.96 3.87 1.63 3-37 Prob > |t| 0.3426 0.0120 0.0003 0.0735 0.0022 0.1298 significant Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY (b) Analyze the residuals from this experiment. There is nothing unusual about the residual plots. Residuals vs. Predicted Normal plot of residuals 2.16667 99 0.841667 90 80 70 Res iduals Norm al % probability 95 -0.483333 50 30 20 10 -1.80833 5 1 -3.13333 -3.13333 -1.80833 -0.483333 0.841667 51.13 2.16667 52.57 54.01 55.46 Predicted R es idual Residuals vs. Catalyst 2.16667 Res iduals 0.841667 -0.483333 -1.80833 -3.13333 1 3 2 4 Catalys t (c) Construct a 99 percent confidence interval estimate of the mean response for catalyst 1. y i . ± tα 2 ,N − a MS E n Catalyst 1: 56.9 ± 3.055 2.88 5 56.9 ± 2.3186 54.5814 ≤ µ1 ≤ 59.2186 3-38 56.90 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 3.28. 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 1 2 3 4 5 110 1 880 495 7 Failure Time (minutes) 157 194 2 4 1256 5276 7040 5307 5 29 178 18 4355 10050 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 Model 1.032E+008 4 2.580E+007 6.19 A 1.032E+008 4 2.580E+007 6.19 Residual 6.251E+007 15 4.167E+006 Lack of Fit 0.000 0 Pure Error 6.251E+007 15 4.167E+006 Cor Total 1.657E+008 19 Prob > F 0.0038 0.0038 significant 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 Treatment 1 vs 2 1 vs 3 1 vs 4 1 vs 5 2 vs 3 2 vs 4 2 vs 5 3 vs 4 3 vs 5 4 vs 5 Mean Difference 153.50 -2782.00 -5563.25 149.00 -2935.50 -5716.75 -4.50 -2781.25 2931.00 5712.25 DF 1 1 1 1 1 1 1 1 1 1 Standard Error 1443.44 1443.44 1443.44 1443.44 1443.44 1443.44 1443.44 1443.44 1443.44 1443.44 t for H0 Coeff=0 0.11 -1.93 -3.85 0.10 -2.03 -3.96 -3.118E-003 -1.93 2.03 3.96 Prob > |t| 0.9167 0.0731 0.0016 0.9192 0.0601 0.0013 0.9976 0.0732 0.0604 0.0013 (b) Plot the residuals versus the predicted response. Construct a normal probability plot of the residuals. What information do these plots convey? 3-39 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Residuals vs. Predicted Normal plot of residuals 4327 99 95 Norm al % probability Res iduals 1938.25 -450.5 -2839.25 90 80 70 50 30 20 10 5 1 -5228 6.25 1435.44 2864.62 4293.81 5723.00 -5228 -2839.25 Predicted -450.5 1938.25 4327 Res idual 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: ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean Source Squares DF Square Model 165.06 4 41.26 A 165.06 4 41.26 Residual 16.44 15 1.10 Lack of Fit 0.000 0 Pure Error 16.44 15 1.10 Cor Total 181.49 19 Natural log Constant: 0.000 F Value 37.66 37.66 Prob > F < 0.0001 < 0.0001 significant 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 Treatment 1 vs 2 1 vs 3 1 vs 4 1 vs 5 2 vs 3 2 vs 4 Mean Difference 3.81 -2.66 -3.16 3.15 -6.47 -6.97 DF 1 1 1 1 1 1 Standard Error 0.74 0.74 0.74 0.74 0.74 0.74 t for H0 Coeff=0 5.15 -3.60 -4.27 4.25 -8.75 -9.42 3-40 Prob > |t| 0.0001 0.0026 0.0007 0.0007 < 0.0001 < 0.0001 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 2 vs 3 vs 3 vs 4 vs 5 4 5 5 -0.66 -0.50 5.81 6.31 1 1 1 1 0.74 0.74 0.74 0.74 -0.89 -0.67 7.85 8.52 0.3856 0.5116 < 0.0001 < 0.0001 There is nothing unusual about the residual plots when the natural log transformation is applied. Residuals vs. Predicted Normal plot of residuals 1.64792 99 0.733576 90 80 70 Res iduals Norm al % probability 95 -0.180766 50 30 20 10 -1.09511 5 1 -2.00945 -2.00945 -1.09511 -0.180766 0.733576 1.24 1.64792 4.73 2.99 6.47 8.21 Predicted R es idual Residuals vs. Material 1.64792 Res iduals 0.733576 -0.180766 -1.09511 -2.00945 1 2 3 5 4 Material 3.29. 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: Method 1 2 3 31 62 58 10 40 27 Count 21 24 120 (a) Do all methods have the same effect on mean particle count? 3-41 4 30 97 1 35 68 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 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 Source Squares DF Square Model 8963.73 2 4481.87 A 8963.73 2 4481.87 Residual 6796.00 12 566.33 Lack of Fit 0.000 0 Pure Error 6796.00 12 566.33 Cor Total 15759.73 14 F Value 7.91 7.91 Prob > F 0.0064 0.0064 significant 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 Treatment 1 vs 2 1 vs 3 2 vs 3 Mean Difference -24.80 -59.60 -34.80 DF 1 1 1 Standard Error 15.05 15.05 15.05 t for H0 Coeff=0 -1.65 -3.96 -2.31 Prob > |t| 0.1253 0.0019 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. Residuals vs. Predicted Normal plot of residuals 47 99 95 Norm al % probability Res iduals 23.75 0.5 -22.75 90 80 70 50 30 20 10 5 1 -46 13.40 28.30 43.20 58.10 73.00 -46 Predicted -22.75 0.5 23.75 Res idual (c) Based on your answer to part (b) conduct another analysis of the particle count data and draw appropriate conclusions. 3-42 47 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 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 ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean Source Squares DF Square Model 63.90 2 31.95 A 63.90 2 31.95 Residual 38.96 12 3.25 Lack of Fit 0.000 0 Pure Error 38.96 12 3.25 Cor Total 102.86 14 Constant: 0.000 F Value 9.84 9.84 Prob > F 0.0030 0.0030 significant 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 Treatment 1 vs 2 1 vs 3 2 vs 3 Mean Difference -2.84 -5.04 -2.21 DF 1 1 1 Standard Error 1.14 1.14 1.14 t for H0 Coeff=0 -2.49 -4.42 -1.94 Prob > |t| 0.0285 0.0008 0.0767 3.30 A manufacturer suspects that the batches of raw material furnished by his supplier differ significantly in calcium content. There are a large number of batches currently in the warehouse. Five of these are randomly selected for study. A chemist makes five determinations on each batch as obtains the following data: Batch 1 23.46 23.48 23.56 23.39 23.40 Batch 2 23.59 23.46 23.42 23.49 23.50 Batch 3 23.51 23.64 23.46 23.52 23.49 Batch 4 23.28 23.40 23.37 23.46 23.39 Batch 5 23.29 23.46 23.37 23.32 23.38 (a) Is there significant variation in the calcium content from batch to batch? Use α=0.05. The computer output below shows that for the random effects model there is batch to batch variation. Based on the ANOVA in the JMP output below, the batches differ significantly. JMP Output Summary of Fit RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) 0.525399 0.430479 0.066182 23.4436 25 3-43 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Analysis of Variance Source DF Model 4 Error 20 C. Total 24 Sum of Squares 0.09697600 0.08760000 0.18457600 Effect Tests Source Batch DF 4 Nparm 4 Mean Square 0.024244 0.004380 Sum of Squares 0.09697600 F Ratio 5.5352 F Ratio 5.5352 Prob > F 0.0036* Prob > F 0.0036* (b) Estimate the components of variance. ˆ = MS E 0.004380 σ2 = ˆ = σ τ2 MSTreatments − MS E 0.024244 − 0.004380 = = 0.003973 n 5 This is verified in the JMP REML analysis shown below. JMP Output Parameter Estimates Term Intercept Estimate 23.4436 REML Variance Component Estimates Random Var Ratio Var Component Effect Batch 0.907032 0.0039728 Residual 0.00438 Total 0.0083528 Std Error 0.031141 DFDen 4 t Ratio 752.82 Prob>|t| <.0001* Std Error 95% Lower 95% Upper Pct of Total 0.0034398 0.0013851 -0.002769 0.0025637 0.0107147 0.0091338 47.562 52.438 100.000 Covariance Matrix of Variance Component Estimates Random Effect Batch Residual Batch 1.1832e-5 -3.837e-7 Residual -3.837e-7 1.9184e-6 (c) Find a 95 percent confidence interval for σ τ2 (σ 2 τ +σ 2 )  1  0.024244 1 1  MSTreatments 1  = − 1 = 0.1154 −1   5  0.004380 3.51  n  MS E Fα 2,a −1, N − a     1  0.024244 1 1  MSTreatments 1  U = = − 1 = 9.2780 −1   MS E  5  0.004380 0.1168  n F1−α 2,a −1, N − a   2 ˆ σ L U ≤ 2 τ 2≤ ˆ ˆ 1 + L στ + σ 1+U = L ˆ σ2 9.2780 0.1154 ≤ 2 τ 2≤ ˆ ˆ 1 + 0.1154 σ τ + σ 1 + 9.2780 0.1035 ≤ ˆ σ τ2 ≤ 0.9027 ˆ ˆ σ τ2 + σ 2 (d) Analyze the residuals from this experiment. Are the analysis of variance assumptions are satisfied? 3-44 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY The plot of residuals vs. predicted show no concerns. 0.1 Residual Calcium content 0.0 0.0 -0. -0. 23.25 23.35 23.45 23.55 2 Calcium content Predicted The residuals used in the plot below are based on the REML analysis and shows no concerns. Note, normality is not a concern for this analysis. 0.2 -1.64 -1.28 0.0 -0.67 1.28 1.64 0.67 0.1 0.1 0.0 0 -0. -0. -0. -0. 0.1 0.2 0.5 0.8 0.9 0.95 Normal Quantile Plot 3.31. Several ovens in a metal working shop are used to heat metal specimens. All ovens are supposed to operate at the same temperature, although it is suspected that this may not be true. Three ovens selected at random, and their temperatures on successive heats are noted. The data collected are as follows: Oven Temperature 1 491.50 498.30 498.10 493.50 2 488.50 484.65 479.90 477.35 3 480.10 484.80 488.25 473.00 493.60 471.85 (a) Is there significant variation in temperature between ovens? Use α=0.05. The computer output below shows that there is oven to oven variation. 3-45 478.65 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Minitab Output ANOVA: Temp versus Oven Factor Oven Type random Levels 3 Values 1, 2, 3 Analysis of Variance for Temp Source Oven Error Total DF 2 12 14 S = 5.14224 SS 705.10 317.31 1022.41 R-Sq = 68.96% MS 352.55 26.44 F 13.33 P 0.001 R-Sq(adj) = 63.79% (b) Estimate the components of variation for this model. a   ∑ ni2  1  77  1 a  ∑ ni − i =1 =  = 15 − = 4.9333 n0 a 3 −1  15  a − 1  i =1   ni  ∑  i =1   ˆ = MS E 26.44 σ2 = ˆ = σ τ2 MSTreatments − MS E 352.55 − 26.44 = = 66.10 4.9333 n0 (c) Analyze the residuals from this experiment and draw conclusions about model adequacy. There are no concerns with the residual plots below. Residual Plots for Temp Normal Probability Plot Versus Fits 5 Residual 10 90 Percent 99 50 10 1 -10 -5 0 Residual 5 0 -5 -10 10 480 484 Histogram 496 Versus Order 2 Residual Frequency 492 10 3 1 0 488 Fitted Value -8 -4 0 Residual 4 5 0 -5 -10 8 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Observation Order 3-46 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 3.32. An article in the Journal of the Electrochemical Society (Vol. 139, No. 2, 1992, pp. 524-532) describes an experiment to investigate low-pressure vapor deposition of polysilicon. The experiment was carried out in a large capacity reactor at Sematech in Austin, Texas. The reactor has several wafer positions, and four of these positions are selected at random. The response variable is film thickness uniformity. Three replicates of the experiment were run, and the data are as follows: Wafer Positions Uniformity 1 2.76 5.67 4.49 2 1.43 1.70 2.19 3 2.34 1.97 1.47 4 0.94 1.36 1.65 (a) Is there a difference in the wafer positions? Use Use α=0.05. The JMP output below identifies a difference in the wafer positions. JMP Output Summary of Fit RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) Analysis of Variance Source DF Model 3 Error 8 C. Total 11 Effect Tests Source Wafer Position Nparm 3 0.756617 0.665349 0.807579 2.330833 12 Sum of Squares 16.219825 5.217467 21.437292 DF 3 Mean Square 5.40661 0.65218 Sum of Squares 16.219825 F Ratio 8.2900 Prob > F 0.0077* F Ratio 8.2900 Prob > F 0.0077* (b) Estimate the variability due to wafer position. The JMP REML output below identifies the variance component for the wafer position as 1.5848. JMP Output Parameter Estimates Term Intercept Estimate 2.3308333 REML Variance Component Estimates Random Effect Var Ratio Wafer Position 2.4300043 Residual Total Std Error 0.671231 DFDen 3 Var Component 1.5848083 0.6521833 2.2369917 Std Error 1.4755016 0.3260917 Covariance Matrix of Variance Component Estimates Random Effect Wafer Position Wafer Position 2.177105 Residual -0.035445 Residual -0.035445 0.1063358 3-47 t Ratio 3.47 95% Lower -1.307122 0.2975536 Prob>|t| 0.0403* 95% Upper 4.4767383 2.393629 Pct of Total 70.846 29.154 100.000 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY (c) Estimate the random error component. The JMP REML output above identifies the random error variance component as 0.6522.. (d) Analyze the residuals from this experiment and comment on model adequacy. The plot of residuals vs. predicted shows some uniformity concerns. 1.5 1.0 Residual Uniformity 0.5 0.0 -0. -1. -1. -2. 1 2 3 4 5 6 Uniformity Predicted The residuals used in the plot below are based on the REML analysis. The normal plot shows some concerns with the normality assumption; however, the normality is not important for this analysis. -1.28 0.0 -0.67 1.28 0.67 3 2 1 0 -1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Normal Quantile Plot Uniformity data often requires a transformation, such as a log transformation, and should be considered for this experiment. 3.33. Consider the vapor-deposition experiment described in Problem 3.32. (a) Estimate the total variability in the uniformity response. The JMP REML output shown in part (b) of Problem 3.32 identifies the total variability as 2.2370. 3-48 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY (b) How much of the total variability in the uniformity response is due to the difference between positions in the reactor? From the JMP REML output shown in part (b) of Problem 3.32, the differences between positions represents 70.846% of the total variability. (c) To what level could the variability in the uniformity response be reduced if position-to-position variability in the reactor could be eliminated? Do you believe this is a significant reduction? The variability could be reduced to 29.154% of the current total variability. Based on the 95% confidence intervals calculated below, this is not significant. An increase in sample size might reverse this decision. = L  1  5.40661 1  MSTreatments 1 1  = − 1 = 0.073623 −1    4  0.65218 6.059467  n  MS E Fα 2,a −1, N − a    U =  1  5.40661 1  MSTreatments 1 1  = − 1 = 65.08093 −1   4  0.65218 0.025398  n  MS E F1−α 2,a −1, N − a     ˆ σ2 L U ≤ 2 τ 2≤ ˆ ˆ 1 + L στ + σ 1+U ˆ σ2 65.08093 0.073623 ≤ 2 τ 2≤ ˆ ˆ 1 + 0.073623 σ τ + σ 1 + 65.08093 0.068575 ≤ ˆ σ τ2 ≤ 0.984867 2 ˆ ˆ στ + σ 2 3.34. An article in the Journal of Quality Technology (Vol. 13, No. 2, 1981, pp. 111-114) describes and experiment that investigates the effects of four bleaching chemicals on pulp brightness. These four chemicals were selected at random from a large population of potential bleaching agents. The data are as follows: Chemicals Brightness 1 77.199 74.466 92.746 76.208 82.876 2 80.522 79.306 81.914 80.346 73.385 3 79.417 78.017 91.596 80.802 80.626 4 78.001 78.358 77.544 77.364 77.386 (a) Is there a difference in the chemical types? Use Use α=0.05. From the analysis below, there does not appear to be a difference in chemical types. JMP Output Summary of Fit RSquare RSquare Adj Root Mean Square Error Mean of Response Observations (or Sum Wgts) Analysis of Variance Source 0.123254 -0.04114 4.898921 79.90395 20 DF Sum of Squares Mean Square 3-49 F Ratio Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Source Model Error C. Total DF 3 16 19 Effect Tests Source Chemical Sum of Squares 53.98207 383.99085 437.97292 Nparm 3 DF 3 Mean Square 17.9940 23.9994 Sum of Squares 53.982073 F Ratio 0.7498 Prob > F 0.5383 F Ratio 0.7498 Prob > F 0.5383 (b) Estimate the variability due to chemical types. The JMP REML output below identifies the variance component for chemical types as -1.201081. This negative value is a concern. One solution would be to convert this to zero, but this has concerns as identified in Section 3.9.3 of the textbook. Another course of action is to re-estimate this variance component using a method that always provides a non-negative value. Another alternative is to assume that the underlying model is non-linear and re-examine the problem. JMP Output Parameter Estimates Term Intercept Estimate 79.90395 Std Error 0.948526 REML Variance Component Estimates Random Effect Var Ratio Var Component Chemical -0.050046 -1.201081 Residual 23.999428 Total 22.798347 DFDen 3 Std Error 3.3932473 8.4850792 95% Lower -7.851723 13.312053 t Ratio 84.24 95% Upper 5.4495617 55.589101 Prob>|t| <.0001* Pct of Total -5.268 105.268 100.000 Covariance Matrix of Variance Component Estimates Random Effect Chemical Residual Chemical 11.514127 -14.39931 Residual -14.39931 71.996569 (c) Estimate the variability due to random error. From the JMP REML output shown above, the variance component due to random error is 23.999428. (d) Analyze the residuals from this experiment and comment on model adequacy. Examination of the residuals identified two outliers. These outliers correspond to the Brightness values of 92.746 and 91.596. The experimenter should resolve these outliers. Residual Brightness 10 5 0 -5 75 80 85 90 95 Brightness Predicted 3-50 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 15 -1.64 -1.28 -0.67 0.0 0.67 1.28 1.64 10 5 0 -5 0.1 0.2 0.5 0.8 0.9 0.95 Normal Quantile Plot 3.35. Consider the single-factor random effects model discussed in this chapter. Develop a procedure for finding a 100(1-α)% confidence interval on the ratio σ τ2 (σ 2 τ + σ 2 ) . Assume that the experiment is balanced. The procedure shown below is based on the guidelines presented in Section 1.4 of the textbook. Rather than repeat the details of the seven steps, only additional information is provided below that is specific to the single-factor random effects and the confidence interval. 1. Recognition of and statement of the problem. 2. Selection of the response variable. 3. Choice of factors, levels, and range. For this case, one factor is chosen. However, the number of levels chosen and the number of replicates determines the degrees of freedom for the F value used in the confidence interval calculations. Because the levels are random, it is important to choose an adequate representation of this effect. 4. Choice of experimental design. For this case, the experimental design is a single factor experiment. As mentioned above, the number of replicates is important in the estimation of the confidence intervals. The value for α should also be identified as this could influence the number of replicates chosen. 5. Performing the experiment. 6. Statistical analysis of the data. Perform the analysis of variance in the same manner as a fixed effects case. Identify the MS τ and MS E from the ANOVA. Select the F α /2,a-1,N-a and F 1-α /2,a-1,N-a . Perform the calculations as identified in Equations 3.59a, 3,59b, and 3.60. 7. Conclusions and recommendations. 3.36. 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. t0 = y1. − y 2. Sp 2 n ~ t 2 n − 2 assuming n 1 = n 2 = n 3-51 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY ∑ (y ) ∑ (y2 j − y2. )2 ∑∑ (yij − y1. )2 n n 1j Sp = j =1 j =1 (y1. − y2. )2  n  = ∑ yi . − y..  2 n 2 i =1 i =1 j =1 = 2n − 2 2 Furthermore, 2 − y1. 2 + 2n ≡ MS E for a=2 2 n 2n − 2 , which is exactly the same as SS Treatments in a one-way 2 classification with a=2. Thus we have shown that t 0 = SS Treatments 2 . In general, we know that t u = F1,u so MS E 2 that t 0 ~ F1,2 n − 2 . 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. a 3.37. Show that the variance of the linear combination ∑ c i y i . is σ 2 i =1 a  V  ci yi .  =   i =1   ∑ a ∑ V (ci yi . ) = i =1 a ∑ i =1 = ∑ ∑c ∑n c 2 ii . i =1  ni  ci2V  yij  =  j =1    a a ni ∑ ∑ V (y ), V (y )= σ a ci2 i =1 2 ij ij . j =1 2 2 i ni σ i =1 3.38. In a fixed effects experiment, suppose that there are n observations for each of four treatments. Let 2 2 Prove that Q12 , Q2 , Q3 be single-degree-of-freedom components for the orthogonal contrasts. 2 2 SS Treatments = Q12 + Q 2 + Q3 . C1 = 3 y1. − y 2. − y 3. − y 4. C 2 = 2 y 2. − y 3. − y 4. C 3 = y 3. − y 4. SS C1 = Q12 2 SS C 2 = Q 2 2 SS C 3 = Q3 ( 3 y1. − y 2. − y 3. − y 4. ) 2 12n ( 2 y 2. − y 3. − y 4. ) 2 2 Q2 = 6n ( y − y 4. ) 2 2 Q 3 = 3. 2n 4   y i2. − 6 yi. y j.  9   i =1 i< j   2 2 and since Q12 + Q 2 + Q3 = 12n Q12 = ∑ ∑∑ 4 ∑∑ i< j 1 2 yi . y j . =  y.. − 2  4 ∑ i =1  2 2 yi2.  , we have Q12 + Q 2 + Q3 =   for a=4. 3-52 12 ∑y 2 i. 2 − 3 y .. i =1 12n 4 = ∑ i =1 2 y i2. y .. − = SS Treatments n 4n Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 3.39. Use Bartlett's test to determine if the assumption of equal variances is satisfied in Problem 3.24. Use α = 0.05. Did you reach the same conclusion regarding the equality of variance by examining the residual plots? 2 χ 0 = 2.3026 2 q = (N − a )log 10 S p − q , where c a ∑ (n i − 1)log 10 S i2 i =1  1  (ni − 1)−1 − (N − a )−1    3(a − 1)  i =1  a ∑ c = 1+ a 2 Sp = S12 2 S2 2 S3 ∑ (n − 1)S i2 i =1 N −a = 11.2 = 14.8 i 2 Sp = (5 − 1)11.2 + (5 − 1)14.8 + (5 − 1)20.8 15 − 3 = 20.8 c = 1+ a  1  (5 − 1)−1 − (15 − 3)−1    3(3 − 1)  i =1  c = 1+ = 15.6 1 3 1   +  = 1.1389 3(3 − 1)  4 12  ∑ a 2 q = (N − a )log10 S p − ∑ (ni − 1)log10 Si2 i =1 q = (15 − 3)log10 15.6 − 4 (log10 11.2 + log10 14.8 + log10 20.8 ) q = 14.3175 − 14.150 = 0.1673 q c χ 02 = 2.3026 = 2.3026 0.1673 = 0.3382 1.1389 2 χ 0.05,2 = 5.99 Cannot reject null hypothesis; conclude that the variance are equal. This agrees with the residual plots in Problem 3.24. 3.40. Use the modified Levene test to determine if the assumption of equal variances is satisfied on Problem 3.24. 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: yij −  yi Brand 1 4 0 4 0 4 Brand 2 4 0 5 4 2 3-53 Brand 3 8 0 4 2 0 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY The analysis of variance indicates that there is not a difference between the different brands and therefore the assumption of equal variances is satisfied. Design Expert Output Response: Mod Levine ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean Source Squares DF Square Model 0.93 2 0.47 A 0.93 2 0.47 Pure Error 80.00 12 6.67 Cor Total 80.93 14 F Value 0.070 0.070 Prob > F 0.9328 0.9328 3.41. Refer to Problem 3.22. 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 σ 2 ? Φ2 = nD 2 2 aσ 2 , use MS E from Problem 3.20 to estimate σ 2 . Φ2 = n( )2 10 = 0.986n 2(3)( .9 ) 16 Letting α = 0.05 , P(accept) = 0.1 , υ1 = a − 1 = 2 Trial and Error yields: n υ2 5 6 7 12 15 18 Φ 2.22 2.43 2.62 P(accept) 0.17 0.09 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 3.20) 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…”. 3.42. Refer to Problem 3.26. (a) If we wish to detect a maximum difference in mean battery life of 10 hours 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 MS E from Problem 3.26. 3-54 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY nD 2 n (10 ) 2 = 1.0684n 2 (3)(15.60 ) Letting α = 0.05 , P(accept) = 0.1 , υ1 = a − 1 = 2 Φ= 2 2 aσ 2 Φ= 2 Trial and Error yields: n υ2 4 5 6 9 12 15 Φ 2.067 2.311 2.532 P(accept) 0.25 0.12 0.05 Choose n ≥ 6, therefore N ≥ 18 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? υ1 = a − 1 = 2 [ υ 2 = N − a = 15 − 3 = 12 ] [ α = 0.05 ] P( accept ) ≤ 0.1 1 1 λ = 1 + n ( + 0.01P )2 − 1 = 1 + n ( + 0.01(25))2 − 1 = 1 + 0.5625n Trial and Error yields: n υ2 8 9 10 21 24 27 Φ 2.12 2.25 2.37 P(accept) 0.16 0.13 0.09 Choose n ≥ 10, therefore N ≥ 30 3.43. Consider the experiment in Problem 3.26. 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? α = 0.05 MS E = 15.6 width = t 0.025 ,N − a 2MS E n Trial and Error yields: n υ2 t width 5 10 31 32 12 27 90 93 2.179 2.05 1.99 1.99 5.44 3.62 1.996 1.96 Choose n ≥ 31, therefore N ≥ 93 3-55 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 3.44. Suppose that four normal populations have means of µ 1 =50, µ 2 =60, µ 3 =50, and µ 4 =60. How many observations should be taken from each population so that the probability or rejecting the null hypothesis of equal population means is at least 0.90? Assume that α=0.05 and that a reasonable estimate of the error variance is σ 2 =25. µ i = µ + τ i ,i = 1,2 ,3,4 4 ∑µ i 220 µ= = = 55 4 4 τ 1 = −5,τ 2 = 5,τ 3 = −5,τ 4 = 5 i =1 Φ= 2 n ∑τ aσ 2 2 i = 100n =n 4(25) Φ= n 4 ∑τ 2 i = 100 i =1 υ1 = 3,υ 2 = 4(n − 1),α = 0.05 , From the O.C. curves we can construct the following: n 4 5 υ2 12 16 Φ 2.00 2.24 β 0.18 0.08 1-β 0.82 0.92 Therefore, select n=5 3.45. Refer to Problem 3.44. (a) How would your answer change if a reasonable estimate of the experimental error variance were σ 2 = 36? Φ2 = n ∑τ 2 i = aσ 2 100n = 0.6944n 4(36 ) Φ = 0.6944n υ1 = 3,υ 2 = 4(n − 1),α = 0.05 , From the O.C. curves we can construct the following: n 5 6 7 υ2 16 20 24 Φ 1.863 2.041 2.205 β 0.24 0.15 0.09 1-β 0.76 0.85 0.91 Therefore, select n=7 (b) How would your answer change if a reasonable estimate of the experimental error variance were σ 2 = 49? Φ2 = n ∑τ aσ 2 2 i = 100n = 0.5102n 4(49 ) Φ = 0.5102n 3-56 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY υ1 = 3,υ 2 = 4(n − 1),α = 0.05 , From the O.C. curves we can construct the following: n 7 8 9 Φ 1.890 2.020 2.142 υ2 24 28 32 β 0.16 0.11 0.09 1-β 0.84 0.89 0.91 Therefore, select n=9 (c) Can you draw any conclusions about the sensitivity of your answer in the particular situation about how your estimate of σ affects the decision about sample size? As our estimate of variability increases the sample size must increase to ensure the same power of the test. (d) Can you make any recommendations about how we should use this general approach to choosing n in practice? 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…”. 3.46. Refer to the aluminum smelting experiment described in Section 3.8.3. Verify that ratio control methods do not affect average cell voltage. Construct a normal probability plot of residuals. Plot the residuals versus the predicted values. Is there an indication that any underlying assumptions are violated? Design Expert Output Response: Cell Average ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean Source Squares DF Square Model 2.746E-003 3 9.153E-004 A 2.746E-003 3 9.153E-004 Residual 0.090 20 4.481E-003 Lack of Fit 0.000 0 Pure Error 0.090 20 4.481E-003 Cor Total 0.092 23 F Value 0.20 0.20 Prob > F 0.8922 0.8922 not significant The "Model F-value" of 0.20 implies the model is not significant relative to the noise. There is a 89.22 % chance that a "Model F-value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error 1-1 4.86 0.027 2-2 4.83 0.027 3-3 4.85 0.027 4-4 4.84 0.027 Mean Treatment Difference 1 vs 2 0.027 1 vs 3 0.013 1 vs 4 0.025 2 vs 3 -0.013 2 vs 4 -1.667E-003 3 vs 4 0.012 DF 1 1 1 1 1 1 Standard Error 0.039 0.039 0.039 0.039 0.039 0.039 t for H0 Coeff=0 0.69 0.35 0.65 -0.35 -0.043 0.30 The following residual plots are satisfactory. 3-57 Prob > |t| 0.4981 0.7337 0.5251 0.7337 0.9660 0.7659 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Residuals vs. Predicted Normal plot of residuals 0.105 99 2 0.05125 90 80 70 Res iduals Norm al % probability 95 50 30 20 10 3 -0.0025 -0.05625 5 1 -0.11 -0.11 -0.05625 0.05125 -0.0025 4.833 0.105 4.840 4.847 4.853 4.860 Predicted R es idual Residuals vs. Algorithm 0.105 2 Res iduals 0.05125 3 -0.0025 -0.05625 -0.11 1 2 3 4 Algorithm 3.47. Refer to the aluminum smelting experiment in Section 3.8.3. Verify the ANOVA for pot noise summarized in Table 3.16. Examine the usual residual plots and comment on the experimental validity. Design Expert Output Response: Cell StDev Transform: Natural log ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean Source Squares DF Square Model 6.17 3 2.06 A 6.17 3 2.06 Residual 1.87 20 0.094 Lack of Fit 0.000 0 Pure Error 1.87 20 0.094 Cor Total 8.04 23 Constant: 0.000 F Value 21.96 21.96 Prob > F < 0.0001 < 0.0001 The Model F-value of 21.96 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 3-58 significant Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 1-1 2-2 3-3 4-4 Mean -3.09 -3.51 -2.20 -3.36 Treatment 1 vs 2 1 vs 3 1 vs 4 2 vs 3 2 vs 4 3 vs 4 Error 0.12 0.12 0.12 0.12 Mean Difference 0.42 -0.89 0.27 -1.31 -0.15 1.16 DF 1 1 1 1 1 1 Standard Error 0.18 0.18 0.18 0.18 0.18 0.18 t for H0 Coeff=0 2.38 -5.03 1.52 -7.41 -0.86 6.55 Prob > |t| 0.0272 < 0.0001 0.1445 < 0.0001 0.3975 < 0.0001 The following residual plots identify the residuals to be normally distributed, randomly distributed through the range of prediction, and uniformly distributed across the different algorithms. This validates the assumptions for the experiment. Residuals vs. Predicted Normal plot of residuals 0.512896 99 2 0.245645 90 80 70 Res iduals Norm al % probability 95 -0.0216069 50 30 20 10 3 2 2 -0.288858 5 2 1 -0.55611 -0.55611 -0.288858 -0.0216069 0.245645 -3.51 0.512896 Residuals vs. Algorithm 0.512896 2 Res iduals 0.245645 3 2 2 -0.288858 2 -0.55611 1 3 2 -2.85 Predicted R es idual -0.0216069 -3.18 4 Algorithm 3-59 -2.53 -2.20 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY 3.48. Four different feed rates were investigated in an experiment on a CNC machine producing a component part used in an aircraft auxiliary power unit. The manufacturing engineer in charge of the experiment knows that a critical part dimension of interest may be affected by the feed rate. However, prior experience has indicated that only dispersion effects are likely to be present. That is, changing the feed rate does not affect the average dimension, but it could affect dimensional variability. The engineer makes five production runs at each feed rate and obtains the standard deviation of the critical dimension (in 10-3 mm). The data are shown below. Assume that all runs were made in random order. Feed Rate (in/min) 10 12 14 16 1 0.09 0.06 0.11 0.19 Production 2 0.10 0.09 0.08 0.13 Run 3 0.13 0.12 0.08 0.15 4 0.08 0.07 0.05 0.20 5 0.07 0.12 0.06 0.11 (a) Does feed rate have any effect on the standard deviation of this critical dimension? Because the residual plots were not acceptable for the non-transformed data, a square root transformation was applied to the standard deviations of the critical dimension. Based on the computer output below, the feed rate has an effect on the standard deviation of the critical dimension. Design Expert Output Response: Run StDev Transform: Square root Constant: ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean F Source Squares DF Square Value Model 0.040 3 0.013 7.05 A 0.040 3 0.013 7.05 Residual 0.030 16 1.903E-003 Lack of Fit 0.000 0 Pure Error 0.030 16 1.903E-003 Cor Total 0.071 19 0.000 Prob > F 0.0031 0.0031 significant The Model F-value of 7.05 implies the model is significant. There is only a 0.31% chance that a "Model F-Value" this large could occur due to noise. Treatment Means (Adjusted, If Necessary) Estimated Standard Mean Error 1-10 0.30 0.020 2-12 0.30 0.020 3-14 0.27 0.020 4-16 0.39 0.020 Mean Treatment Difference 1 vs 2 4.371E-003 1 vs 3 0.032 1 vs 4 -0.088 2 vs 3 0.027 2 vs 4 -0.092 3 vs 4 -0.12 DF 1 1 1 1 1 1 Standard Error 0.028 0.028 0.028 0.028 0.028 0.028 t for H0 Coeff=0 0.16 1.15 -3.18 0.99 -3.34 -4.33 Prob > |t| 0.8761 0.2680 0.0058 0.3373 0.0042 0.0005 (b) Use the residuals from this experiment of investigate model adequacy. Are there any problems with experimental validity? The residual plots are satisfactory. 3-60 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY Residuals vs. Predicted Normal plot of residuals 0.0584817 99 2 0.028646 90 80 70 Res iduals Norm al % probability 95 2 -0.00118983 50 30 20 10 -0.0310256 5 1 -0.0608614 -0.0608614 -0.0310256 -0.00118983 0.028646 0.27 0.0584817 0.30 0.33 0.36 0.39 Predicted R es idual Residuals vs. Feed Rate 0.0584817 2 Res iduals 0.028646 2 -0.00118983 -0.0310256 -0.0608614 1 4 3 2 Feed Rate 3.49. Consider the data shown in Problem 3.22.  (a) Write out the least squares normal equations for this problem, and solve them for µ and τ using the i, usual constraint    ∑ 3 ö τ i =1 i = 0  . Estimate τ 1 − τ 2 .   ö 15µ ö 5µ ö 5µ ö +5τ 1 ö +5τ 1 ö +5τ 2 3 ∑ τö i =207 =54 ö +5τ 2 ö 5µ Imposing ö +5τ 3 =111 ö +5τ 3 =42 ö ö ö ö = 0 , therefore µ = 13.80 , τ 1 = −3.00 , τ 2 = 8.40 , τ 3 = −5.40 i =1 3-61 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY öö τ1 − τ 2 = −3.00 − 8.40 = −11.40 ö ö ö (b) Solve the equations in (a) using the constraint τ 3 = 0 . Are the estimators τ i and µ the same as you found in (a)? Why? Now estimate τ 1 − τ 2 and compare your answer with that for (a). What statement can you make about estimating contrasts in the τ i ? ö ö ö Imposing the constraint, τ 3 = 0 we get the following solution to the normal equations: µ = 8.40 , τ 1 = 2.40 ö ö , τ 2 = 13.8 , and τ 3 = 0 . These estimators are not the same as in part (a). However, öö τ1 − τ 2 = 2.40 − 13.80 = −11.40 , is the same as in part (a). The contrasts are estimable. (c) Estimate µ + τ 1 , 2τ 1 − τ 2 − τ 3 and µ + τ 1 + τ 2 using the two solutions to the normal equations. Compare the results obtained in each case. 1 Contrast µ + τ1 Estimated from Part (a) 10.80 Estimated from Part (b) 10.80 2 2τ 1 − τ 2 − τ 3 -9.00 -9.00 3 µ +τ1 +τ 2 19.20 24.60 Contrasts 1 and 2 are estimable, 3 is not estimable. 3.50. Apply the general regression significance test to the experiment in Example 3.5. Show that the procedure yields the same results as the usual analysis of variance. From the etch rate table: y.. = 12355 from Example 3.5, we have: ö µ = 617.75 τö = −66.55 τö2 = −30.35 1 τö = 7.65 τö4 = 89.25 3 4 5 ∑∑ y i =1 j =1 2 ij = 7, 704,511 , with 20 degrees of freedom. 5 ö R (µ ,τ ) = µ y.. + ∑τöyi. i =1 = (617.75 )(12355 ) + (−66.55 )(2756 ) + (−30.35 )(2937 ) + (7.65 )(3127 ) + (89.25 )(3535 ) = 7, 632,301.25 + 66, 870.55 = 7, 699,171.80 with 4 degrees of freedom. 4 5 2 SS E = ∑∑ yij − R (µ ,τ ) = 7, 704,511 − 7, 699,171.80 = 5339.2 i =1 j =1 with 20-4 degrees of freedom. This is identical to the SS E found in Example 3.5. The reduced model: 3-62 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY ö R (µ ) = µ y.. = (617.75 )(12355 ) = 7, 632,301.25 , with 1 degree of freedom. R (τ µ ) = R (µ ,τ ) − R (µ ) = 7, 699,171.80 − 7, 632,301.25 = 66,870.55 , with 4-1=3 degrees of freedom. Note: R( µ )= SS Treatment from Example 3.1. τ Finally, R (τ µ ) F0 = 3 SS E 16 66,870.55 22290.8 3 = = = 66.8 5339.2 333.7 16 which is the same as computed in Example 3.5. 3.51. Use the Kruskal-Wallis test for the experiment in Problem 3.23. Are the results comparable to those found by the usual analysis of variance? From Design Expert Output of Problem 3.21 Response: Life in in h ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean Source Squares DF Square Model 30.17 3 10.06 A 30.16 3 10.05 Residual 65.99 20 3.30 Lack of Fit 0.000 0 Pure Error 65.99 20 3.30 Cor Total 96.16 23 H= F Value 3.05 3.05 Prob > F 0.0525 0.0525 not significant  a Ri2  12 12 . [4060.75]− 3 (24 + 1) = 6.22  ∑  − 3 (N + 1) = N (N + 1)  i =1 ni  24 (24 + 1) 2 χ 0.05 ,3 = 7.81 Accept the null hypothesis; the treatments are not different. This agrees with the analysis of variance. 3.52. Use the Kruskal-Wallis test for the experiment in Problem 3.24. Compare conclusions obtained with those from the usual analysis of variance? From Design Expert Output of Problem 3.22 Response: Noise ANOVA for Selected Factorial Model Analysis of variance table [Partial sum of squares] Sum of Mean Source Squares DF Square Model 12042.00 3 4014.00 A 12042.00 3 4014.00 Residual 2948.80 16 184.30 Lack of Fit 0.000 0 Pure Error 2948.80 16 184.30 Cor Total 14990.80 19 F Value 21.78 21.78 3-63 Prob > F < 0.0001 < 0.0001 significant Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY H=  a Ri2  12 12 . [2726.8]− 3 (20 + 1) = 14.91  ∑  − 3 (N + 1) = N (N + 1)  i =1 ni  20 (20 + 1) 2 χ 0.05,3 = 7.81 Reject the null hypothesis because the treatments are different. This agrees with the analysis of variance. 3.53. Consider the experiment in Example 3.5. Suppose that the largest observation on etch rate is incorrectly recorded as 250A/min. What effect does this have on the usual analysis of variance? What effect does it have on the Kruskal-Wallis test? The incorrect observation reduces the analysis of variance F 0 from 66.8 to 0.50. It does change the value of the Kruskal-Wallis test statistic but not the result. Minitab Output One-way ANOVA: Etch Rate 2 versus Power Analysis of Variance for Etch Rat Source DF SS MS Power 3 15927 5309 Error 16 168739 10546 Total 19 184666 F 0.50 P 0.685 3.54 A textile mill has a large number of looms. Each loom is supposed to provide the same output of cloth per minute. To investigate this assumption, five looms are chosen at random, and their output is noted at different times. The following data are obtained: Loom 1 2 3 4 5 14.0 13.9 14.1 13.6 13.8 14.1 13.8 14.2 13.8 13.6 Output 14.2 13.9 14.1 14.0 13.9 14.0 14.0 14.0 13.9 13.8 14.1 14.0 13.9 13.7 14.0 (a) Explain why this is a random effects experiment. Are the looms equal in output? Use α=0.05. (b) Estimate the variability between looms. (c) Estimate the experimental error variance. (d) Find a 95 percent confidence interval for sdfdsfseererrw(need equation typed here) (e) Analyze the residuals from this experiment. assumptions are satisfied? Do you think that the analysis of variance (f) Use the REML method to analyze this data. Compare the 95 percent confidence interval on the error variance from REML with the exact chi-square confidence interval. 3.55 A manufacturer suspects that the batches of raw material furnished by his supplier differ significantly in calcium content. There are a large number of batches currently in the warehouse. Five of these are 3-64 Solutions from Montgomery, D. C. (2012) Design and Analysis of Experiments, Wiley, NY randomly selected for study. A chemist makes five determinations on each batch as obtains the following data: Batch 1 23.46 23.48 23.56 23.39 23.40 Batch 2 23.59 23.46 23.42 23.49 23.50 Batch 3 23.51 23.64 23.46 23.52 23.49 Batch 4 23.28 23.40 23.37 23.46 23.39 Batch 5 23.29 23.46 23.37 23.32 23.38 This is the same as question 3.30 except for (e). (a) Is there significant variation in the calcium content from batch to batch? Use α=0.05. (b) Estimate the components of variance. (c) Find a 95 percent confidence interval for sdfdsfseererrw(need equation typed here) (d) Analyze the residuals from this experiment. satisfied? Are the analysis of variance assumptions are (e) Use the REML method to analyze this data. Compare the 95 percent confidence interval on the error variance from REML with the exact chi-square confidence interval. 3-65

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Homework 5The problem numbers are from the 8th edition of the textbook and maybe dierent from those in earlier editions.11.5, 11.9, 11.18, 11.19, 11.311
Shanghai Jiao Tong University - IE - 410
Homework 6The problem numbers are from the 8th edition of the textbook and maybe dierent from those in earlier editions.13.4, 13.14, 13.21, 14.2, 14.7, 14.13, 14.141
Shanghai Jiao Tong University - IE - 410
3.7.(a)One-way ANOVA: Tensile Strength versus Mixing TechniqueSourceMixing TechniqueErrorTotalS = 113.3Level1234N4444DF31215SS489740153908643648R-Sq = 76.09%Mean2971.03156.32933.82666.3StDev120.6136.0108.381.0MS163247
Shanghai Jiao Tong University - IE - 410
IE410 HW#6Qiubo Xu13.4(a)General Linear Model: Response versus Row, ColumnFactor Type Levels ValuesRow random3 1, 2, 3Column random4 1, 2, 3, 4Analysis of Variance for Response, using Adjusted SS for TestsSource DF Seq SS Adj SS Adj MS F PRow
Shanghai Jiao Tong University - IE - 410
Insourcing/OutsoucingThe FlexCon Piston DecisionThis case addresses many issues that affect insourcing/outsourcing decisions. A complex andimportant topic facing businesses today is whether to produce a component, assembly, orservice internally (insour
Shanghai Jiao Tong University - IE - 410
IE 410 HW#4Qiubo Xu8.6According to the Minitab output, the effects appear to be much lesssignificant than those found in Example 6.6.8.23(a)The treatment combinations given in this problem are correct.The effects are shown in the Minitab output an
Wartburg - BA - 447
WHAT?WHAT?WHO?HOW?WHAT?WHO?WHAT?HOW?$?WHO??WHAT?HOW?$?WHO??CLIENTCLIENTCLIENTSEGMENTSEGMENTSEGMENTSSSWHAT?HOW?$?WHO??VALUECLIENTCLIENTPROPOSITISEGMENTSSEGMENTSONCLIENTCLIENTCLIENTSEGMENTSEGMENTSEGMENTSSS
Wartburg - BA - 447
Product Life CycleEntrepreneurshipStages in the Product Life CycleIntroduction GrowthMaturityDeclineDigitalcamerasMini-discElectric carsDVDVR**= virtual realityThe time at each stage variesgreatlyStages in the Product Life CycleIntroductio
Wartburg - BA - 447
Marketing PLANEwestThe Product Adoption/Diffusion Curve Groups customers into categories based on how quicklythey adopt a new product Innovators Early adopters Early majority Late majority LaggardsFigure 15.2: New ProductAdoption/Diffusion Curv
Wartburg - BA - 447
Business ModelGenerationOnly Entrepreneurs get to workwith Models!A Business Model is . . .A business model describes the rationale of how anorganization creates, delivers, and captures valueCustomer SegmentsThe Customer SegmentsBuilding Block de
Wartburg - BA - 447
Ideas Worth Acting ON!Idea feasibility analysisCreativity and InnovationCreativity the ability to develop newideas and to discover new ways oflooking at problems and opportunities;thinking new things.Innovation the ability to apply creativesolutio
Wartburg - BA - 447
CreatingaSuccessfulCreatingaSuccessfulFinancialPlanFinancialPlanCHapter 11: Financial Plan Copyright 2008 Prentice Hall Publishing1FinancialReportingssssCommon mistake among business owners:CommonFailing to collect and analyze basic financial
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15Connectwithme.30TheCustomerSituation(SocialNeed,MarketOpportunity,etc.)45Product,Service,Solution,NewInnovation.60ValueProposition(valuepropositionisananalysisandquantifiedreviewofthebenefits,costsandvalue).Isthepersonhookedorcurious?75Tot
Wartburg - BA - 447
Concept/mind mappingTraditional outlining:First itemSecond itemI.II.sub itemsub itemA.B.A.B.III.sub sub itemsub sub itemThird item5/23/02COTF(1)5/23/02COTFThe alternative: Mapping!Advanced organizers David Ausubel 1960Meaningfullear
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By Timothy EwestBA 447 EntrepreneurshipBATo Build a Fire:Lessons For The EntrepreneurLearning ObjectivesLearning1. To develop a comparative understanding of the need toenter into partnerships in business ventures.2. To cultivate an awareness of t
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Background and MotivationDespite their impact on the economy, relatively little is knownabout entrepreneurs' backgrounds and motivations. A studyreleased by the Kauffman Foundation in July, titled &quot;The Anatomyof an Entrepreneur,&quot; led by co-authors Viv
Wartburg - BA - 447
Chapter1and2Entreprenurship1TheIndividualCharacteristicsofanEntrepreneurResearchpointstoentrepreneurstypicallyhavingcertaincharacteristics Riskingtaking. Theyarenotbigrisktakers,rathermoderate,calculatedrisktakerswhodefinetherisksinvolvedwiththe
Wartburg - BA - 447
The Book and the Authors JOHN ABBOTTProf Chan Kim JOHN ABBOTTProf ReneeMauborgneAccolades Over 1 million copies sold in 2005 Translated into over 32 foreignlanguages a world record Taught as the major theory of strategyat leading business schoo
Wartburg - BA - 447
Dr.TimothyEwestWhatisit?Thesummationofthecorebusinessdecisionsandtradeoffsemployedbyacompanytoearnprofit.ThereareFOURgroups:1. RevenueSources.2. KeyExpenses.3. InvestmentSize.4. CriticalSuccessFactors.RevenueSources:Howmanydifferentrevenuestream
Wartburg - BA - 447
Option Price InputsCurrent PriceStrike PriceDiv YieldTime until expiration (% of year)Nominal Annual Risk-Free RatePeriodic Rate63265050.00%0.60%1%0.10%0.00%D1-0.537659N(d1)0.2954N(-d1)0.7046D2-0.587659N(d2)0.2784N(-d2)0.7216exp(
Wartburg - BA - 447
Stock Index Futures There are a number of futures contracts on stock marketindexes. Important ones include: The S&amp;P 500 The Dow Jones Industrial Average Because of the difficulty of actual delivery, stock indexfutures are usually cash settled. That
Wartburg - BA - 447
Spot-Future Parity From previous notes, spot-futures parity: In general:F0 = S0 (1+ rf)n For commodities with carrying costs:F0 = S0 (1+ rf + c)nwhere c is % carrying costs in decimals For stocksF0 = S0 (1+ rf - d)nwhere d is % dividend yield in
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xArbitrage Pricing TheoryModern FinancePortfoliosArbitrage Pricing TheoryxDeveloped as a way to get around restrictiveassumptions of CAPM and its dependence on singlefactor (market)focus on arbitrage and welldiversified portfolios (if Beta-return r
Wartburg - BA - 447
xIntroduction to Single Factor ModelsModern FinancePortfoliosQuick ReviewxrP = wB rB + wS rSRate of return on the portfolio:-Use for finding past return in individual year for a portfolioxExpected rate of return on the portfolio:E (r ) = w E (rB
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xThe Capital Asset Pricing ModelModern FinancePortfoliosCapital Asset Pricing Model (CAPM)xxxEquilibrium, single-factor model that underliesall modern financial theoryDerived using principles of diversification with(many) simplifying assumptions
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xDiversification and Efficient PortfoliosModern FinancePortfoliosPortfoliosxPortfolios are groups of assets, such as stocks andbonds, that are held by an investor.xOne convenient way to describe a portfolio is bylisting the proportion of the tota
Wartburg - BA - 447
xRisk and Return: The BasicsModern FinancePortfoliosCalculating ReturnsFor StocksPercentage Return=(Pt+1+Dt+1-Pt)/PtThis can be broken into:Dividend Yield = Dt+1 / PtCapital Gain/Loss = (Pt+1 - Pt)/ PtPercentage Return = Dividend Yield + Capital
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Pricing StrategiesPricing StrategiesChapter10:PricingCopyright 2008 Prentice Hall Publishing Company1Price Conveys Image Price sends important signals tocustomers quality, prestige,uniqueness, and others. Common small business mistake: Failureto
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GatheringCompetitiveIntelligenceCultivating Your CompetitiveIntelligence SourcesInternet Marketing Intelligence:Research Tools, Techniques, andResources (Paperback) , ~2003by Edward Forrestested beatelinDsi t esW ebelated&amp;riceser vfo-S
Wartburg - BA - 447
Models of e-BusinessTimothy EwestModels of e-Business What is an e-business model? A representation of the way a business operates (specifically the fundamentalcomponents or the primary form of commerce the business takes). Represents a change from
University of Toronto - IMM - 250
Todays Lecture, February 15th, 2012:Dr. Jen GommermanGrowing an adaptive immune systemSevere Combined Immunodeficiency Syndrome(SCID)Bubble BoyDavid Vetter poses inside of his bubble inhis Houston home in this Dec. 17, 1976 filephoto. Vetter was b
University of Toronto - IMM - 250
Volunteer note-taker requiredAccessibility Services requires your support to recruit a volunteer notetaker in IMM250H1 course as anaccommodation for a students documented disability. These notes are important for this students fullparticipation in your
University of Toronto - IMM - 250
When things go wrongDiseases and infections causedby innate immune mutation ordysfunctionDr. Dana PhilpottOutline of lecture When the innate immune system over-reacts: Sepsis SARS/H5N1 Mutations in the innate immune system impacting oninflammati
University of Toronto - IMM - 250
In review.Bruce Beutler: TLR4 Activated by LPSNormal mice die after being injectedwith hi doses of LPSA strain of mice, C3H/HeJ, were knownsince the 1960s to have defectiveresponses to LPS and surviveBeutlers group identifies a mutation inthe C3H/
University of Toronto - IMM - 250
Being a B cell: how antibodiesneutralize pathogensDr. Michael RatcliffeLecture outline1.2.3.4.5.6.7.8.9.An introduction to HIV/AIDSHIV as a retrovirusHIV life cycleInterfering with the HIV life cycleAntibody structure and interaction with
University of Toronto - IMM - 250
Once the barrier is breached how arepathogens detected and dealt with?Innate immune detection (this lecture)Cellular and humoral innate factors(next week)Innate immunity Innate denotes a property of some thingor action which is essential and specif
University of Toronto - IMM - 250
University of Toronto - IMM - 250
Todays Lecture, February 21 2012:Dr. Jen GommermanFine-tuning the troops: priming the T cell responseLecture Outline:-Review of T cells and how they are triggered by Antigen-Review of the TCR and its components-The multi-step process of T cell activ
University of Toronto - IMM - 250
Thean'genBcellreceptorrearrangementoccursinA.B.C.D.E.GerminalCenterCor'calareasoflymphnodes(LN)BonemarrowAllsecondarylymphoidorgansBothAandCAns.CTCRsignalsthroughA.B.C.D.E.CD28CD3CD4CD8BothTCRandchainsAns.BOneimportanteectorcytokin
University of Toronto - IMM - 250
pg. 1 UNIVERSITY OF TORONTOFaculty of Arts and ScienceGLG205Midterm Examination 2 for Spring 2011(Confronting Global Change)Duration 6:10pm to 8:00pmYou are permitted only a pencil, pen and an eraser!FORM AStudent Name:_Student ID#:_Instru
University of Toronto - IMM - 250
Todays Lecture, April 4th, 2012:Dr. Jen GommermanHorrorAutotoxicus: mechanisms of autoimmune diseaseEhrlich was first torecognize autoimmunityPaul EhrlichMore people affected in westerncountries than developing countries,most chronic and you live
University of Toronto - IMM - 250
MC Questions(One best answer)The rst production of attenuated chicken cholera vaccinewas achieved byA.B.C.D.E.I. MetchnikoffE. JennerL. PasteurR. KochP. EhrlichAns. CThe term variolation refers toA. The inoculation of individuals with cowp
University of Toronto - IMM - 250
MC Questions(One best answer)The rst production of attenuated chicken cholera vaccinewas achieved byA.B.C.D.E.I. MetchnikoffE. JennerL. PasteurR. KochP. EhrlichAns. CThe term variolation refers toA. The inoculation of individuals with cowp
University of Toronto - IMM - 250
TestI322374IDload::89IDstTeerwnTestID:2332DoDoloadnloDowwneradeID:r ID:89741. WhichofthefollowingstatementsaboutcommensalmicrofloraisNOTcorrectTestID:23324897r ID:adenloDowA.B.C.D.E.4897Ans.CProvidenut
University of Toronto - IMM - 250
TestI322374IDload::89IDstTeerwnTestID:2332DoDoloadnloDowwneradeID:r ID:89741. WhichofthefollowingstatementsaboutcommensalmicrofloraisNOTcorrectTestID:23324897r ID:adenloDowA.B.C.D.E.4897Ans.CProvidenut
University of Toronto - IMM - 250
IMM250 Term Test PrepLecture1. Overview/History of ImmunologyNameContributionThucydidesReported the concept of immunityZakariya RaziFirst diagnosed smallpox versus measlesDescribed allergic asthmaUnderstand that fever is a mechanism for fighting
University of Toronto - IMM - 250
UNIVERSITY OF TORONTOFaculty of Arts and ScienceMidterm Exam (v. 1), Spring 2012IMM250H1SWednesday, February 29th, 2012Duration - 2 hoursNo aids allowed. Please use pencil.After completing this exam,you may NOT take the exam paper with you.This t