Stat 8120 Spring 2011 Assignment 3 Solutions 1: Answer the first question in problem 13.18 on page 540 of the textbook. As we discussed in class, the F test statistics for testing the interaction effects are relatively straightforward. MSABC MSAB MSAC MSBC FABC , FAB , FAC and FBC MSE MSABC MSABC MSABC The test statistics for testing the main effects are more complicated. There are no simple ratios of mean squares that allow for testing the main effects. The approximate F test statistics for testing the main effects are MSA MSABC MSB MSABC MSC MSABC FA , FB , and FC MSAB MSAC MSAB MSBC MSAC MSBC Questions 2 through 5 all refer to the following experiment: The first four columns in the data set are the results of a factorial experiment run by Starbucks to identify the best setting to reduce leakage in the bags the coffee is packed in for retail sales. The experiment is real, but the data have been changed to avoid disclosing any proprietary information. The predictor variables are the viscosity of the adhesive used to seal the bag, the pressure at which the bag is sealed and the spacing (gap) of the sealing apparatus. The gap is measured in millimeters larger (positive) or smaller (negative) than the nominal setting. For this experiment, 20 bags are produced at each combination of settings and the bags are then tested for air leakage. The response variable is the proportion of the twenty bags that leaked. 2: Analyze the data and identify any significant effects. Also identify the best combination of viscosity, pressure and gap - the combination that minimizes the proportion of bags that leak. The ANOVA table below indicates the presence of a significant interaction between the viscosity and gap variables. There is no significant effect of pressure at all. The Tukey pairwise comparisons find the treatment combinations with the lowest mean leakage to be those with gap = 2 and viscosity equal to 320 or 380. Pressure can be set at either of its values with no significant effect. Analysis of Variance for Leakage, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Viscosity 1 0.002500 0.002500 0.002500 0.50 0.500 Pressure 1 0.010000 0.010000 0.010000 2.00 0.195 Gap 1 0.140625 0.140625 0.140625 28.13 0.001 Viscosity*Pressure 1 0.000625 0.000625 0.000625 0.12 0.733 Viscosity*Gap 1 0.090000 0.090000 0.090000 18.00 0.003 Pressure*Gap 1 0.010000 0.010000 0.010000 2.00 0.195 Viscosity*Pressure*Gap 1 0.005625 0.005625 0.005625 1.13 0.320 Error 8 0.040000 0.040000 0.005000 Total 15 0.299375 Grouping Information Using Tukey Method and 95.0% Confidence Viscosity Gap N Mean Grouping 380 -2 4 0.5 A 320 -2 4 0.3 B 320 2 4 0.2 B C 380 2 4 0.1 C The residual plots indicate reasonable Normality and one treatment combination with rather larger variance than the others. The treatment combination with the largest variability is the pressure = 170, viscosity = 380, gap = 2 combination. Because the viscosity = 380, gap = 2 combination is one of those that yields the lowest mean leakage value, I might suggest using the pressure = 190 for producing the coffee bags. It appears that even though pressure does not affect the mean value of leakage, pressure may affect the variability. 3: There had to be 320 total bags produced for the entire experiment (20 at each of the 16 combinations). Suppose it was impossible to produce all 320 bags in one day. Each replicate of the experiment was run on a different day identified by the variable “Day” in the data set. Analyze the data again taking into account the fact that the experiment took two days to run. Comment on whether it was important to account for the different days. Comment on whether the results differ from those in question 2. While the values of the sums of squares, F statistics and p-values changed slightly, there are no substantial differences between the results here and those of question 2. The F value for the blocking variable, Day, is 4.49, indicating that blocking was worthwhile. The Normal probability plot of the residuals shows the tails to be a bit longer than in question 2. The pressure = 170, viscosity = 380, gap = 2 results (circled on the graph) still have rather large variability, so I would suggest using pressure = 190 for producing the coffee bags. 4: Suppose the pressure settings in the original experiment had been chosen randomly from a large possible number of settings. Analyze the data again, estimating the variance components of all random effects. Comment on any differences between these results and those of question 2. The results are now quite different from those in question 2. The following ANOVA table shows there are no significant effects at all. Factor Type Levels Values Viscosity fixed 2 320, 380 Pressure random 2 170, 190 Gap fixed 2 -2, 2 Analysis of Variance for Leakage, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Viscosity 1 0.002500 0.002500 0.002500 4.00 0.295 Pressure 1 0.010000 0.010000 0.010000 2.00 0.707x Gap 1 0.140625 0.140625 0.140625 14.06 0.166 Viscosity*Pressure 1 0.000625 0.000625 0.000625 0.11 0.795 Viscosity*Gap 1 0.090000 0.090000 0.090000 16.00 0.156 Pressure*Gap 1 0.010000 0.010000 0.010000 1.78 0.410 Viscosity*Pressure*Gap 1 0.005625 0.005625 0.005625 1.13 0.320 Error 8 0.040000 0.040000 0.005000 Total 15 0.299375 x Not an exact F-test. The residual plots are identical to those in question 2. The reasons for the loss of significant of the viscosity*gap interaction term lies in the test statistic now that the model has a random effect. In question 2, the F test statistic for the viscosity*gap interaction was MSVG FVG with 8 degrees of freedom for the denominator. MSE In this question, the F test statistic for the viscosity*gap interaction is MSVG FVG with only 1 degree of freedom for the denominator. Not only have we lost MSVPG 7 degrees of freedom for the denominator, which equates to a substantial loss of power, but the denominator has increased from MSE=0.005 to MSVPG=0.005625. The table below shows that none of the random terms contribute much to the overall variability. Variance Components, using Adjusted SS Estimated Source Value Pressure 0.00063 Viscosity*Pressure -0.00125 Pressure*Gap 0.00109 Viscosity*Pressure*Gap 0.00031 Error 0.00500 5: Suppose that in the original experiment the gap had been hard to change. The engineers are willing to change the gap three times throughout the experiment (that is one original setting and three changes). Design the appropriate experiment and perform the analysis. Compare the results with those of question 2. The ANOVA table indicates a significant viscosity*gap interaction, as in question 2. Fixed Effect Tests Source DF DFDen F Ratio Prob > F Viscosity 1 6 0.8000 0.4055 Pressure 1 6 3.2000 0.1238 Gap 1 2 13.2353 0.0679 Viscosity*Pressure 1 6 0.2000 0.6704 Viscosity*Gap 1 6 28.8000 0.0017* Pressure*Gap 1 6 3.2000 0.1238 Viscosity*Pressure*Gap 1 6 1.8000 0.2283 The Fisher’s LSD pairwise tests have an unusual pattern. It appears the tests can distinguish a statistically significant difference between the means (actually the least squares means, i.e. the fitted means) of the 320, 2 and 380, 2 treatment combinations (with least squares means of 0.2375 and 0.1125), but not the 320, -2 and 380, 2 treatment combinations (with least squares means of 0.275 and 0.1125). Viscosity*Gap LSMeans Differences Student's t α=0.050 Level Least Sq Mean 380,-2 A 0.45000000 320,-2 B C 0.27500000 320,2 B 0.23750000 380,2 C 0.11250000 The reason for the unusual pattern can be seen in the pairwise tests that make up the Fisher’s LSD. In the comparisons below, we are first comparing across levels of the whole plot factor (from a gap = 2 setting to gap = -2 setting). Note the degrees of freedom of the t-test (3.255605). Because the whole plots are a random effect, this comparison is done with an approximate test. Comparing 320,2 with 320,-2 Difference -0.03750 t Ratio -0.6396 Std Err Dif 0.05863 DF 3.255605 Upper CL Dif 0.14107 Prob > |t| 0.5646 Lower CL Dif -0.21607 Prob > t 0.7177 Confidence 0.95 Prob < t 0.2823 The next test is done comparing treatment combinations within the same level of the whole plot factor (gap = -2). Note the degrees of freedom of the t-test (df = 6). This is an “exact” t-test. Comparing 380,-2 with 320,-2 Difference 0.175000 t Ratio 4.427189 Std Err Dif 0.039528 DF 6 Upper CL Dif 0.271723 Prob > |t| 0.0044* Lower CL Dif 0.078277 Prob > t 0.0022* Confidence 0.95 Prob < t 0.9978 The test that led to the unusual pattern in the Fisher’s LSD is a test done across levels of the whole pot factor (from gap = -2 to gap = 2). The reduced degrees of freedom of the test is a reflection of the fact that we imposed a restriction on the randomization of the whole plot factor. Because we did not allow the gap variable to be changed as often as it might have been under complete randomization, we have lost some power in tests across different levels of gap. Comparing 380,2 with 320,-2 Difference -0.16250 t Ratio -2.77161 Std Err Dif 0.05863 DF 3.255605 Upper CL Dif 0.01607 Prob > |t| 0.0633 Lower CL Dif -0.34107 Prob > t 0.9683 Confidence 0.95 Prob < t 0.0317* You can see the comparison above is “nearly” significant (p-value = 0.0633). If we are willing to be a little flexible with our level, we can decide that the 380, 2 combination yields results that are, on average, significantly different from the results of the 320, -2 combination. Our conclusion is that the 380, 2 combination of viscosity and gap gives the best overall result. Using the profiler, we can see the margin of error of the 95% confidence interval for the predicted leakage percentage to be ± 0.124, a rather large margin of error. The whole plots contribute 37.5% of the total variance. Random Effect Var Ratio Var Component Std Error 95% Lower 95% Upper Pct of Total Whole Plots 0.6 0.001875 0.0026943 -0.003406 0.0071558 37.500 Residual 0.003125 0.0018042 0.0012976 0.0151534 62.500 Total 0.005 100.000 The residual lots were much like those in question 2, including the larger variability for the pressure = 170, viscosity = 380, gap = 2 combination. 6: The next three columns in the data set are from an experiment to test how well the temperature of an industrial oven is controlled. The target temperature for the oven is 475° F. The experiment is run in four plants using four operators within each plant. For the purposes of this experiment, the operators are fixed effects. Analyze the data from this experiment and identify which plant(s), if any are doing a satisfactory job of maintaining a furnace temperature near the target of 475° F. The ANOVA table indicates a significant plant effect and a significant effect of the operators nested in the plants. Analysis of Variance for Temp, using Adjusted SS for Tests Source DF Seq SS Adj SS Adj MS F P Plant 3 731.52 731.52 243.84 13.74 0.000 Operator(Plant) 12 499.81 499.81 41.65 2.35 0.008 Error 176 3122.92 3122.92 17.74 Total 191 4354.24 The Tukey pairwise comparisons show that plants 1 and 4 are doing the best job of holding the mean temperature at 475° F. It could even be argued that plant 2 is not doing badly. Plant 3 is clearly the doing the worst job. Grouping Information Using Tukey Method and 95.0% Confidence Plant N Mean Grouping 2 48 477.3 A 4 48 475.8 A B 1 48 474.5 B 3 48 472.0 C

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