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**Unformatted text preview: **Stat 512 – 2 Solutions to Homework #8 Dr. Simonsen Due October 26, 2005 by 4:30pm The first three problems use the Filling Machines dataset from Problem 16.11 of KNNL described on page 725, and continue the analysis begun on Homework #7. 1. Use the Tukey multiple comparison method to determine which pairs of machines differ significantly. Summarize the results. The Tukey comparison method shows that machines 3 and 4 are significantly different from machines 1, 2, 5, and 6 (on a pairwise basis). The GLM Procedure Tukey's Studentized Range (HSD) Test for wtdev NOTE: This test controls the Type I experimentwise error rate, but it generally has a higher Type II error rate than REGWQ. Alpha 0.05 Error Degrees of Freedom 114 Error Mean Square 0.03097 Critical Value of Studentized Range 4.09949 Minimum Significant Difference 0.1613 Means with the same letter are not significantly different. Tukey Grouping Mean N machine A 0.46000 20 3 A A 0.36550 20 4 B 0.19050 20 2 B B 0.15150 20 6 B B 0.12500 20 5 B B 0.07350 20 1 2. Suppose you want to compare the average of the first two machines with the average of the last four. Use the ESTIMATE and CONTRAST statements in PROC GLM to test the appropriate hypothesis. Report the estimated value of this contrast with its standard error; state the null and alternative hypotheses, the test statistic with degrees of freedom, the P- value and your conclusion. The contrast is ( ) ( ) 3 4 5 6 1 2 2 4 L μ + μ + μ + μ μ + μ = − . The estimated value of this contrast is with standard error ˆ 0.1435 L = − { } ˆ 0.03408 s L = . We test the null hypothesis H : ( ) ( ) 3 4 5 6 1 2 2 4 μ + μ + μ + μ μ + μ − = vs H a : ( ) ( ) 3 4 5 6 1 2 2 4 μ + μ + μ + μ μ + μ − ≠ . [These could also be correctly written as H : L = 0 vs. Ha: L ≠ 0 or H : ( ) ( ) 3 4 5 6 1 2 2 4 μ + μ + μ + μ μ + μ = vs. Ha: ( ) ( ) 3 4 5 6 1 2 2 4 μ + μ + μ + μ μ + μ ≠ ]. The test statistic is either F = 17.73 with 1,114 df or t = –4.21 with 114 df, and the P-value is P < 0.0001. We reject H and conclude that the mean for the first two machines is not the same as the mean for the last four machines. The GLM Procedure Dependent Variable: wtdev Contrast DF Contrast SS Mean Square F Value Pr > F prob2 1 0.54912667 0.54912667 17.73 <.0001 Standard Parameter Estimate Error t Value Pr > |t| prob2 -0.14350000 0.03407905 -4.21 <.0001 3. Check assumptions using the residuals. Turn in the plots/output you used to check the assumptions and state your conclusions. resi d-0. 4-0. 3-0. 2-0. 1 0. 0 0. 1 0. 2 0. 3 0. 4 machi ne 1 2 3 4 5 6 The assumptions of normality and constant variance appear to be satisfied, as shown by the residual plot and qqplot. [Grader: the residual plot may be plotted vs. machine or vs. predicted value. Only one is necessary, plus the qqplot.] The remaining problems use the Helicopter Service dataset from Problem 18.15 on page 804 of KNNL. ...

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