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Unformatted text preview: STAT 4220 HW5 Solution 1 Problem 1 (a) See the appendix for R code.-15-10-5 5 10 15 4-3 4-2 3-2 4-1 3-1 2-1 99% family-wise confidence level Differences in mean levels of x We see that at the 99% confidence level, only treatments 3 and 4 are significantly different. (b) Recall in homework 3 that only treatments 3 and 4 were found to be different at the 99% level. This agrees exactly with what we have found here. R just finds the raw differences and generates the confidence intervals, as opposed to calculating the test statistics and comparing them to a critical value. (c) At the 95% level, we find that (3,1), (4,2), and (4,3) are all significantly different.-10-5 5 10 4-3 4-2 3-2 4-1 3-1 2-1 95% family-wise confidence level Differences in mean levels of x 1 2 Problem 2 (a) Fitting the model yields the following ANOVA table. > summary(aov(g)) Df Sum Sq Mean Sq F value Pr(>F) factor(treatment) 9 1.82549 0.202832 23.6429 < 2.2e-16 *** factor(block) 8 0.49958 0.062448 7.2791 4.902e-07 *** Residuals 72 0.61769 0.008579--- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 (b) The hypothesis under consideration is H : β j = 0, for all j , vs. H 1 : β j 6 = 0, for some j , where β k denotes the effect of the k th level of the blocking factor. With p ¿ . 0001, H is rejected. We find that blocking was, indeed, effective for this experiment. This is in contrast to what was found in homework 4, where the blocking factor was not found to be significant at all. (c) By careful inspection of the intervals, we can determine which treatment pairs are significantly different by finding the intervals that do not contain zero....
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This note was uploaded on 06/06/2011 for the course STAT 4220 taught by Professor Smith during the Spring '08 term at UGA.
- Spring '08