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Unformatted text preview: Solutions to Assignment 2 Stat 512, Spring 2010 1. (a) Here is R code to enter the data and fit the model. > gain < c( + 4.1, 3.3, 3.1, 4.2, 3.6, 4.4, + 5.2, 4.8, 4.5, 6.8, 5.5, 6.2, + 6.3, 6.5, 7.2, 7.4, 7.8, 6.7, + 6.5, 6.8, 7.3, 7.5, 6.9, 7.0, + 9.5, 9.6, 9.2, 9.1, 9.8, 9.1) > > group < rep(c(1,2,3,4,5), each=6 ) > group < factor(group) > > result < lm( gain ~ group ) > summary(result) Call: lm(formula = gain ~ group) Residuals: Min 1Q Median 3Q Max1.0000 0.2958 0.0500 0.3917 1.3000 Coefficients: Estimate Std. Error t value Pr(>t) (Intercept) 3.7833 0.2293 16.501 5.97e15 *** group2 1.7167 0.3242 5.294 1.74e05 *** group3 3.2000 0.3242 9.869 4.18e10 *** group4 3.2167 0.3242 9.921 3.77e10 *** group5 5.6000 0.3242 17.271 2.09e15 *** Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 0.5616 on 25 degrees of freedom Multiple Rsquared: 0.9289, Adjusted Rsquared: 0.9175 Fstatistic: 81.67 on 4 and 25 DF, pvalue: 5.598e14 The omnibus Fstatistic (81.67 on 4 and 25 df) allows us to strongly reject the null hypothesis that all of the group means are equal. (b) The four contrasts are L 1 = 2 1 L 2 = 3 1 L 3 = 4 1 L 4 = 5 1 Placing the contrast weights into the rows of a matrix A =  1 1 0 0 0 1 0 1 0 0 1 0 0 1 0 1 0 0 0 1 and computing AA T = 2 1 1 1 1 2 1 1 1 1 2 1 1 1 1 2 , we see that these contrasts are not orthogonal. In the dummycoding scheme used in part (a), these contrasts correspond to the coefficients labeled group2 , group3 , group4 , and group5 . All of them are highly significant at the .05 level. Using a Bonferroni adjustment, we can maintain a familywise error rate of .05 by splitting .05 into four equal parts, .05/4=.0125, and declaring a contrast significant if its pvalue is less than .0125. All four of these are significant at that more stringent level. (c) The four contrasts are L 1 = 1 + 2 + 3 + 4 + 5 4 L 2 = 2 + 3 2 + 4 + 5 2 L 3 = 2 + 3 L 4 = 4 + 5 Placing the contrast weights into the rows of a matrix A =  1 1 / 4 1 / 4 1 / 4 1 / 4 1 / 2 1 / 2 1 / 2 1 / 2 1 1 1 1 and computing AA T = 5 / 4 0 0 0 0 1 0 0 0 0 2 0 0 0 0 2 , we see that they are orthogonal. We can test these contrasts in R by computing each L and test statistic manually. But here is a trick that makes it easier. Suppose we append a row of equal weights 1/5 to...
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This note was uploaded on 01/19/2012 for the course STAT 512 taught by Professor John during the Spring '11 term at Peru State.
 Spring '11
 John

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