# hw4a - Solutions to Assignment 4 Stat 512 Spring 2010 1(a...

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Unformatted text preview: Solutions to Assignment 4 Stat 512, Spring 2010 1. (a) No, it doesn’t. The old display was used during the first three-week period, and the new display (Treatment 1) was used during the second three-week period. Even if the average change is statistically significant, we do not know if the effect is due to the change in display, a trend over time, or some other variables that have not been measured. In this experiment, Treatments 1, 2 and 3 were randomly assigned, so these data can be used to make causal comparisons among Treatments 1, 2 and 3. But the old display was not randomly assigned, and so we cannot include the old display in our causal comparisons. (b) > # read in data and extract variables > Drugstore <- read.table("Drugstore.txt", header=T) > treatment <- factor(Drugstore\$treatment) > contrasts(treatment) <- contr.sum > post <- Drugstore\$post > pre <- Drugstore\$pre > # perform ANOVA > result <- lm( post ~ treatment ) > anova(result) Analysis of Variance Table Response: post Df Sum Sq Mean Sq F value Pr(>F) treatment 2 1524.40 762.20 6.6899 0.01117 * Residuals 12 1367.20 113.93--- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Yes, the treatment effect is significant. > # define contrast weights > c1 <- c(-1,1,0) > c2 <- c(-1,0,1) > c3 <- c(0,-1,1) > # compute estimated contrasts > Ybari <- tapply( post, treatment, mean) > Lhat1 <- sum(c1*Ybari) > Lhat2 <- sum(c2*Ybari) > Lhat3 <- sum(c3*Ybari) > # compute SE’s > MSE <- sum( result\$residuals^2 ) / result\$df.residual > a <- 3 > n <- 5 > SE1 <- sqrt( MSE * sum(c1^2)/n ) > SE2 <- sqrt( MSE * sum(c2^2)/n ) > SE3 <- sqrt( MSE * sum(c3^2)/n ) > # compute t-statistics and F’s > T1 <- Lhat1 / SE1 > T2 <- Lhat2 / SE2 > T3 <- Lhat3 / SE3 > F1 <- T1^2 > F2 <- T2^2 > F3 <- T3^2 > # compute pvals > p1 <- 1 - pf(F1, 1, 12) > p2 <- 1 - pf(F2, 1, 12) > p3 <- 1 - pf(F3, 1, 12) > # create table > result <- cbind( + Lhat = c(Lhat1, Lhat2, Lhat3), + SE = c(SE1, SE2, SE3), + T = c(T1, T2, T3), + F = c(F1, F2, F3), + p = c(p1, p2, p3)) > rownames( result ) <- c( "2 versus 1", "3 versus 1", "3 versus 2" ) > result Lhat SE T F p 2 versus 1 23.8 6.750802 3.5255068 12.4291984 0.00418084 3 versus 1 17.6 6.750802 2.6070975 6.7969573 0.02292398 3 versus 2 -6.2 6.750802 -0.9184093 0.8434757 0.37649542 Treatment 2 is significantly different from 1, and 3 is significantly different from 1, but 3 is not significantly different from 2. (c) First, fit the unequal slopes model to see if the slopes are different. > # center the covariate > preC <- pre - mean(pre) > # unequal slopes model > result <- lm( post ~ preC + treatment + preC:treatment) > anova(result) Analysis of Variance Table Response: post Df Sum Sq Mean Sq F value Pr(>F) preC 1 1317.79 1317.79 81.6808 8.251e-06 *** treatment 2 1397.28 698.64 43.3040 2.409e-05 *** preC:treatment 2 31.33 15.66 0.9709 0.4151 Residuals 9 145.20 16.13--- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 The interactions are not significant, so there is no convincing evidence that the slopes...
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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.

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hw4a - Solutions to Assignment 4 Stat 512 Spring 2010 1(a...

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