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Unformatted text preview: Department of Statistics STATS351: Design and Analysis of Experiments Model Answers, Assignment 5, Semester 2 2008 1. [ 15 marks ] (a) [ 5 marks ] i. In R: > site<-factor(rep(c("I","II","III","IV"),4)) > day<-factor(rep(1:4,c(4,4,4,4))) > inhibitor<-factor(LETTERS[c(1:4,4,1:3,3,4,1,2,2:4,1)]) > counts<-c(193,522,235,103,40,88,386,83,19,26,82,254,130,35,58,124) > insect.df<-data.frame(site,day,inhibitor,counts) > insect.aov<-aov(log(counts)~inhibitor+Error(site*day)) > summary(insect.aov) Error: site Df Sum Sq Mean Sq F value Pr(>F) Residuals 3 1.65592 0.55197 Error: day Df Sum Sq Mean Sq F value Pr(>F) Residuals 3 4.1585 1.3862 Error: site:day Df Sum Sq Mean Sq F value Pr(>F) inhibitor 3 7.4524 2.4841 44.269 0.0001738 *** Residuals 6 0.3367 0.0561 ii. There is very strong evidence that expected numbers of moths caught is not the same for all of the different treatments. iii. The ANOVA table indicates that blocking on both site and day was sensible as the Residual Mean Square for both the site stratum and the day stratum are much larger than that for the site:day stratum. (b) [ 5 marks ] The output for model.tables is: > model.tables(insect.aov,type="means") Tables of means Grand mean 4.60036 inhibitor inhibitor A B C D 4.742 5.655 4.095 3.911 > lsd<-qt(.975,6)* sqrt(2*.0561/4) > lsd  0.4098121 > tsr<-qtukey(.95,4,6)/sqrt(2)*sqrt(2*.0561/4) > tsr  0.5797721 In this case we arrive at exactly the same conclusions whether we use the LSD or the...
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- Spring '09