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6166 UNIT 3 Section 1 Answers

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From problem 9.24,page 464 in Ott and Longnecker. Researchers conducted a study of the effects of three heartworm drugs on the fat content of the shoulder muscles in Labrador retrievers. Twenty (20) dogs were randomly divided into four treatment groups of five (5) dogs each. Dogs in group A were assigned to the untreated control. The remaining three groups identify which of the heartworm drug all dogs in the group received. At the end of two years of treatment, the percentage fat content of the shoulder muscles was determined and is given here.
A B C D 2.64 3.24 2.90 2.91 2.56 3.71 3.02 2.89 3.30 2.95 3.78 3.21 2.19 3.01 2.96 2.89 2.45 3.08 2.87 2.68
Answers:
1. y_{ij} = m + a_{i} + e_{ij}
2. This problem has analysis that follows closely that described in the Engineering example. Look at that problem for a detailed writeup. The SAS program for this analysis can be found here. The full output can be found here. Here I will only provide the final tables and graphs.
Dependent Variable: fat Sum of Source DF Squares Mean Square F Value Pr > F Model 3 0.95052000 0.31684000 2.85 0.0703 Error 16 1.77920000 0.11120000 Corrected Total 19 2.72972000 Level of fat treatment N Mean Std Dev A 5 2.62800000 0.41227418 B 5 3.19800000 0.30605555 C 5 3.10600000 0.38115614 D 5 2.91600000 0.18942017 Bartlett's Test for Homogeneity of fat Variance Source DF ChiSquare Pr > ChiSq treatment 3 2.2137 0.5293 Brown and Forsythe's Test for Homogeneity of fat Variance ANOVA of Absolute Deviations from Group Medians Sum of Mean Source DF Squares Square F Value Pr > F treatment 3 0.0579 0.0193 0.26 0.8513 Error 16 1.1752 0.0734 Tests for Normality Test Statistic p Value ShapiroWilk W 0.840287 Pr < W 0.004 KolmogorovSmirnov D 0.245416 Pr > D <0.010 Cramervon Mises WSq 0.240502 Pr > WSq <0.005 AndersonDarling ASq 1.351514 Pr > ASq <0.005 KruskalWallis one way analysis of variance for 2year dog fat data for Problem 9.24 The NPAR1WAY Procedure Wilcoxon Scores (Rank Sums) for Variable fat Classified by Variable treatment Sum of Expected Std Dev Mean treatment N Scores Under H0 Under H0 Score  A 5 28.0 52.50 11.452131 5.60 B 5 75.0 52.50 11.452131 15.00 C 5 61.0 52.50 11.452131 12.20 D 5 46.0 52.50 11.452131 9.20 Average scores were used for ties. KruskalWallis Test ChiSquare 6.9824 DF 3 Pr > ChiSquare 0.0725
Figure 1. Histogram of residuals with overlay of normal density curve.
Figure 2. Normal probability plot of residuals.
The formal normality test and the probability plot suggest that possibly these data could benifit from a transformation. But what transformation? Plotting the treatment sample means versus the sample variances or even standard deviations suggests that none of the standard transformation would work.
Figure 3. Plot of treatment sample means versus variances.
Since the tests of variances do not reject the null hypothesis of equal variances and there seems no obvious way to transform the data that would simultaneously maintain homogeneity while providing for more normally distributed residuals, we will tentatively accept the analysis of variance on the untransformed observations and conclude that there are no treatment differences (pvalue = 0.07 which is greater than our a=0.05 typical test level.)
We compute the nonparameteric KruskalWallis test statistics as a further check on our decision. The results of the test suggest further that there are no statistically significant differences among the treatments (pvalue = 0.0725).