ST4241: Design and Analysis of Clinical Trials - The Department of ...

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a2 a3 a4 a5 a6 1.750000 1.083333 3.333333 3.416667 1.416667 a2 a3 a4 a5 a6 a2 4.640741 2.320370 2.320370 2.320370 2.320370 a3 2.320370 4.640741 2.320370 2.320370 2.320370 a4 2.320370 2.320370 4.640741 2.320370 2.320370 a5 2.320370 2.320370 2.320370 4.640741 2.320370 a6 2.320370 2.320370 2.320370 2.320370 4.640741 By using the above information, the test statistics for the three contrasts are computed as L 1 = −0.6633, L 2 = −1.1010, L 3 = 0.0734. The Schiffe criterion should be used. The critical value is given by √ 5F 5,15,0.05 = 3.8087. None of the contrasts is significant. 3. The data form the study comparing four formulations considered in the lecture notes is given at the end of this question. Using R, fit a linear model to the data. Using the information in the fitted object, do the following: (i) By computing the means for each formulation, it is obvious form that the estimated effect of Formulation 3 is different from all the others. Check that the value of the contrast C = a 3 − (a 1 + a 2 + a 4 )/3 is −0.6489 with an estimated standard error 0.0883. Test the significance of the contrast at level 0.05 using Scheffe’s criterion. (Why should Scheffe’s criterion be used?) The following code is used to extract the information from the fitted object lm.fit and compute the required quantities: b=lm.fit$coef[14:16] v=vcov(lm.fit)[14:16,14:16] c1 = c(-1/3,1,-1/3) t(c1)%*%b sqrt( t(c1)%*%v%*%c1 ) L = t(c1)%*%b/sqrt( t(c1)%*%v%*%c1 ) The computation yields the given values for C and its standard error. The test statistic equals −7.35599. Its absolute value, compared with √ 3F 3,8,0.05 = 3.492641, is not significant. 6
(ii) Check that, of the six pairwise differences a j − a j ′, only the three differences involving Formulation 3 are statistically significant by the Tukey criterion. The following code is used for the computation: c12 = c(1,0,0) c13 = c(0,1,0) c14 = c(0,0,1) c23 = c(1,-1,0) c24 = c(1,0,-1) c34 = c(0,1,-1) C = rbind(c12,c13,c14,c23,c24,c34) l = C%*%b ss = sqrt( diag( C%*%v%*%t(C)) ) L = l/ss The computed test statistics corresponding the six pairs are: c12 0.76096856 c13 -5.73428051 c14 0.05461109 c23 6.49524907 c24 0.70635747 c34 -5.78889161 The critical value for the statistics using Tukey’s criterion is q 4,8,0.05 / √ 2 = 4.529/ √ 2 = 3.2025. It is seen that c13, c23, c34 are significant. 7
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