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

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.
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