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

ST4241: Design and Analysis of Clinical Trials
2012/2013: Semester I
Tutorial 8
1. The linear model for the balanced incomplete block design with design parameters g, k, r
and λ is as follows:
X ij = µ + s i + α j + ɛ ij ,
where ∑ j α j = 0, s i ’s are i.i.d. random variables with mean zero and variance σ 2 s, ɛ ij ’s are
i.i.d. random errors with mean zero and variance σ 2 e and are independent of s i ’s. Let a j be the
unbiased estimate of α j as defined in the lecture notes.
(i) Show that the a j ’s are least squares estimates of α j ’s; that is they are minimizers of
∑ ∑
(X ij − µ − s i − α j ) 2 , subject to ∑
i j
j
α j = 0.
By Lagrange method, taking derivatives w.r.t. the unknowns of the function and setting
them to zero
∑ ∑
(X ij − µ − s i − α j ) 2 + κ ∑ α j
i j
j
yields the following equations:
(1)
(2)
(3)
(4)
X·· = grµ + k ∑ s i
X i· = k(µ + s i ) + ∑ α j , i = 1, . . . , n
j∈T i
X·j = r(µ + α j ) + ∑ s i + κ, j = 1, . . . , g
i∈S j
∑
α j = 0
j
where T i is the set of j’s such that Treatment j is applied in Block i, S j is the set of i’s
such that Block i receives Treatment j.
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Sum up (3) over j yields
(5) X·· = grµ + k ∑ s i + gκ
(5)-(1) yields that κ = 0. From (2) and (3), we have
(6)
(7)
µ + s i = ¯X i· − 1 k
∑
j ′ ∈T i
α j
′
X·j = rα j + ∑ i∈S j
(µ + s i )
Substituting (6) into (7) yields
It yields
(ii) Show that
X·j = rα j + ∑ ( ¯X i· − 1 ∑
α
k j
′)
i∈S j j ′ ∈T i
= rα j + rM j − 1 ∑ ∑
k
α j
′
i∈S j j ′ ∈T i
= rα j + rM j − r − λ
k α j.
(1 − r − λ
kr )α j = ¯X·j − M j , i.e., α j = 1
eff ( ¯X·j − M j ).
Var(a j ) =
where eff = g(k−1)
k(g−1) .
σ2 e
reff
g − 1
, Cov(a j , a k ) = − σ2 e
g
1
reff g ,
Write
¯X·j − M j = 1 ∑
(X ij −
r
¯X i·)
i∈S j
= 1 ∑
(α j − 1 ∑
α
r k j
′ + ɛ ij − 1 ∑
ɛ
k ij
′).
i∈S j j ′ ∈T i j ′ ∈T i
2

Thus
Var( ¯X·j − M j ) = 1 ∑
Var(ɛ
r 2 ij − 1 ∑
ɛ
k ij
′)
i∈S j j ′ ∈T i
= 1 ∑
[(1 − 1 r 2 k )2 + k − 1 ]σ
k 2 e
2
i∈S j
= k − 1
rk σ2 e.
Hence
Var(a j ) = k − 1
rkeff 2 σ2 e =
σ2 e k − 1
reff keff =
σ2 e
reff
g − 1
.
g
Similarly,
Cov( ¯X·j − M j , ¯X·l − M l )
= 1 ∑ ∑
Cov(ɛ
r 2 ij − 1 ∑
ɛ
k ij
′, ɛ i ′ l − 1 ∑
ɛ ′
k i j
′)
i∈S j i ′ ∈S l j ′ ∈T i j ′ ∈T ′ i
= 1 r λCov(ɛ 2 ij − 1 ∑
ɛ
k ij
′, ɛ il − 1 ∑
ɛ
k ij
′)
j ′ ∈T i j ′ ∈T i
= λ r 2 (−σ2 e
k ).
Hence
Cov(a j , a l ) =
λ
r 2 eff 2 (−σ2 e
k ) = − 1 σe
2
reff g .
(iii) Show that, for any contrast ∑ g
j=1 c ja j ,
Var(
g∑
c j a j ) =
j=1
σ2 e
reff
∑
c 2 j.
j
3

g∑
Var( c j a j )
j=1
= ∑ j
= ∑ j
= ∑ j
∑
c j c l Cov(a j , a l )
c 2 j
c 2 j
l
σ 2
reff
g − 1
g
− ∑ j≠l
σ 2
reff [g − 1 + 1 g g ] = ∑ j
1 σe
2 c j c l
reff g
c 2 j
σ 2
reff .
Note that
0 = ( ∑ j
c j ) 2 = ∑ j
c 2 j + ∑ j≠l
c j c l .
2. The data from the interexaminer reliability study considered in the lecture notes is given on
the next page.
(i) Using the linear model approach, derive the F ratio for testing the significance of the
differences among the examiners.
Examiner
Patient 1 2 3 4 5 6 Mean
1 10 14 10 11.33
2 3 3 1 2.33
3 7 12 9 9.33
4 3 8 5 5.33
5 20 26 20 22.00
6 20 14 20 18.00
7 5 8 14 9.00
8 14 18 15 15.67
9 12 17 12 13.67
10 18 19 13 16.67
Mean 8.6 11.2 13.2 10.6 16.2 14.2 12.33
The data is fitted to the following model
∑10
6∑
X = µ 0 + s i b i + a j t j + ɛ,
i=2 j=2
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where the dummy variables b i ’s and t j ’s are defined conventionally.
The R code
lm.fit1=lm(x~s+a)
produces the following ANOVA table:
Df Sum Sq Mean Sq F value Pr(>F)
s 9 982.00 109.11 11.7558 2.615e-05 ***
a 5 35.44 7.09 0.7638 0.5898
Residuals 15 139.22 9.28
The F ratio for the examiner effect is 0.7638 with a p-value 0.5898. There is no significant
effect of the examiners.
(ii) By looking at the data, it seems that Examiner 5 and 6 have higher mean scores than the
other examiners. One would like to see whether the following contrasts are significant:
α 1 + α 2 + α 3 + α 4
4
α 1 + α 2 + α 3 + α 4
4
α 1 + α 2 + α 3 + α 4
4
− α 5 + α 6
,
2
− α 5 ,
− α 6 .
Using an appropriate multiple comparison criterion, test the significance of the above
contrasts by controlling the overall error rate at 0.05.
In terms of the parameters of the linear model, the contrasts are equivalent to the linear
forms:
a 2 + a 3 + a 4
4
a 2 + a 3 + a 4
4
a 2 + a 3 + a 4
4
− a 5 + a 6
,
2
− a 5 ,
− a 6 .
The estimated a j ’s and their variance matrix are as follows:
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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.
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(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.
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Formulation
Patient 1 2 3 4
1 -1.0894 (A) -1.3200 (B)
2 -1.7577 (B) -0.9817 (A)
3 -1.0771 (B) -1.7531 (A)
4 -0.9381 (A) -1.6769 (B)
5 -1.2044 (B) -0.7795 (A)
6 -1.0395 (B) -1.0426 (A)
7 -1.0991 (B) -0.8092 (A)
8 -2.0245 (A) -1.3374 (B)
9 -0.9846 (A) -1.4712 (B)
10 -1.1395 (B) -1.6683 (A)
11 -0.8069 (A) -1.1913 (B)
12 -0.7789 (A) -1.1694 (B)
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