ST 520 Statistical Principles of Clinical Trials - NCSU Statistics ...

CHAPTER 4 ST 520, A. TSIATIS and D. Zhang
is denoted by µ2. The n1 and n2 patients that are to be randomized to treatments 1 and 2 are
thought of as independent random samples chosen from this hypothetical super-population.
An estimator for the treatment difference is given by
¯Y2 − ¯ Y1,
where ¯ Y1 is the sample average response of the n1 patients assigned treatment 1 and ¯ Y2 is the
sample average response of the n2 patients assigned treatment 2. Let us also assume that the
population variance of response is the same for the two treatments; i.e. σ 2 1 = σ 2 2 = σ 2 . This may
be reasonable if we don’t have any a-priori reason to think these variances are different. The
variance of our estimator is given as
var( ¯ Y2 − ¯ Y1) = var( ¯ Y2) + var( ¯ Y1) = σ 2
�
1
n1
+ 1
�
.
n2
Question: Subject to the constraint that n = n1+n2, how do we find the most efficient estimator
for µ2 − µ1? That is, what treatment allocation minimizes the variance of our estimator? i.e.
σ 2
�
1
+
n1
1
�
= σ
n2
2
�
1
nπ +
�
1
=
n(1 − π)
σ2
� �
1
.
n π(1 − π)
The answer is the value of 0 ≤ π ≤ 1 which maximizes π(1 −π) = π −π 2 . Using simple calculus,
we take the derivative with respect to π and set it equal to zero to find the maximum. Namely,
d(π − π 2 )
dπ
= 1 − 2π = 0; π = .5.
This is guaranteed to be a maximum since the second derivative
d 2 (π − π 2 )
dπ 2
= −2.
Thus, to get the most efficient answer we should randomize with equal probability. However, as
we will now demonstrate, the loss of efficiency is not that great when we use unequal allocation.
For example, if we randomize with probability 2/3, then the variance of our estimator would be
σ2 � �
1
=
n (2/3)(1/3)
4.5σ2
n
instead of
σ 2
n
�
1
(1/2)(1/2)
�
PAGE 54
= 4σ2
n ,