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

CHAPTER 6 ST 520, A. TSIATIS and D. Zhang
then we can find the value n which satisfies (6.3).
Consider the previous example of normally distributed response data where we use the t-test
to test for treatment differences in the mean response. If we randomize patients with equal
probability to the two treatments so that n1 = n2 ≈ n/2, then substituting (6.1) and (6.2) into
(6.3), we get
or
∆A
�
4
σY n
n 1/2 =
n =
� 1/2 = (Zα + Zβ),
� �
(Zα + Zβ)σY × 2
∆A
� (Zα + Zβ) 2 σ 2 Y
∆ 2 A
Note: For two-sided tests we use Zα/2 instead of Zα.
Example
�
× 4
.
Suppose we wanted to find the sample size necessary to detect a difference in mean response of
20 units between two treatments with 90% power using a t-test (two-sided) at the .05 level of
significance. We expect the population standard deviation of response σY to be about 60 units.
In this example α = .05, β = .10, ∆A = 20 and σY = 60. Also, Zα/2 = Z.025 = 1.96, and
Zβ = Z.10 = 1.28. Therefore,
n = (1.96 + 1.28)2 (60) 2 × 4
(20) 2
or about 189 patients per treatment group.
6.3 Comparing two response rates
≈ 378 (rounding up),
We will now consider the case where the primary outcome is a dichotomous response; i.e. each
patient either responds or doesn’t respond to treatment. Let π1 and π2 denote the population
response rates for treatments 1 and 2 respectively. Treatment difference is denoted by ∆ = π1−π2.
We wish to test the null hypothesis H0 : ∆ ≤ 0 (π1 ≤ π2) versus HA : ∆ > 0 (π1 > π2). In
some cases we may want to test the null hypothesis H0 : ∆ = 0 against the two-sided alternative
HA : ∆ �= 0.
PAGE 88