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

CHAPTER 7 ST 520, A. TSIATIS and D. Zhang
Remark: In the special case where the response data are exactly normally distributed with equal
treatment-specific variances, the test statistic Tn given by (7.10), under the null hypothesis, has a
distribution which is exactly equal to K −1 times a central F distribution with K −1 numerator
degrees of freedom and n − K denominator degrees of freedom. That is
Tn/(K − 1) =
� Kj=1 nj( ¯ Yj − ¯ Y ) 2 /(K − 1)
has an F distribution under the null hypothesis. This is exactly the test statistic that is used
for testing the equality of means in a one-way ANOVA model.
However, when the sample size n is large, we can use the test statistic given by (7.9), based on an
asymptotic chi-square distribution, to test H0 without making either the normality assumption
or the assumption of equal variances.
7.5 Sample size computations for continuous response
Let us consider the case where patients are allocated equally to the K treatments so that
s 2 Y
n1 = . . . = nK = n/K,
and, for design purposes, we assume that the treatment specific variances are all equal which we
posit to be the value σ2 Y . The question is how do we compute the sample size that is necessary to
have power (1−β) to detect an alternative where any two treatments population mean responses
may differ by ∆A or more? Using considerations almost identical to those used for testing equality
of response rates for dichotomous outcomes, we can find the least favorable configuration which
after substituting into (7.12), yields the non-centrality parameter
n∆2 A
2Kσ2 Y
Hence, to obtain the desired power of (1 − β), we need
or
Example:
n∆2 A
2Kσ2 Y
. (7.13)
= φ 2 (α, β, K − 1),
n = 2Kσ2 Y φ2 (α, β, K − 1)
∆2 . (7.14)
A
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