Power calculation testing interaction effect for Cox proportional hazards regression with two covariates for Epidemiological Studies. Both covariates should be binary variables. The formula takes into account the correlation between the two covariates.
powerEpiInt.default0(n,
theta,
p,
psi,
G,
rho2,
alpha = 0.05)The power of the test.
integer. total number of subjects.
numeric. postulated hazard ratio.
numeric. proportion of subjects taking the value one for the covariate of interest.
numeric. proportion of subjects died of the disease of interest.
numeric. a factor adjusting the sample size. The sample size needed to
detect an effect of a prognostic factor with given error probabilities has
to be multiplied by the factor G when an interaction of the
same magnitude is to be detected.
numeric. square of the correlation between the covariate of interest and the other covariate.
numeric. type I error rate.
This is an implementation of the power calculation formula derived by Schmoor et al. (2000) for the following Cox proportional hazards regression in the epidemiological studies: $$h(t|x_1, x_2)=h_0(t)\exp(\beta_1 x_1+\beta_2 x_2 + \gamma (x_1 x_2)),$$ where both covariates \(X_1\) and \(X_2\) are binary variables.
Suppose we want to check if the hazard ratio of the interaction effect \(X_1 X_2=1\) to \(X_1 X_2=0\) is equal to \(1\) or is equal to \(\exp(\gamma)=\theta\). Given the type I error rate \(\alpha\) for a two-sided test, the power required to detect a hazard ratio as small as \(\exp(\gamma)=\theta\) is $$power=\Phi\left(-z_{1-\alpha/2}+\sqrt{\frac{n}{G}[\log(\theta)]^2 p (1-p) \psi (1-\rho^2)}\right),$$ where \(z_{a}\) is the \(100 a\)-th percentile of the standard normal distribution, \(\psi\) is the proportion of subjects died of the disease of interest, and $$\rho=corr(X_1, X_2)=(p_1-p_0)\times \sqrt{\frac{q(1-q)}{p(1-p)}},$$ and \(p=Pr(X_1=1)\), \(q=Pr(X_2=1)\), \(p_0=Pr(X_1=1|X_2=0)\), and \(p_1=Pr(X_1=1 | X_2=1)\), and $$G=\frac{[(1-q)(1-p_0)p_0+q(1-p_1)p_1]^2}{(1-q)q (1-p_0)p_0 (1-p_1) p_1}.$$
If \(X_1\) and \(X_2\) are uncorrelated, we have \(p_0=p_1=p\) leading to \(1/[(1-q)q]\). For \(q=0.5\), we have \(G=4\).
Schmoor C., Sauerbrei W., and Schumacher M. (2000). Sample size considerations for the evaluation of prognostic factors in survival analysis. Statistics in Medicine. 19:441-452.
powerEpiInt.default1, powerEpiInt2
# Example at the end of Section 4 of Schmoor et al. (2000).
powerEpiInt.default0(n = 184,
theta = 3,
p = 0.61,
psi = 139 / 184,
G = 4.79177,
rho2 = 0.015^2,
alpha = 0.05)
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