powerSurvEpi (version 0.1.3)

powerEpiInt.default0: Power Calculation Testing Interaction Effect for Cox Proportional Hazards Regression

Description

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.

Usage

powerEpiInt.default0(n, 
		     theta, 
		     p, 
		     psi, 
		     G, 
		     rho2, 
		     alpha = 0.05)

Arguments

n

integer. total number of subjects.

theta

numeric. postulated hazard ratio.

p

numeric. proportion of subjects taking the value one for the covariate of interest.

psi

numeric. proportion of subjects died of the disease of interest.

G

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.

rho2

numeric. square of the correlation between the covariate of interest and the other covariate.

alpha

numeric. type I error rate.

Value

The power of the test.

Details

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 $$power=\mathrm{\Phi }(-z_{1-\alpha /2}+\sqrt{\frac{n}{G}[\log (\theta ){]}^{2}p(1-p)\psi (1-{\rho }^2)}),$$ where $z_a$ is the $100a$-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 $G=4$.

References

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.

See Also

powerEpiInt.default1, powerEpiInt2

Examples

# NOT RUN {
  # 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)
  
# }