powerCT.default0 {powerSurvEpi} | R Documentation |
Power Calculation in the Analysis of Survival Data for Clinical Trials
Description
Power calculation for the Comparison of Survival Curves Between Two Groups under the Cox Proportional-Hazards Model for clinical trials.
Usage
powerCT.default0(k,
m,
RR,
alpha = 0.05)
Arguments
k |
numeric. ratio of participants in group E (experimental group) compared to group C (control group). |
m |
integer. expected total number of events over both groups. |
RR |
numeric. postulated hazard ratio. |
alpha |
numeric. type I error rate. |
Details
This is an implementation of the power calculation method described in Section 14.12 (page 807) of Rosner (2006). The method was proposed by Freedman (1982).
Suppose we want to compare the survival curves between an experimental group () and
a control group (
) in a clinical trial with a maximum follow-up of
years.
The Cox proportional hazards regression model is assumed to have the form:
Let be the number of participants in the
group
and
be the number of participants in the
group.
We wish to test the hypothesis
versus
not equal to 1,
where
underlying hazard ratio
for the
group versus the
group. Let
be the postulated hazard ratio,
be the significance level. Assume that the test is a two-sided test.
If the ratio of participants in group
E compared to group C
, then the power of the test is
where
is the
-th percentile of
the standard normal distribution
,
is the cumulative distribution function (CDF)
of
.
Value
The power of the test.
Note
The power formula assumes that the central-limit theorem is valid and hence is appropriate for large samples.
References
Freedman, L.S. (1982). Tables of the number of patients required in clinical trials using the log-rank test. Statistics in Medicine. 1: 121-129
Rosner B. (2006). Fundamentals of Biostatistics. (6-th edition). Thomson Brooks/Cole.
See Also
Examples
# Example 14.42 in Rosner B. Fundamentals of Biostatistics.
# (6-th edition). (2006) page 809
powerCT.default0(k = 1,
m = 171.9,
RR = 0.7,
alpha = 0.05)