array chip wrote:
Dear R users, I am not asking questions specifically on R, but I know there are 
many statistical experts here in the R community, so here it goes my questions:

Freedman (1982) propose an approximation of sample size/power calculation based 
on log-rank test using the formula below (This is what nQuery does):
             (Z(1-α/side)+Z(power))^2*(hazard.ratio+1)^2
      N  =  ---------------------------------------------
                     (2-p1-p2)*(hazard.ratio-1)^2

Where Z is the standard normal cumulative distribution. p1 and p2 are the 
survival probability of the 2 groups at a given time, say t.

As you can see, the sample size depends on the survival probabilities, p1 and 
p2. This is where my question lies. Let’s say we have 2 survival curves. I can 
choose p1 and p2 at time 1 year, and calculate a sample size. I can also choose 
p1 and p2 at time 5 years (still the same hazard ratio since the same 2 
survival curves), and calculate a different sample size. How to interpret the 2 
estimates of sample size?

This problem doesn’t occur when we calculate the number of events required 
using this formula:
               4*( Z(α/side)+Z(power))^2
              --------------------------
                 (log(hazard.ratio))^2

Because number of events required only depends on hazard ratio.

Thanks for any suggestions.

John

As I recall, the survival probability used in Freedman is not at some arbitrary time of your choosing, but rather at the average length of follow-up time anticipated in the study.

Kevin

--
Kevin E. Thorpe
Biostatistician/Trialist, Knowledge Translation Program
Assistant Professor, Dalla Lana School of Public Health
University of Toronto
email: kevin.tho...@utoronto.ca  Tel: 416.864.5776  Fax: 416.864.3016

______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Reply via email to