- Mark as New
- Bookmark
- Subscribe
- Mute
- RSS Feed
- Permalink
- Report Inappropriate Content
I have been given a hazard ratio = 0.64 and a control median of 8 months. How can I use that information to estimate the treatment median survival time at 80% power and show 38% improvement in hazard ratios?
- Mark as New
- Bookmark
- Subscribe
- Mute
- RSS Feed
- Permalink
- Report Inappropriate Content
Check out PROC POWER, and look at the syntax for TWOSAMPLESURVIVAL. I am pretty sure that you just need to plug in the values, and exponentiate the hazard rate for the reference group. There is an example 70.6 in the SAS/STAT 9.3 documentation to get started.
Steve Denham
- Mark as New
- Bookmark
- Subscribe
- Mute
- RSS Feed
- Permalink
- Report Inappropriate Content
I don't think that you can do both of those from the little information you have provided. See
http://www.statsdirect.com/help/sample_size/sample_size_for_survival_analysis.htm
for an easy to use survival analysis sample size computation.
Given the ratio and one rate, you can compute the sample size required for 80% power.
You can compute a 38% "improvement in hazard ratios" (though it is not to me well defined) to get a new HR, and then do the computation above.
I believe the SAS Power and Sample Size product can do the can computations.
Doc Muhlbaier
Duke
- Mark as New
- Bookmark
- Subscribe
- Mute
- RSS Feed
- Permalink
- Report Inappropriate Content
I guess I should have specified that we already have a sample size of 200. I hope that clears up a little bit.
- Mark as New
- Bookmark
- Subscribe
- Mute
- RSS Feed
- Permalink
- Report Inappropriate Content
I guess I don't understand the question. If you have a known sample size and known power, do you wish to solve for the minimum detectable difference? You have specified what the difference is, so that seems out. More info would help here.
Thanks,
Steve Denham