4
$\begingroup$

Suppose one has 2 placebo-controlled clinical drug trials (drug X vs placebo; and drug Y vs placebo).

And suppose one creates a Kaplan-Meier plot for time to some event (eg, disease progression) for drug X vs placebo, and one creates another similar K-M plot for drug Y vs placebo.

Would it then be valid to compare the 2 K-M graphs (at least qualitatively, by eye) and say: drug X seems to separate from placebo faster than drug Y separates from placebo, so drug X seems to have a more rapid effect?

I think something like this is sometimes done in the literature, but I've seen a paper which stated that one cannot make such inferences from Kaplan-Meier graphs.

$\endgroup$
2
  • $\begingroup$ Is there censoring (dropout, other disease)? Do you assume that it is the same for both trials? $\endgroup$
    – martin
    Commented Aug 7, 2013 at 9:28
  • $\begingroup$ Yes, I'm envisaging a situation with censoring, since that's quite common in such clinical trials. And the censoring would not be the same in each study arm $\endgroup$
    – user2503
    Commented Aug 8, 2013 at 15:05

2 Answers 2

1
$\begingroup$

It's fairly common for authors to notice and comment in the text of an article that the median time to failure or relapse is a particular interval. Of course, not all studies get to the point where the KM curves cross the 0.50 line. And the usual statistical tests are based on the number of events to a single duration, so that is really comparing differences in the "vertical direction" rather than across the horizontal (time) dimension.

$\endgroup$
1
$\begingroup$

In general this is an unsafe practice. What would be needed is a simultaneous confidence interval for the difference in two survival curves. It is easy to get non-simultaneous (pointwise) confidence bands (e.g., in R rms package survdiffplot function). An example is in http://biostat.mc.vanderbilt.edu/wiki/pub/Main/StatGraphCourse/graphscourse.pdf p. 15.

To approximate a simultaneous 0.95 confidence band you might try using pointwise 0.99 confidence limits.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.