4
$\begingroup$

I used the "survival" package in R to calculate a Kaplan Meier estimate for survival.

An example of my output is like this:

Call:
survdiff(formula = Surv(dat$Survival, dat$Event) ~ dat$Allele, 
    rho = 1.5)

                   N Observed Expected (O-E)^2/E (O-E)^2/V
dat$Allele=Variant 8     3.25     4.11     0.181     0.639
dat$Allele=WTHomo  8     4.45     3.59     0.207     0.639

I don't think this particular test is valid, given the small number of individuals in the test. Should the expected values both be >5 for the test to be statistically sound? Or is based on number of observed values...Should that be >5? If anyone had a reference or direction I could look in, I would appreciate it.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

You are asking two different questions. The power of a 2-sample logrank test depends on the total number of events. Typically the number should be in the hundreds or greater. For estimating the entire single survival curve using the Kaplan-Meier estimator, depending on the timing of censored observations, the number of subjects required to reach satisfactory precision in the estimate is about 184.

Improve this answer
$\endgroup$
10
  • $\begingroup$ Thank you, I understand. Re:power of the 2-sample log rank test. I have heard that one way I could work out if my logrank test is valid is to do a "power calculation". I have found a VERY basic example of such a calculation: statstodo.com/SSizSurvival_Pgm.php (the section: Single calculation; power estimation). I assume that in the example, the survival rate for group 1 is the final survival rate. I looked for a similar program, in R/python. I came across your cpower in Hmisc.... $\endgroup$
    – user1288515
    Commented May 24, 2016 at 9:34
  • $\begingroup$ ....But I'm not sure this is suitable, since the aim is not to "allow estimation of the expected number of events in the two groups". Would you know of a way to statistically calculate whether the log rank test has enough power? The example of code I put up is just an extremely small example to illustrate my point, I do have other data sets with more data. $\endgroup$
    – user1288515
    Commented May 24, 2016 at 9:34
  • $\begingroup$ The expected number of events would require every subject to have the same follow-up time unless you use a parametric model, and would require a minimum of 96 subjects. Estimation of the expected number of events is not the goal of the log rank test. But sometimes for study planning (as with cpower and spower functions in R Hmisc) we estimate event probabilities at a fixed time using published or historical data to plug into the power calculations/simulations. $\endgroup$
    – Frank Harrell
    Commented May 24, 2016 at 12:14
  • $\begingroup$ Thank you. I should not be focussing on the expected values to decide if the dataset is big enough. I apologise that I sound confused, I'm a biologist.I did a survival analysis using different genotypes, but I want to make sure my test is statistically accurate, and there is enough data for the test to give "real" results... $\endgroup$
    – user1288515
    Commented May 24, 2016 at 13:01
  • $\begingroup$ So my question would be, if a reviewer came to you and said "you did this kaplan meier analysis, but you have not many individuals in the data set, how can you be sure that the data set was big enough for the log rank test to accurately calculate survival differences between two groups when you say that the difference in survival is statistically significant between two groups"? This is the question that I am trying to anticipate; and come up with a calculation/software that I can use to answer this. I appreciate your help. $\endgroup$
    – user1288515
    Commented May 24, 2016 at 13:02

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.