[R-meta] Meta-Analysis: Proportion in overall survival rate

[Viechtbauer, Wolfgang (SP)]

Dear Nelly,

Your equation for the SE assumes that the p behaves like a 'regular' proportion computed from a binomial distribution. I am not sure if this is correct when using the Kaplan-Meier estimator to derive such a proportion.

As far as your input to rma() is concerned - that is correct. However, I would consider not meta-analyzing the proportions directly, but doing a logit transformation on p, so using qlogis(p) for yi and sqrt(1/(p*n) + 1/((1-p)*n)) for the SE.

Best,
Wolfgang

>-----Original Message-----
>From: R-sig-meta-analysis [mailto:r-sig-meta-analysis-bounces using r-project.org]
>On Behalf Of ne gic
>Sent: Wednesday, 27 May, 2020 20:02
>To: Dr. Gerta Rücker
>Cc: r-sig-meta-analysis using r-project.org
>Subject: Re: [R-meta] Meta-Analysis: Proportion in overall survival rate
>
>Dear Michael, Gerta and List,
>
>I would like to cross-check with you what I have done.
>
>I have restricted myself to Kaplan-Meier studies which gave the number at
>risk at 2 years, and also n_0 at baseline.
>
>I then estimated the absolute number of those surviving as *n_t *= n_0*S(t)
>following Gerta's idea. I took the reported proportions at 2 years to
>represent the S(t).
>
>I calculated the standard error (SE) using the formula: *se *= square root (
>*p*(1-*p*)/n). Where *p* = proportion at 2 years i.e. S(t)
>, n = *n_t*, the estimated number of of those surviving.
>
>I then used the random effects model in metafor as follows:
>rma(yi = *p*, sei = *se*, data=mydata, method="REML")
>
>The resulting estimate seems reasonable to me. But I want to confirm with
>you if this is the way one would input SE and the proportion to the
>function.
>
>Welcome any comments.
>
>Sincerely,
>nelly