[R-meta] Meta-Analysis: Proportion in overall survival rate
ken @end|ng |rom kjbe@th@com@@u
Thu May 28 13:17:11 CEST 2020
It won’t be correct. To calculate CI for Kaplan-Meier what is usually done is to use log(-log(S)) and there is then a formula for the SE of this which means that 95% CI can be calculated. If you have 95% CI calculated this way then it should be possible to work back to log(-log(S) and SE which can then be used for a meta-analysis. What also is sometimes used is Greenwoods estimate of the SE, which has the problem that it doesn’t guarantee that confidence intervals are constrained between 0 and 1. Most modern programs don’t use it except to give a SE on the survival but use log(-log(S)) for confidence intervals.
> On 28 May 2020, at 4:16 am, Viechtbauer, Wolfgang (SP) <wolfgang.viechtbauer using maastrichtuniversity.nl> wrote:
> 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.
>> -----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
>> Welcome any comments.
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