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

[ne gic]

Dear Wolfgang,

A quick follow up. After meta-analyzing the proportions using a logit
transformation using qlogis(p), how I can back transform the proportion to
fit the normal range as I get some values below 0 on the forest plot when I
directly use the rma object.

forest(pes.da_30plus, xlab = "2-year survival (%)")

Sincerely,
nelly

On Wed, May 27, 2020 at 8:17 PM 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.
>
> 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
>

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