[R-meta] Meta-Analysis: Proportion in overall survival rate
Viechtbauer, Wolfgang (SP)
wo||g@ng@v|echtb@uer @end|ng |rom m@@@tr|chtun|ver@|ty@n|
Thu May 28 13:13:07 CEST 2020
Use atransf=plogis or transf=plogis. help(forest.rma) to read about the purpose of these arguments. But this won't give you percent, but proportions. If you really want percent, then use atransf=function(x) plogis(x)*100 or the same for transf.
Best,
Wolfgang
>-----Original Message-----
>From: ne gic [mailto:negic4 using gmail.com]
>Sent: Thursday, 28 May, 2020 13:02
>To: Viechtbauer, Wolfgang (SP)
>Cc: r-sig-meta-analysis using r-project.org
>Subject: Re: [R-meta] Meta-Analysis: Proportion in overall survival rate
>
>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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