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

[ne gic]

Dear Gerta,

Thank you, I take note. For simplicity, I restricted the meta-analysis to
proportions estimated from Kaplan-Meier curves, which was about 96% of the
studies.

I greatly appreciate the clarification.

Sincerely,
nelly

On Wed, May 27, 2020 at 10:16 PM Dr. Gerta Rücker <
ruecker using imbi.uni-freiburg.de> wrote:

> Dear Nelly,
>
> I forgot to add that my suggestion was thought as an approach when
> mixing directly reported proportions and proportions estimated from
> Kaplan-Meier curves (with the caveat I mentioned before in my first post).
>
> However, if you really restrict your analysis to those studies that
> offer a Kaplan-Meier estimator, you may not need this proportion
> approach at all, if standard errors/confidence intervals for the
> Kaplan-Meier curves at time point 2 years are available. Then you could
> take these standard errors as se in the rma call.
>
> Best,
>
> Gerta
>
> Am 27.05.2020 um 22:05 schrieb Dr. Gerta Rücker:
> > Dear Nelly, dear all,
> >
> > I have another problem with this approach, and this is the
> > denominator, "n". You insert n_t for n, but n_t = n_0*S(t) estimates
> > the number of patients surviving at time t which is the numerator of
> > the ratio. The denominator is n_0, not n_t. And n_0 should be used as n.
> >
> > (Suppose a study without any censoring. Then the Kaplan-Meier
> > estimator in t gives exactly p = n_t/n_0 which is the same as if the
> > study reports simply the 2-year survival proportion relative to the
> > sample size, n_0.)
> >
> > Thus the n in your calculation should be n_0, irrespectively of the
> > transformation method (I agree with Wolfgang that the logit
> > transformation is preferable).
> >
> > I don't see a principle problem with this approach, because n_t is an
> > unbiased estimate of the numerator and sample size n_0 is - for sure -
> > the denominator.
> >
> > Best,
> >
> > Gerta
> >
> > Am 27.05.2020 um 21:07 schrieb ne gic:
> >> Many thanks for the insights, WoIfgang!
> >>
> >> I concur, the proportion is likely not from a binomial distribution,
> >> so I
> >> take your advice.
> >>
> >> 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
> >>     [[alternative HTML version deleted]]
> >>
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