[R-meta] Difference in Proportions Meta-Analysis
neg|c4 @end|ng |rom gm@||@com
Mon Jun 8 15:09:30 CEST 2020
Thanks. I am using metafor (or rather planning to for this).
Initially I had performed two separate single arm meta-analysis of
proportion as follows to get an estimate for each of the arms:
rma(yi = arm1_prop_plogis, sei = arm1_se, slab=author_year ,
rma(yi = arm2_prop_plogis, sei = arm2_se, slab=author_year ,
But then it was pointed out that it would be more interesting to
meta-analyze the difference in proportions from both arms, and hence my
So what I have done is:
1. Calculate the raw proportion differences i.e. before using the R
2. Calculate a single SE from the SE of both arms using the equation
provided by Wolfgang.
Then the two outputs are what I hope to provide as inputs to rma exactly as
I had done before for the single arm analysis.
Is there a more direct way to do this? or am I missing something?
Thanks for your help,
On Mon, Jun 8, 2020 at 2:48 PM Michael Dewey <lists using dewey.myzen.co.uk>
> Dear Nelly
> I am not sure what software you use but both meta and metafor provide
> analysis of risk differences (which is what differences in proportions
> are) so you may get what you want directly there.
> On 08/06/2020 11:36, ne gic wrote:
> > Dear List,
> > I aim to perform a meta-analysis and present a forest plot of the
> > difference in proportions between two groups.
> > For each group I have proportion and standard error (SE).
> > It's straightforward to get the difference in the proportions, but am not
> > sure how to handle the SEs, I presume I can't just subtract them. Is
> > a formula on how to handle SEs?
> > Sincerely,
> > nelly
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