[R-meta] Difference in Proportions Meta-Analysis

Dear Michael,

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 ,
data=gastric_data, method="REML")

rma(yi = arm2_prop_plogis, sei = arm2_se, slab=author_year ,
data=gastric_data, method="REML")

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
question.

So what I have done is:


   1. Calculate the raw proportion differences i.e. before using the R
   function "qlogis"
   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,

Sincerely,
nelly







On Mon, Jun 8, 2020 at 2:48 PM Michael Dewey <lists at dewey.myzen.co.uk>
wrote:

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.

Michael

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

there

a formula on how to handle SEs?

Sincerely,
nelly

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