[R-meta] Difference in Proportions Meta-Analysis

ne gic neg|c4 @end|ng |rom gm@||@com
Mon Jun 8 15:09:30 CEST 2020

```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?

Sincerely,
nelly

On Mon, Jun 8, 2020 at 2:48 PM Michael Dewey <lists using 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
> >
> >       [[alternative HTML version deleted]]
> >
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> >
>
> --
> Michael
> http://www.dewey.myzen.co.uk/home.html
>

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```