[R-meta] Difference in Proportions Meta-Analysis
Sorry, that was a typing mistake: I transformed the proportions using qlogis, then back transformed them using the plogis. Here is literally what I have done: On excel: 1. Calculated the difference in proportions between arm1 and arm2 (arm1 - arm2) to get arm12_prop_diff 2. Used the formula you provided to calculate the "joint" standard error (se_diff) In R: # Import the excel data. gastric_data$arm12_prop_diff <- as.numeric(gastric_data$arm12_prop_diff) gastric_data$se_diff <- as.numeric(gastric_data$se_diff) # Log transformation gastric_data$arm12_prop_qlogs_diff <- qlogis(gastric_data$arm12_prop_diff) # fit random-effects model pes.da=rma(yi = arm12_prop_qlogs_diff, sei = se_diff, slab=author , data=gastric_data, method="REML") forest(pes.da, xlab = "2-year survival difference (%)", order=order(gastric_data$arm12_prop_diff), atransf=plogis) Is this still not a reasonable way to go about this? On Mon, Jun 8, 2020 at 5:27 PM Viechtbauer, Wolfgang (SP) <
wolfgang.viechtbauer at maastrichtuniversity.nl> wrote:
I am not sure I really understand what you did. arm1_prop_plogis and
arm2_prop_plogis actually sound like they are plogis()-transformed
proportions, which doesn't make sense (plogis() is the inverse-logit
transformation). The logit transformation is qlogis(). Not sure how you
computed arm1_se and arm2_se. Why not use escalc(measure="PLO", ...), which
will do things correctly for you?
But if you want to compare the two groups within studies directly, then
you need to use measures such as the risk difference (measure="RD"), the
log transformed risk ratio ("RR"), or the log transformed odds ratio
("OR"). In fact, the difference between two logit-transformed proportions
*IS* the log transformed odds ratio.
So, just use escalc(measure="OR", ...) and then pass the 2x2 table counts
via arguments 'ai', 'bi', 'ci', 'di' (or the 'event' counts via arguments
'ai' and 'ci' and the group sizes via 'n1i' and 'n2i'). See help(escalc)
and search for "OR".
Best,
Wolfgang
-----Original Message----- From: R-sig-meta-analysis [mailto:
r-sig-meta-analysis-bounces at r-project.org]
On Behalf Of ne gic Sent: Monday, 08 June, 2020 15:09 To: Michael Dewey Cc: r-sig-meta-analysis at r-project.org Subject: Re: [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