[R-meta] Difference in Proportions Meta-Analysis

ne gic neg|c4 @end|ng |rom gm@||@com
Mon Jul 6 17:08:29 CEST 2020


Dear Wolfgang & list,

This is what I have done, so I cross check with you again.

Following your suggestion, I computed the number events for both groups
with p*n and called the resulting event counts ai and ci. The n1i and n2i
were already available from the risk tables.

To get a RR as the measure I did this: Is this correct?

mydat <- escalc(measure="RR", ai = ai, n1i = n1i, ci = ci, n2i = n2i, data
= bmi_gc, slab=paste(author))

res <- rma(yi, vi, data=mydat, method="REML")

forest(res, digits = 2, atransf = exp )

### add text with Q-value, dfs, p-value, and I^2 statistic
text(-1, -1, pos=4, cex=0.8, bquote(paste(" (Q = ",
                                            .(formatC(res$QE, digits=2,
format="f")), ", df = ", .(res$k - res$p),
                                            ", p = ", .(formatC(res$QEp,
digits=2, format="f")), "; ", I^2, " = ",
                                            .(formatC(res$I2, digits=1,
format="f")), "%)")))

To get an OR as the measure, does it suffice to just change the RR to OR or
do I need to change e.g. atransf to something else downstream of the code?
especially the back-transformation.

Apologies if this is trivial or repetitive, but I really haven't got the
hang of this yet.

Sincerely,
nelly


On Wed, Jun 10, 2020 at 4:13 PM Viechtbauer, Wolfgang (SP) <
wolfgang.viechtbauer using maastrichtuniversity.nl> wrote:

> Hi Nelly,
>
> Don't you have the sample size on which the proportions are based? If not,
> the standard error of a proportion is SE[p] = sqrt[p(1-p)/n], so given p
> and SE[p], one can easily recover the sample size with n = p(1-p) / SE^2.
> Do this for both groups. Call the resulting sample sizes n1i and n2i. Also
> compute the number events with p*n. Do this for both groups. Call the
> resulting event counts ai and ci. Now you have everything to use
> escalc(measure="OR", ai, n1i, ci, n2i).
>
> Best,
> Wolfgang
>
> >-----Original Message-----
> >From: ne gic [mailto:negic4 using gmail.com]
> >Sent: Wednesday, 10 June, 2020 13:13
> >To: Viechtbauer, Wolfgang (SP)
> >Cc: r-sig-meta-analysis using r-project.org
> >Subject: Re: [R-meta] Difference in Proportions Meta-Analysis
> >
> >Dear Wolfgang,
> >
> >I have taken time to read the entire escalc documentation
> >(https://wviechtb.github.io/metafor/reference/escalc.html ) and I am
> sorry
> >to say I don't understand how I can do that in a single step with
> >escalc(measure="OR", ...) as you previously mentioned. I would like to do
> >this correctly following your suggestion so as to minimize any errors as I
> >redo steps that can be done in fewer steps.
> >
> >For each of my two groups I have proportion and standard error. Could you
> >kindly show me how I can do this using one of the available datasets
> please?
> >
> >Sincerely,
> >nelly
> >
> >On Mon, Jun 8, 2020 at 6:48 PM Viechtbauer, Wolfgang (SP)
> ><wolfgang.viechtbauer using maastrichtuniversity.nl> wrote:
> >No, this is not correct. You should not apply the logit-transformation to
> >the difference and the SE is not correct either. You should apply the
> logit-
> >transformation to the two proportions individually, compute the correct SE
> >of the logit-transformed proportions, take the difference between the two
> >logit-transformed proportions, and then you can use the equation to get
> the
> >SE of this difference. But all of this can just be done in a single step
> >with escalc(measure="OR", ...). As I mentioned, the difference between two
> >logit-transformed proportions is the log odds ratio.
> >
> >Best,
> >Wolfgang
>

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