[R-meta] From Risk Difference to Odds Ratio
Viechtbauer, Wolfgang (SP)
wo||g@ng@v|echtb@uer @end|ng |rom m@@@tr|chtun|ver@|ty@n|
Fri Feb 21 20:03:22 CET 2020
If you have a risk difference (RD = pT - pC) and then assume a base risk for the control group (i.e., pC), then pT = RD + pC. Then you have pT and pC from which you can directly compute the OR. Doing this via the NNT seems like an unnecessary complication to me.
And yes, if you do this, then I would do a sensitivity analysis with varying values of pC.
>From: R-sig-meta-analysis [mailto:r-sig-meta-analysis-bounces using r-project.org]
>On Behalf Of Bernard Fernou
>Sent: Friday, 21 February, 2020 16:05
>To: r-sig-meta-analysis using r-project.org
>Subject: [R-meta] From Risk Difference to Odds Ratio
>I am conducting a meta analysis regarding the relationship between the
>presence of organophosphates and obesity in metafor. Our IV is mostly
>reported as dichotomous so, as obesity is also dichotomous, we have decided
>to perform a meta-analysis of odds ratio.
>However, some authors reported adjusted associations using risk difference.
>We have contacted these authors to access raw data but we have not received
>any response yet. Therefore, our plan was to try to convert this risk
>difference in odds ratio.
>To do so, we started from the fact that risk difference is equal to 1/NNT.
>Then, we convert this NNT in odds ratio using the formula provided in the
>Cochrane Handbook (link is provided below). The assumed control risk was
>estimated using the proportion of cases with obesity in the "control" group
>(we believe this is feasible since the study is cross sectional). We plan
>to perform sensitivity analyses with different assumed control risk.
>Would it make sense or should we simply exclude this study?
>Thank you for your help
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