I have a systematic review to compare incidence of an illness (Y) with or without an intervention (so groups A and B). These groups are relatively homogeneous, so aside from unmeasured confounders I think it is a reasonable (enough) assumption to compare groups A and B to each other. In a meta-analysis, I would do a random effects model based on trials of A vs B on outcome Y to pool the individual RR with appropriate weights.

The issue is that the intervention I'm interested in does not have RCTs or direct comparison studies; I have a bunch of studies with incidence of Y for group A, and a bunch of studies with incidence of Y for group B.

Is it reasonable to report a relative risk for A vs B by doing a simple pooling of the data? E.g. (sum(events in A)/sum(patients in A)) / (sum(events in B)/sum(patients in B)) with the appropriate confidence intervals? (i.e. "there were 200 events (2%) in group A, compared to 50 events (1%) in group B, RR 2.0 95% CI aa-bb).

Or am I limited to presenting an independent pooled estimate for each group individually? (i.e. "there were 200 events (2%) in group A, compared to 50 events (1%) in group B)."



1 Answer 1


What you are looking for is meta-regression. You set out your data-set to include for each study the incidence rate, its standard error, and an indicator of which treatment was implemented, and any other variables you might need. You then run a meta-regression using the treatment indicator as a moderator variable. You do not say what software you use but I believe in Stata this is called metareg and in R you can use one of the packages meta and metafor both of which are available on CRAN. If you do use R and you have questions about implementation then best to direct them to the mailing list https://stat.ethz.ch/mailman/listinfo/r-sig-meta-analysis// (you have to register first) where the authors of both those packages listen.

  • $\begingroup$ Thank you! I am familiar with metareg but have not applied it to studies with only one intervention group. Appreciate the help. $\endgroup$
    – stensy
    Mar 16, 2018 at 14:45

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