I have a compilation of following data on 10 studies, all of which had 2 groups: a control group which received usual treatment and an active group which received an intervention. The subjects were followed up for one year and any adverse events were counted. The data is as follows:

study   events_control  total_control   events_active       total_active
'First et al'       25              100         38                  200
'Second et al'      30              150         45                  400

Hence, in the first study, 25 of 100 in control group had events while 38 of 200 had events in active group. And so on.

How can I combine all these data to determine if events in active group were more or less than those in control group? Thanks for your help.

Edit: I was thinking of this option: since I have all the numbers, can I combine all 4 columns and get events in control and active groups and also total number of subjects with all studies taken together. Then I can compute overall odds ratio and its confidence interval.

Using this method 55/250 events occurred in the control group and 83/600 in active group, giving the overall OR is 0.57 with 95% CI of 0.39 and 0.83. These are not too different from the estimate of 0.536 (CI: 0.156, 0.916) using metafor package, as detailed in the excellent answer by @user33. Can this be a reasonable strategy? If not, what are the drawbacks?

Edit2: To take into account that data has come from different studies, can we use a mixed effects anova as follows:

aov(event_rate ~ treatment_type + Error(study_id/treatment_type), data=mydata)

1 Answer 1


In order to be able to conduct a meta-analysis based on your data (this seems to me what you want to do), you first have to select what kind of effect size you want to use. If all your data has this structure you could for instance select log odds ratios as effect size measure. For more information about effect sizes see Borenstein, Hedges, Higgins, and Rothstein (2009). Introduction to meta-analysis, Chapter 5.

Once you selected your effect size you have to compute for each study this effect size and its sampling variance. You could do this easily with for instance the metafor package in R.

To give an example:

ai = events_control

bi = no_events_control

ci = events_active

di = no_events_active

Assuming that your effect sizes of interest are log odds ratios you can compute these and its sampling variances by


ai <- c(25, 30)
bi <- c(75, 120)
ci <- c(38,45)
di <- c(162, 355)

es <- escalc(ai = ai, bi = bi, ci = ci, di = di, measure = "OR")

> es yi vi 1 0.3514 0.0858 2 0.6792 0.0667

You could use then these log odds ratios (yi) and the sampling variances (vi) for conducting different types of meta-analyses. The code below gives a standard random-effects meta-analysis:

> rma(yi = es$yi, vi = es$vi)

Random-Effects Model (k = 2; tau^2 estimator: REML)

tau^2 (estimated amount of total heterogeneity): 0 (SE = 0.1079)
tau (square root of estimated tau^2 value):      0
I^2 (total heterogeneity / total variability):   0.00%
H^2 (total variability / sampling variability):  1.00

Test for Heterogeneity: 
Q(df = 1) = 0.7043, p-val = 0.4013

Model Results:

estimate       se     zval     pval    ci.lb    ci.ub          
  0.5358   0.1937   2.7657   0.0057   0.1561   0.9155       ** 

Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Another option to get the same output as presented above is not first computing the log odds ratios and its sampling variances via the escalc() function, but directly running the meta-analysis by

rma(ai = ai, bi = bi, ci = ci, di = di, measure = "OR")

For more information about meta-analysis please see the help functions of the metafor package or use the earlier mentioned book.

  • $\begingroup$ Thanks for a great answer. I will certainly look at metafor package. Please see the edit in my question above regarding a simpler option. Is it valid? $\endgroup$
    – rnso
    Jun 11, 2015 at 16:17
  • $\begingroup$ Also why did you choose a random effects rather than fixed effects here? $\endgroup$
    – rnso
    Jun 11, 2015 at 16:48
  • $\begingroup$ I would not apply the analysis strategy as you proposed in your edit because then you ignore the fact that the data come from different studies. To answer you question about random-effects or fixed-effect meta-analysis... I did not really choose for a random-effects meta-analysis, but this is the default option. Based on theoretical reasons you have to select either a fixed-effect meta-analysis or a random-effects meta-analysis $\endgroup$
    – User33
    Jun 12, 2015 at 7:16
  • 1
    $\begingroup$ Good answer, but one point is misleading: The analysis is done with log-transformed odds ratios, not the odds ratios themselves. So, the aggregated value you get is also a log-OR (which of course can be easily back-transformed via exponentiation). $\endgroup$
    – Wolfgang
    Jun 14, 2015 at 7:27
  • 1
    $\begingroup$ Yes, 0.5358 and the confidence interval are of course log odds ratios. Sorry for the confusion from my side ;-) $\endgroup$
    – User33
    Jun 14, 2015 at 14:29

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