I am trying to compute Cohen's in some studies that I gathered in a Systematic Literature Review.

Unfortunately, almost none of them report the standard deviation (nor df/t, or something for back-computing Cohen's d). Thus, I cannot neither compute Cohen's nor calculate the standard error of each study.

However, the studies commonly report the means and the sample sizes.

I was thinking about using the response ratio (mean1/mean2) as the effect size, and weighting each study according to its sample size (instead of its standard error as usual).

Question 1: do you know any reference where they have done exactly that, and where they acknowledge to have obtained similar results to those with standard error weighting?

Question 2: do you know whether it is posible doing that with network meta analysis in R? I have studies comparing A vs B, and studies comparing A vs C.

  • $\begingroup$ Sample size can be a reliable weighting factor only for categorical variables, but you might proceed with using point effect estimates and sample sizes nonetheless, with a hypothesis-generating scope. You can surely use them as you suggest in R with netmeta or mvmeta packages. $\endgroup$ – Joe_74 Sep 27 '18 at 10:53

Using the ratio of means implies a very different effect measure than Cohen's D or some other difference in absolute values. That does not necessarily mean it's bad, but simply that you think that e.g. blood pressure being 130 vs. 120 (ratio 1.083) is a bigger effect than 140 vs. 130 (1.077). I suspect it also has implication on what error distribution you assume, but the exact implications are not clear to me (the more usual approach for relative effect measures would be to analyze at values on the log-scale, but of course you cannot change the analysis that was done).

In principle weighting by the sample size makes sense for a difference between groups (or for Cohen's D), if you assume the standard deviation to be the same across studies. This is because the SD of the sampling distribution of a difference between two means in both groups is $\sqrt{\text{SD}_1^2/n_1+\text{SD}_2^2/n_2}$ and if $n_1=n_2=N/2$ and $\text{SD}_1=\text{SD}_2=\text{SD}$, then that simplifies $\text{SD} \sqrt{4/N}$. So, to get inverse variance weights, you would want something close to $N/(4\times\text{SD}^2)$. If the SD is the same in all trials and the trials have 1:1 allocation, then weighting by N gets you very close (there's of course issues like the SD being estimated in practice etc.) to inverse variance weights.

If you have the same variable in each study, you are probably much better off doing a network meta-analysis of the data from each arm using the approach of Piepho et al. (2012) using sample size per arm weights. If you have very different variables in each study, then perhaps meta-analyzing these rather incoherent studies is non-ideal?

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