 Solution

A fixed-effect meta-analysis models the studies as estimating a common underlying treatment effect. In a generic inverse-variance version, independent log odds ratios $y_i$ have approximate distributions $N(\delta ,v_i)$ and are pooled with weights $1/v_i$. The supplied forest plot instead has the Mantel–Haenszel pooled odds ratio weights: for event/nonevent cells $a_i,b_i,c_i,d_i$ and total $n_i$, the normalized weights are proportional to $b_i{c}_i/n_i$. They reproduce the printed percentages and yield a pooled odds ratio about $2.338$. This is still a common-effect analysis; the two weighting methods should not be silently identified.

A random-effects meta-analysis allows different true effects: $y_i\mid {\delta }_i\sim N({\delta }_i,v_i)$ with ${\delta }_i\sim N(\mu ,{\tau }^2)$. The between-study heterogeneity variance ${\tau }^2$ adds to sampling variance, giving marginal variance $v_i+{\tau }^2$. Its pooled estimate targets the mean of the study-effect distribution rather than one identical effect. Uncertainty about heterogeneity must also be considered.