 Solution

Let $y_i$ be each trial's estimated log risk ratio, with estimated sampling variance $s_i^2$. A fixed-effect meta-analysis models the independent estimates as $y_i\simeq N(\mu ,s_i^2)$, with one common true log risk ratio $\mu$. Differentiating the Gaussian log-likelihood, or minimizing ${\sum }_i(y_i-\mu {)}^2/s_i^2$, gives inverse-variance weights $w_i=1/s_i^2$ andThe variance follows directly by adding independent variances of the weighted estimates: ${\sum }_i{w}_i^2{s}_i^2/({\sum }_i{w}_i{)}^2=1/{\sum }_i{w}_i$.

Using the printed rounded estimates and standard errors, ${\sum }_i{w}_i=50.2923$, soThe pooled risk ratio is about $0.902$. An interval using the same quantile two is $(-0.3848,0.1792)$ on the log scale, so it includes no effect. Recomputing all trial estimates from the event counts would change the last digits because the printed log estimates and errors are rounded.

The key assumptions are independent trials, a common true treatment effect on this chosen log scale, approximately unbiased and approximately normal trial estimates, and suitable sampling-variance estimates. The common-effect assumption is stronger than merely studying the same named treatment: systematic population, treatment or design differences may produce genuine heterogeneity. For an unbiased interpretation of the pooled evidence, inclusion of studies must also not depend selectively on their results. Treating the plug-in variances as fixed is the usual approximation in the displayed variance formula.