= Solution
A <random-effects meta-analysis> allows differences in true treatment effects across countries and eligibility criteria. For trial $i=1,\ldots,6$, let $y_i$ be its estimated <log odds ratio> and $v_i$ its estimated within-trial <variance>. A usual approximate <statistical model> is
$$
y_i\mid\delta_i\sim N(\delta_i,v_i),\qquad \delta_i=\mu+b_i,\qquad b_i\sim N(0,\tau^2),
$$
independently across trials, with $b_i$ <independent> of sampling errors. Here $\mu$ is the mean true <log odds ratio> in the population of comparable trials, and $\tau^2\geq0$ is between-trial heterogeneity, not additional sampling error. Therefore $y_i\sim N(\mu,v_i+\tau^2)$, and for a fitted heterogeneity value,
$$
\boxed{\widehat\mu=\frac{\sum_iw_i y_i}{\sum_iw_i},\qquad w_i=(v_i+\widehat\tau^2)^{-1}.}
$$
One can estimate $\tau^2$ by <restricted maximum likelihood> or another justified method; $(\sum_iw_i)^{-1}$ is a plug-in conditional <variance> and does not fully account for estimating heterogeneity. With only six studies that uncertainty matters. Different eligibility rules motivate this <random effect> but do not by themselves establish exchangeability or remove <bias>; known <effect modifiers> may warrant stratification or regression.
Use <leave-one-study-out influence analysis>: fit all six trials, then omit Cohen 1989 and refit both $\mu$ and $\tau^2$. Report $\widehat\mu-\widehat\mu_{(-C)}$, the corresponding change in the <odds ratio>, and changes in the <confidence interval> and heterogeneity. As a diagnostic with heterogeneity held fixed, writing $W=\sum_iw_i$ gives
$$
\widehat\mu-\widehat\mu_{(-C)}=\frac{w_C}{W-w_C}(y_C-\widehat\mu).
$$
The fitted Cohen weight would be $(0.15+\widehat\tau^2)^{-1}$ and $y_C=-1.95$. A precise but discordant trial can have large influence; refitting assesses additional influence through heterogeneity. The other five trials' data are absent, so a numerical six-trial influence assessment is not identifiable from the displayed table.
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