= Solution
For an <independent subgroup contrast in meta-analysis>, the difference of disjoint <independent> subgroup estimates has estimated <variance> equal to the sum of the two variances. The subgroup contrast and <standard error> are
$$
D=0.45-0.16=0.29,\qquad
\operatorname{SE}(D)=\sqrt{0.11+0.05}=0.4.
$$
The <Wald test> of equal subgroup mean <log odds ratios> therefore gives
$$
\boxed{Z=\frac{0.29}{0.4}=0.725,\qquad
Z^2=0.525625,\qquad p\approx0.468.}
$$
Under the null, $Z$ is approximately <standard normal> and $Z^2$ has a <chi-squared distribution> with one degree of freedom. A quantile-2 <confidence interval> for the difference is $(0.29-0.8,0.29+0.8)=(-0.51,1.09)$. \b[There is no statistically persuasive evidence of a subgroup difference.] The result does not establish equal effects. Nor does a between-trial subgroup comparison, by itself, identify a causal advantage of one technique: other study characteristics may differ.
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