= Solution
A <fixed-effect meta-analysis> assigns <inverse-variance weights>. Using the unrounded within-study <variance> from part (b), this trial has $w=8$, so \b[its percentage weight is]
$$
\boxed{100\frac8{400}=2\%.}
$$
Using the printed rounded <standard error> of $0.35$ instead gives approximately $2.04\%$; the raw cell calculation yields the more precise answer.
For a <random-effects meta-analysis>, the corresponding weight and percentage weight are
$$
\boxed{w^*=\frac1{0.125+\widehat\tau^2},\qquad
100\frac{(0.125+\widehat\tau^2)^{-1}}{\sum_{i=1}^{21}(v_i+\widehat\tau^2)^{-1}}.}
$$
With $\widehat\tau^2=0.05$, its unnormalised weight is $40/7\approx5.714$. The supplied sums of fixed-effect weights do not determine the denominator of the random-effects percentage; the individual $v_i$ are needed. Positive heterogeneity makes the weight distribution more even. Percentage weight measures the trial's contribution to a pooled mean with fixed weights; total influence can also involve re-estimation of heterogeneity.
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