 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 its percentage weight isUsing 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 areWith ${\hat{\tau }}^2=0.05$, its unnormalised weight is $40/7\approx 5.714$. 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.