Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 44 / 4 / c / Solution

Assume equal allocation of $n$ independent offenders to each arm and a two-sided statistical test of significance level $\alpha$. Write $p_0=0.60$ for P30 and $p_1=0.56$ for initial community-service assignment, with $\Delta =p_0-p_1=0.04$ and $\bar{p}=(p_0+p_1)/2=0.58$. The two event counts have independent binomial distributions. Under the stated alternative, the difference of sample proportions is approximately a normal distribution with mean $\Delta$ and varianceThe pooled two-proportion test has approximate upper rejection threshold $z_{1-\alpha /2}\sqrt{2\bar{p}(1-\bar{p})/n}$ at the planning rates, where $z_q$ is the standard normal quantile. To obtain statistical power $0.80$ against a positive difference, place the alternative mean $z_{0.80}$ alternative-standard-errors above that threshold:Solving yields the usual sample size for comparing two proportions:per arm, rounded upwards, with total sample size $2n$. This targets the dominant rejection tail under the planning alternative; an exact power calculation can include the very small opposite-tail probability or use the binomial distributions directly.

For illustration, at $\alpha =0.05$, $z_{0.975}\simeq 1.960$ and $z_{0.80}\simeq 0.842$, giving $n\simeq 2388.8$. ThusFor a prespecified one-sided test, replace $z_{1-\alpha /2}$ by $z_{1-\alpha }$. The calculation assumes complete endpoint ascertainment, independent individual allocation and that the two initial-policy rates are appropriate; losses or clustered allocation would require corresponding design adjustments.