Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 32 / 2 / d / Solution

For equal independent samples, take $p_B=0.0002$, $p_C=0.0006$ and $\bar{p}=0.0004$ in the two-proportion formula, with two-sided $\alpha =0.05$ and power $1-\beta =0.80$. It givesThus the usual normal planning estimate isThe combined approximate design therefore checks about eight blocks, with about eight positive batches from B and twenty-four from C at the rounded design.

Those expected counts are low enough for test discreteness to matter. If a genuinely exact two-sided Fisher exact test is specified, the null allocation of the positive batches between two equal-size suppliers is hypergeometric, conditional on their total. Under the proposed alternative, its power is obtained by summing the independent binomial probabilities over the exact rejection region:An independent calculation of the exact power of Fisher's exact test gives about $0.7750$ at $n=40000$ and $0.8728$ at $n=50000$. Consequently five whole 10,000-batch blocks per supplier suffice for at least 80% power with this conservative exact test. Four blocks are the intended normal-approximation answer, not an exact 80% guarantee irrespective of the chosen test. The distinction is a property of rare-event discreteness, not a change in the proposed effect size.