Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 46 / 2

Use equal allocation and a conventional two-sided 5% comparison of independent proportions. The assumed risks are $p_C=0.005$ and $p_I=0.005(1-0.30)=0.0035$, giving target difference $\delta =0.0015$. Put $\overline{p}=(p_C+p_I)/2=0.00425$ and let $m$ be the number per arm.

The null critical difference is approximately $z_{0.975}\sqrt{2\overline{p}(1-\overline{p})/m}$. Under the alternative, the difference has standard deviation $\sqrt{[p_C(1-p_C)+p_I(1-p_I)]/m}$. For 80% statistical power, the target difference must exceed the critical difference by $z_{0.8}$ alternative standard deviations. Solving gives the sample size for comparing two proportionsThis is about $29,500$ per arm, or $59,000$ altogether. Rounding for planning:The rare-event Poisson approximation gives a similar answer. This assumes the stated baseline risk applies, no endpoint loss, and no attenuation by treatment contamination. The two-sided convention is explicit: a one-sided 5% test would require fewer participants.

Use the assumed 30% risk reduction for recipients and one-to-one transfers to distinct participants in the equal-sized opposite arm. Receipt fractions become 80% under intervention assignment and 20% under control assignment. ThusFor 30,000 participants per arm, the expected counts areThe original benefit $1.5$ per thousand becomes $0.9$ per thousand: symmetric treatment crossover attenuates a risk difference by $1-2(0.2)=0.6$. Expected counts still sum to 255, since intact transfers preserve the total number of recipients. These predictions assume unchanged receipt-specific risks and distinct control recipients. Loss, sharing or repeated transfers would require a different receipt model.

Define the difference as control minus intervention, positive for fewer deaths under intervention assignment. The observed difference per thousand isFor independent binomial distributions, its unpooled standard error per thousand isThe approximate 95% confidence interval is thereforeThe rare-event Poisson standard error $\sqrt{148+123}/30$ gives almost the same interval. A reduction of 1.5 per thousand is consistent with the data, since it lies inside the interval. Zero also lies inside, so a nonzero benefit is not established at the two-sided 5% significance level. This does not establish equivalence or prove the proposed effect size, and cannot by itself establish treatment contamination.

Measure receipt and transfers directly rather than inferring them only from the outcome difference. Use identifiable pack codes and confidential early follow-up to record whether packs were retained, lost, taken, given away or administered, and to whom. Where feasible, corroborate administration reports with care records and establish whether recipients belonged to the opposite randomized group.

A process-evaluation sample across both arms can estimate the frequency and direction of treatment contamination. Separate receipt tracking from blinded outcome ascertainment where possible. Keep randomized assignments for the primary intention-to-treat analysis: excluding transfers or reclassifying participants by receipt can introduce confounding, since receipt behavior is not randomized. Receipt data can inform a supplementary analysis whose additional assumptions are explicit.