Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 29 / 3

Treat each release as an independent Bernoulli observation with a two-week event probability, giving two independent binomial distributions. Under the specified effect, the probabilities are $p_0=0.0016$ and $p_1=0.0008$ with $n=15000$ in each period. The expected event counts are therefore 24 and 12. For a conventional two-sided 5% comparison of proportions, let $D={\hat{p}}_0-{\hat{p}}_1$, $\delta =p_0-p_1=0.0008$ and $\bar{p}=(p_0+p_1)/2=0.0012$. UseThe approximate null rejection region is $|D|>1.96s_0$, and under the alternative $D$ is approximately $N(\delta ,s_1^2)$. Thus the power of a two-sample rare-event comparison isHere $\Phi$ is the standard normal distribution function. Hence the normal approximation gives $\boxed{\text{two-sided power approximately }52\%}$: detecting the proposed halving is far from assured.

For the small counts, a discrete calculation is also appropriate. Approximate the two counts by independent Poisson distributions with means 24 and 12. Conditional on their total $K$, the pre-intervention count is $\mathrm{Binomial}(K,1/2)$ under equal rates and $\mathrm{Binomial}(K,2/3)$ under the halving alternative; $K\sim \mathrm{Poisson}(36)$ under that alternative. If ${\mathcal{R}}_k$ is the rejection set of the symmetric exact two-sided 5% binomial test, its power isThis conservative discrete test has about 45% power, lower than the normal approximation because its achieved level can be below 5%. If a decrease-only one-sided 5% test had been specified in advance, replace 1.96 by 1.645; the normal-approximation power is about 64%. The tail convention and the chosen test must accompany any quoted statistical power.

Use a concurrent randomized comparison of the new leaflet with the existing release information. Random allocation makes the groups comparable in expectation and avoids attributing a secular change in drug supply, release populations or post-release services to the leaflet. Both arms should use the same outcome ascertainment and two-week follow-up. This improves the causal interpretation over a historical before-and-after comparison; adequate sample size remains necessary given the low statistical power calculated above. If contamination prevents individual allocation, a carefully designed cluster randomized controlled trial is an alternative, with its sample-size calculation allowing for within-cluster dependence.

The one-day slaughter-inspection yield isor 500 detections per million inspected adult cattle. The routine field detections correspond, using a stable herd size of five million, toper adult-cattle day, or 8 detections per million adult-cattle days. Therefore the numerical one-day comparison isThe annual field rate is $14600/(5\times 10^6)=0.00292$ per cattle-year; comparing the inspection probability directly with this annual proportion would mix observation periods. Applying the random inspection-sample proportion to all 150000 slaughtered adults would predict about 75 detected cases in that slaughter cohort. These arithmetic comparisons describe detection yields, not a 62.5-fold biological disease-risk difference: a cross-sectional inspection estimates a prevalence yield, whereas daily field reporting is an ascertainment rate over animal-time.

Two reasons are selection bias and unequal ascertainment.