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

Label batches before consulting the random-number table, so their labels are independent of the test results. A uniformly random permutation of $1,\dots ,16$, restricted to a smaller label set, induces a uniformly random ordering of that set. This provides a simple rejection-and-restriction scheme:

With independent random digits instead, rejection sampling gives the same uniform choices: for A keep only digits $1,\dots ,6$; for C keep only $1,\dots ,5$. For B, use uniform two-digit numbers, retain labels $01,\dots ,15$, and reject repeats within a three-batch sample. Taking residues modulo six or fifteen from a table whose range is not divisible by that number would give unequal probabilities. These are respectively day-stratified simple random sampling, morning/afternoon stratified sampling, and a one-day design using cluster sampling. The printed word “rest” for A is interpreted as “test”, consistently with the surveillance task. If it instead meant leaving one batch untested, choose that omitted batch uniformly and test the other five; this alternative would test 25 batches per week, with inclusion probability $5/6$, and has a different testing budget.

Under the one-test-per-day interpretation, A spreads surveillance across all five days and is operationally straightforward, but uses only five tests per week and does not guarantee morning/afternoon representation. Within each day its uniform choice avoids systematically selecting a particular batch. B uses six tests, guarantees representation of both production periods, and spreads the selected batches over the week's production, though some weekdays can receive no test. C also uses six tests and may minimize the cost of collecting samples, but concentrates them on one randomly selected day.

If batches produced on the same day share contamination risks, C's six observations are positively correlated and provide less information than six dispersed batches. A simple equal-cluster-size approximation has design effect $1+(6-1)\rho$, where $\rho$ is within-day intraclass correlation; its effective sample size is consequently smaller than six when $\rho >0$. C can also miss intermittent problems occurring on other days.

I would prefer B for estimating and comparing batch contamination rates, assuming positive within-day correlation and no overriding collection-cost advantage for C. It combines slightly more testing than A with explicit production-period coverage and avoids C's concentration on a single day. A is a reasonable alternative when regular daily surveillance is the priority. All three are legitimate probability samples with the stated inclusion probabilities; the preference concerns precision and coverage, not a claim that C is intrinsically biased. Their relative efficiency is not universal without a model for production variability and costs.

A numerical power calculation needs a significance level, desired power, sidedness and allocation; these are not specified in this part. For a concrete planning illustration, assume independent batches, equal samples per supplier, a two-sided 5% test with 80% power, and true rates $p_B=0.02$ and $p_A=0.04$. The observed 2% from B and C is being used as a planning value for B, not as proof that B's population rate is exactly known. There is also a numerical inconsistency in the stated frame: at six batches per working day, B and C together produce only 120 batches in two weeks, and their proposed schemes would test 24. The asserted 3,000 sampled batches cannot literally come from that frame. The calculation treats $60/3000$ as a stipulated planning estimate; its collection would require a larger frame or longer period.

For two independent sample proportions, the approximate null variance of their difference is $2\bar{p}(1-\bar{p})/n$ and its variance at the alternative is $[p_A(1-p_A)+p_B(1-p_B)]/n$, with $\bar{p}=(p_A+p_B)/2$. Separating the null critical value from the alternative mean by the required power quantile gives the sample size for comparing two proportions:With $z_{0.975}=1.960$, $z_{0.8}=0.842$ and $\bar{p}=0.03$, this gives $n\simeq 1140.83$, so the normal-approximation calculation rounds to 1,141 batches per supplier, or approximately 1,150 for a practical planning target. This is per supplier, not the combined total. A different power or a one-sided test changes the answer. Positive clustering of sampled batches requires a cluster-aware calculation or inflation, so the independent-batch calculation should not simply be applied to C's one-day clusters.

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.