Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 37 / 3 / iii

Using the upper stated scrapie-test-positive fraction, its value among slaughtered sheep is $2000/4000000=0.0005$. The stipulated conditional BSE positivity then gives marginal test-positive probabilitybecause the other stratum contributes zero. If the sample consists of independently tested, representative animals, the count $D$ has a binomial distribution with parameters $n,q$. ThereforeThe two results areThe Poisson distribution approximation uses mean counts $0.5$ and $5$, respectively. If fewer than 2000 animals belong to the test-positive stratum, these probabilities of finding nothing are higher; the calculation at 2000 is not a guaranteed detection probability.

There is also a sampling-design distinction. If exactly $K$ positive animals in a fixed annual population of $N$ are known and the sample is without replacement, use the hypergeometric distribution instead:The given expected prevalence does not assert an exactly fixed $K$, so the independent prevalence model is the natural interpretation of the requested calculation.