Past exam of the probability, statistics and options I course of the University of Oxford / 2011 / O / 2

The Neyman-Pearson lemma concerns testing a simple null density $f_0$ against a simple alternative $f_1$. A test that rejects where $f_1/f_0>c$, accepts where the ratio is below $c$, and if necessary randomizes on the boundary to have null rejection probability $\alpha$, is a most powerful test among tests of size at most $\alpha$. Thus a likelihood-ratio threshold maximizes power at the specified simple alternative. The boundary randomization is relevant for discrete distributions.

For $S={\sum }_i{X}_i$, the ratio of the two sample likelihood functions isFor $0<{\mu }_0<{\mu }_1$ it is strictly increasing in $S$. The Neyman-Pearson lemma therefore selects an upper tail:Under the null, $S\sim \mathrm{Poisson}(n{\mu }_0)$, so choose $k$ with this upper-tail probability at most $\alpha$. If exact size $\alpha$ is required, randomize at the adjacent boundary value as needed. The rejection rule is the same for every ${\mu }_1>{\mu }_0$, giving a uniformly most powerful test for the one-sided alternative.

The total $S$ under $H_0$ has Poisson distribution with mean $12$, and the observed total is $16$. Its one-sided p-value is thereforeThe supplied ${\varphi }_{12}(17)\approx 0.1$ also shows that the observed count does not enter the approximate 10% upper-tail rejection region. Thus there is insufficient evidence to reject $\mu =1$ at either the 10% or 5% level. The question specifies no testing level, so the p-value gives the general conclusion: rejection is justified only at levels at least about $0.1556$.

The printed ${\varphi }_1(16)\approx 0$ concerns a Poisson variable with mean one, not the twelve-month total. It is therefore not the null tail for this data; substituting it would incorrectly treat the yearly count as one month's count.