Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 17H / d / Solution

Both alternatives have larger $\theta$, so use upper-tail rejection regions. Let $z_{0.95}={\Phi }^{-1}(0.95)\simeq 1.645$ be the 95th quantile of the standard normal distribution.

For the likelihood estimator,When $S_1$ is large, the normal approximation to the Poisson distribution gives, under $H_0$,An approximate size-$0.05$ test therefore rejects $H_0$ when

When every $x_i$ is large, $Y_i\approx N(\theta x_i,\theta x_i)$ independently. A linear combination of independent normal variables is normal, so under $H_0$,The corresponding approximate size-$0.05$ test rejects whenIn both cases the null rejection probability is approximately $0.05$, while values near the alternative mean $2$ increasingly fall in the rejection region as the total exposure grows. These are the normal-approximation tests for a Poisson exposure model.

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