= Solution
Both alternatives have larger $\theta$, so use upper-tail rejection regions. Let $z_{0.95}=\Phi^{-1}(0.95)\simeq1.645$ be the 95th <quantile> of the standard <normal distribution>.
For the likelihood estimator,
$$
\sum_iY_i\sim\operatorname{Poisson}(\theta S_1).
$$
When $S_1$ is large, the <normal approximation to the Poisson distribution> gives, under $H_0$,
$$
\widehat\theta_{MLE}\approx N\left(1,\frac1{S_1}\right).
$$
An approximate size-$0.05$ test therefore rejects $H_0$ when
$$
\boxed{\widehat\theta_{MLE}
>1+\frac{1.645}{\sqrt{S_1}}}.
$$
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 $H_0$,
$$
\widehat\theta_{LS}\approx
N\left(1,\frac{S_3}{S_2^2}\right).
$$
The corresponding approximate size-$0.05$ test rejects when
$$
\boxed{\widehat\theta_{LS}
>1+1.645\frac{\sqrt{S_3}}{S_2}}.
$$
In 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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