Let $X_1$, $X_2$, ..., $X_{50}$ be a random sample of bacterial counts in 10 one ml random samples of your tap water, and let $Y = \sum_{i=1}^{50} X_i$. It is desired to test the null hypotheses $H_0: \theta = 0.1$ against the alternative hypothesis $H_1: \theta \ge 0.1$, at the $\alpha$ level of significance, using $Y$ as the test statistic. The decision rule is to reject $H_0$ if $y \ge c$ (the critical value).
Given: $Y$ has a Poisson distribution with probability mass function
$$f(y|\theta) = \begin{cases}
\frac{e^{-50\theta}(50\theta)^y}{y!} & \text{if } y = 0, 1, 2, ..., \text{ and } \theta > 0; \\
0 & \text{otherwise}.
\end{cases}$$
Consider testing the null hypotheses $H_0: \theta = 0.1$ against the alternative hypothesis $H_1: \theta = 0.2$, at the $\alpha$ level of significance, using $\bar{X}$ as the test statistic. Show that the best test of the hypotheses rejects $H_0$ if $\bar{X} = \frac{1}{50} \sum_{i=1}^{50} x_i \ge c$ where $c$ satisfies $\alpha = P(\bar{X} \ge c|\theta = 0.1)$.