Let $X_{1}, X_{2}, \ldots, X_{n}$ denote a random sample from a normal distribution $N(\theta, 100) .$ Show that $C=\left\{\left(x_{1}, x_{2}, \ldots, x_{n}\right): c \leq \bar{x}=\sum_{1}^{n} x_{i} / n\right\}$ is a best critical
region for testing $H_{0}: \theta=75$ against $H_{1}: \theta=78$. Find $n$ and $c$ so that
$$P_{H_{0}}\left[\left(X_{1}, X_{2}, \ldots, X_{n}\right) \in C\right]=P_{H_{0}}(\bar{X} \geq c)=0.05$$
and
$$P_{H_{1}}\left[\left(X_{1}, X_{2}, \ldots, X_{n}\right) \in C\right]=P_{H_{1}}(\bar{X} \geq c)=0.90$$