Let $X_1, ..., X_n \sim N(\theta,16)$.
How can I find then the uniformly most powerful (UMP) test of level $0.10$ for the hypothesis
$H_0 : \theta = 25$ vs $H_{\alpha} : \theta < 25$.
So, here are my attempts:
We can consider: $H_0 : \theta = 25$ vs $H_{\alpha} : \theta = \theta_1$, où $\theta_1 < 25$.
We have two simple hypothesis, so we can move on with an Neyman-Pearson Test.
Considering
$$\frac{f(x_1,...,x_n \mid \theta_0)}{f(x_1,...,x_n \mid \theta_1)} = \exp - \frac{1}{2 \cdot 16} \left( 2n \bar{x} (\theta_1 - 25) + n 25^2 - n \theta^2\right),$$
which is a monotone increasing function of the statistic $\bar{x}\, (\text{ since }\, \theta_1 < \theta_0 )$. So, we want to find a constant $k$, s.t.:
$$ \alpha = \mathbb{P} [ \bar{X} < k | H_0 ] = \cdots =\Phi \left(\frac{k -25}{4/ \sqrt{n}}\right).$$
Hence the test of level $\alpha = 0.1$ with the largest power has the critical region
$$\bar{X} < 25 + z(\alpha) \dfrac{\sigma}{\sqrt{n}}$$
This critical region depends on $\theta_0, \sigma, n$ and $\alpha$, but not from $\theta_1$.
Since $\theta_1 < \theta_0 = 25$, this test is also UMP (for all $\theta_1 < \theta_0$).