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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 $\theta_1 < \theta_0$).

Reb2000
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    Please edit your post to add your attempts. – StubbornAtom Apr 11 '20 at 05:58
  • I added my attempts. I was trying to prove this statement by an application of Neyman-Pearson. – Reb2000 Apr 11 '20 at 12:25
  • Everything is fine, except it would probably be $\overline X<25\color\red{-}\frac{4z_{\alpha}}{\sqrt n}$ in the end. – StubbornAtom Apr 11 '20 at 14:05
  • Thanks for answering! can you explain me, why there is a minus instead? The quantil of $z(\alpha)$ is $-1.28$ ... – Reb2000 Apr 11 '20 at 14:49
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    $P_{H_0}(\overline X<k)=P\left(\frac{\sqrt n(\overline X-25)}{4}<\frac{\sqrt n(k-25)}{4}\right)=\alpha\implies P\left(\frac{\sqrt n(\overline X-25)}{4}>\frac{\sqrt n(k-25)}{4}\right)=1-\alpha$, so that $\frac{\sqrt n(k-25)}{4}=z_{1-\alpha}=-z_{\alpha}$. I am not talking about the actual value of $z_{\alpha}$. – StubbornAtom Apr 11 '20 at 14:59
  • where $z(\alpha) = \frac{k-25}{4/ \sqrt{n}}$ – Reb2000 Apr 11 '20 at 15:00
  • I see. Then I can also directly write: $\bar{X} < 25 - 1.28 \cdot \frac{4}{\sqrt{n}}$. – Reb2000 Apr 11 '20 at 15:15
  • Yes, because $z_{0.1}\approx 1.28$. I think you can provide an answer to your question now. – StubbornAtom Apr 11 '20 at 15:20
  • $z_{0.1} \approx - 1.28$, right ? – Reb2000 Apr 11 '20 at 15:28
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    No. This is what I was pointing out all this while. Definition of $z_{\alpha}$ is $P(Z>z_{\alpha})=\alpha$ where $Z$ is standard normal, i.e. $z_{\alpha}$ is the upper $100\alpha%$ point (or the $(1-\alpha)$th quantile) of $N(0,1)$. Note that $z_{1-\alpha}=-z_{\alpha}$ due to symmetry. – StubbornAtom Apr 11 '20 at 15:32

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