# Determining the most powerful test for a simple hypothesis

Let $$X_1,...,X_n$$ be $$N(\mu,\sigma^2)$$ where $$\mu$$ is unknown and $$\sigma$$ is known. Let the null hypothesis $$H_0$$ state that $$\mu=\mu_0$$ and the alternate be that $$\mu=\mu_1$$ where $$\mu_1>|mu_0$$.

Using the Neyman-Pearson Lemma, I plan to reject the null hypothesis for a sufficiently large value of $$\frac{L(x|\mu_1)}{L(x|\mu_0}$$, the likelihood function under the alternate divided by the likelihood function under the null. The question specifies that my response should be in “simplest implementable form” which I will get to in a second.

After having simplified my ratio of likelihood functions, I ultimately determined that the ratio is positively related to $$\sum_{I=1}^nX_i$$, so we will reject the null if there is a sufficiently large value for $$\sum_{I=1}^nX_i$$.

For my answer, I use the fact that $$\sum_{I=1}^nX_i$$ is normally distributed with mean $$n\mu$$ and variance $$n\sigma^2$$, so the most powerful test tin simplest implementable form would be as follows:

Reject the null hypothesis if $$z_{\alpha}<\phi(\frac{\sum_{I=1}^nX_i-n\mu_0}{\sqrt{n}\sigma})$$, where $$z_{\alpha}$$ is the upper $$100\alpha$$ percentile of the standard normal distribution.

I’m not sure at all if this is correct, so I would appreciate some feedback.