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.


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