Using Neyman Pearson Lemma, find a Most Powerful test for $H_0 : \sigma = 2$ vs. $H_1 : \sigma = 1 \ $ at level $\alpha$ based on a random sample $X_1, \dots , X_n \ $ from $N(3, \sigma^2 )$. Also suggest a Uniformly Most Powerful test at level $α$ for $H_0 : \sigma = 2$ vs. $H_1 : \sigma < 2$.
I'm new to this topping. I have an idea with mean unknown, but with variance unknown I don't have any experience.
The likelihood ratio is proportional to $\frac{\left(\frac1{\sqrt{8\pi}}\right)^n\exp(-\sum(x_i-3)^2/8)}{\left(\frac1{\sqrt{2\pi}}\right)^n\exp(-\sum(x_i-3)^2/2)} = \frac1{2^n}\exp\left(\frac38 \sum(x_i-3)^2\right)$ which is an increasing function of $\sum(x_i-3)^2$
so your test is going to reject $H_0$ when $\sum(x_i-3)^2$ is too small and not reject it when $H_0$ is not too small
As you say, what constitutes too small can be found from a chi-squared distribution. Under $H_0$, the distribution of $\frac{X_i-3}{2}$ should be a standard normal distribution with mean $0$ and variance $1$, so $\frac{1}{4}\sum(X_i-3)^2$ should have a chi-squared distribution with $n$ degrees of freedom, and since the rejection region corresponds to "too small", you want to consider the bottom tail
When extending this from $H_1: \sigma=1$ to $H_0: \sigma \lt 2$, the same arguments apply. So again you reject when $\sum(x_i-3)^2$ is too small, and the test is exactly the same