The most powerful test for variance of normal distribution Let $ X_1, \ldots, X_{10} $ be simple samples from normal distribution with expected value${}= 0$ and variance${}=  \sigma^2 $. Write the most powerful test for $ \alpha = 0.05 $ to verify the hypothesis that $ \sigma^2 =1 $ versus $ \sigma^2 = \sigma^2_1 $ for $ \sigma^2_1 >1 $. For which $ \sigma^2_1  $ the power would be greater than 0.95?
I have estimated likelihood ratio as:
$$ \left(\frac{\sigma^2}{\sigma^2_1}\right)^{10}\exp\left(\left(\frac{1}{2\sigma^2}-\frac{1}{2\sigma^2_1}\right)\sum_{i=1}^{10}X_i^2\right) \geq k $$
Given $\ \chi^2(0.95, 10) = 18.3 $ and $\ \chi^2(0.05, 10) = 3.94 $, I write that:
$$ \alpha = P\left(\sum_{i=1}^{10}\frac{X_i^2}{\sigma} \geq \frac{k}{\sigma^2} \,\Big|\, \sigma^2 = 1 \right) $$
Questions:


*

*is it correct?

*is it correct to set $ k=\sigma^2\chi^2(0.05, 10) $? Not to use (0.95, 10)?

*how to move on with this? 

*For which $ \sigma^2_1  $ the power would be greater than 0.95? I am really confused with this question. I have thought that the power is independent regarding $ \sigma^2_1 $.
 A: The rejection region for testing the null $H_0:\sigma^2=1$ against the alternative $H_1:\sigma^2>1$ is of the form $T>k$, where $T=\sum\limits_{k=1}^{10} X_i^2$ and $k$ is so chosen that size of the test is $0.05$. 
You also have $$\frac T{\sigma^2}\sim \chi^2_{10}$$
This implies $$P_{H_0}\left[T>k\right]=P_{H_0}\left[\chi^2_{10}>k\right]=0.05\,,$$
so that in terms of the upper $5\%$ point of a chi-square distribution you have $$k=\chi^2_{10,0.05}\,,$$
whose value can be found from a chi-square table or software.
So the test function is 
$$\varphi=\begin{cases}1&,\text{ if }T>\chi^2_{10,0.05} \\ 0 &,\text{ else }\end{cases}$$
Now the power at $\sigma^2=\sigma_1^2(>1)$ is 
\begin{align}
E_{\sigma_1^2}[\varphi]&=P_{\sigma_1^2}\left[\frac T{\sigma_1^2}>\frac1{\sigma_1^2}\chi^2_{10,0.05}\right]
\\&=P\left[\chi^2_{10}>\frac1{\sigma_1^2}\chi^2_{10,0.05}\right]
\end{align}
This would give you the value of $\sigma_1^2$ for which this power equals $0.95$:
$$\frac1{\sigma_1^2}\chi^2_{10,0.05}=\chi^2_{10,0.95} \implies \sigma_1^2 = \frac{\chi^2_{10,0.05}}{\chi^2_{10,0.95}} = k' \,(\text{say})$$
The nature of the power function $E_{\sigma^2}[\varphi]=P_{\sigma^2}\left[T>\chi^2_{10,0.05}\right]$ (i.e. increasing/decreasing in $\sigma^2$) would then suggest the possible values of $\sigma^2_1$ for which $E_{\sigma_1^2}[\varphi]>0.95$.
