I'm having a little trouble with the following problem:
"Let $X \sim N(0,\sigma^2)$ and consider testing $H_0: \sigma = \sigma_0$ versus $H_A: \sigma = \sigma_1$, where $\sigma_1 > \sigma_0$. The values $\sigma_0$ and $\sigma_1$ are fixed.
For a sample $X_1, \ldots, X_n$:
(1) What is the likelihood ratio?
(2) What is the rejection region of a level $\alpha$ test?"
So far, I've gotten that the likelihood ratio is: \begin{equation*} \Lambda = \bigg( \frac{ \sigma_1 } { \sigma_0 } \bigg)^n \exp\bigg\{ \frac{1}{2}\Big(\frac{1}{\sigma_1^2}-\frac{1}{\sigma_0^2} \Big) \sum\limits_{i = 1}^n X_i^2 \bigg\} \end{equation*} However, the answer that I was given for the likelihood ratio doesn't have $\Big( \frac{ \sigma_1 } { \sigma_0 } \Big)^n$, it has $\frac{ \sigma_1 } { \sigma_0 }$. Is this a mistake in the answer?
For the rejection region, I had done the following: \begin{equation*} - 2 \log(\Lambda) = - 2 n \log\bigg( \frac{ \sigma_1 } { \sigma_0 } \bigg) + \Big(\frac{1}{\sigma_1^2}-\frac{1}{\sigma_0^2} \Big) \sum\limits_{i = 1}^n X_i^2 \end{equation*}
Since $-2\log(\lambda) \sim \chi^2_1$, we see that \begin{equation*} - 2 n \log\bigg( \frac{ \sigma_1 } { \sigma_0 } \bigg) + \Big(\frac{1}{\sigma_1^2}-\frac{1}{\sigma_0^2} \Big) \sum\limits_{i = 1}^n X_i^2 > \chi^2_1(\alpha) \end{equation*}
\begin{equation*} \bigg| \frac{1}{\sigma_1^2}-\frac{1}{\sigma_0^2} \bigg| \geq \frac{2z(\alpha/2) + 4 n \log(\sigma_1/\sigma_0) } {x^2} \end{equation*}
This rejection region doesn't seem right to me though. Could someone explain to me where I went wrong?
You are right that there should be an $n$th power.
You reject the null hypothesis if $\Lambda$ is too small (since the density under the null hypothesis is in the numerator. That happens if $$ \left( \frac 1 {\sigma_1^2} - \frac 1 {\sigma_0^2} \right) \sum_{i=1}^n X_i^2 $$ is too small. If $\sigma_1>\sigma_0$, that happens if $\displaystyle \sum_{i=1}^n X_i^2$ is too big. If $\sigma_1<\sigma_0$, it happens if that sum is too small.
Under the null hypothesis you have $\displaystyle \frac 1 {\sigma_0^2} \sum_{i=1}^n X_i^2 \sim \chi^2_n$.
Suppose we let $\chi_n^2(\alpha)$ the the positive number satisfying $\Pr(\chi_n^2 > \chi_n^2(\alpha)) = \alpha$. If we reject the null hypothesis when our sum of squares is too big, then we reject the null hypothesis if that sum is bigger than $\sigma_0^2 \chi_n^2(\alpha)$. The rejection region is then the interval $(\sigma^2 \chi_n^2(\alpha),\infty)$.
If the opposite inequality on the two hypothetical variances holds, then mutatis mutandis.