Ask George Casella Question 8.12 b This is not a too much difficult question, but I think I have problems on concepts or understandings. The problem is a normal population: with mean $\mu$ and known variance $\sigma^2$. Take $\alpha$=.05. Then do the LRT that is: $H_0: \mu=0 $ versus$H_1:\mu \neq0$.
The solution is based on example 8.2.2. I can understand this example. In example 8.2.2., the variance is 1. So according to our problem here, I find the LRT statistic is
$$\lambda(\textbf{x})=exp[-n\bar{x}^2/(2\sigma^2)]$$
This is similar to (8.2.2), except in our question we don't fix variance to be 1 and our specific value ($\mu_0$) is zero.
Then my idea is $\lambda(\textbf{x})\leq c$, so we have $|\bar{x}|\geq \sqrt{\frac{-2\sigma^2 logc}{n}}$
Now I got my first problem. The solution said For α = .05 take c = 1.96. But I don't know what value c I should plug in. It reminds me when I do differential equations. We see a lot of constants, and each step we have a new c. For example, I can name c above as c. I can also name $logc$ as c. I think I can even name the whole term above as c, since n is fixed and we know $\sigma^2$. When the solution said c = 1.96, I don't know what part is 1.96.
Then I want to get power function. I know power function is $P(|\bar{x}|\geq \sqrt{\frac{-2\sigma^2 logc}{n}})$. I know $\bar{x}$ has $norm(\mu, \sigma^2/n)$. So
$$P(|\bar{x}|\geq \sqrt{\frac{-2\sigma^2 logc}{n}}) =2*P(\bar{x} \geq \sqrt{\frac{-2\sigma^2 logc}{n}})=P(Z\geq \sqrt{-2logc}-\frac{\mu \sqrt{n}}{\sigma})$$
I don't know the above is correct and how to do next.
The solution is below:

 A: The critical region is simply of the form $|\overline X|>k$ where $k(>0)$ is a constant such that $$P_{H_0}(|\overline X|>k)=\alpha\,.$$
Noting that $\frac{\sqrt n(\overline X-\mu)}{\sigma}\sim N(0,1)$, you have
$$P_{H_0}\left(|\overline X|>k\right)=P_{H_0}\left(\left|\frac{\sqrt n\overline X}{\sigma}\right|>\frac{\sqrt nk}{\sigma}\right)=P\left(|Z|>\frac{\sqrt nk}{\sigma}\right)\,,\quad \text{ where }Z\sim N(0,1)\,.$$
Now $P\left(|Z|>\frac{\sqrt nk}{\sigma}\right)=\alpha$ means $\sqrt n k/\sigma=z_{\alpha/2}$, the upper quantile of a standard normal distribution. Therefore $k=\frac{\sigma z_{\alpha/2}}{\sqrt n}$ and the rejection region is written as $$\left|\frac{\sqrt n\overline X}{\sigma}\right|>z_{\alpha/2}$$
For $\alpha=0.05$, we have $z_{\alpha/2}\approx 1.96$.
But I think the power function by definition should be
\begin{align}
\beta(\mu)&=P_{\mu}\left(\left|\frac{\sqrt n\overline X}{\sigma}\right|>z_{\alpha/2}\right)
\\&=1-P_{\mu}\left(-z_{\alpha/2}\le \frac{\sqrt n\overline X}{\sigma}\le z_{\alpha/2}\right)
\\&=1-P_{\mu}\left(-z_{\alpha/2}-\frac{\sqrt n\mu}{\sigma}\le \frac{\sqrt n(\overline X-\mu)}{\sigma} \le z_{\alpha/2}-\frac{\sqrt n\mu}{\sigma}\right)
\\&=1-\left[\Phi\left(z_{\alpha/2}-\frac{\sqrt n\mu}{\sigma}\right)-\Phi\left(-z_{\alpha/2}-\frac{\sqrt n\mu}{\sigma}\right)\right]
\\&=1-\Phi\left(\frac{\sqrt n\mu}{\sigma}+z_{\alpha/2}\right)+\Phi\left(\frac{\sqrt n\mu}{\sigma}-z_{\alpha/2}\right)
\end{align}
