The random variables $\lbrace X_k: 1 \leq k \leq n \rbrace$ are independent and identical distributed. The average is $\mu$.
$H_0:$ the datas $\tilde{X_k}=(X_k-\mu)$ are standard normal distributed
$H_1:$ the datas are bilaterally exponential distributed with density $f(x)=\exp\lbrace - \sqrt{2} |x| \rbrace / \sqrt{2}$.
How to show that the critical region of $H_0$ against $H_1$ is $K= \lbrace \sum_{k=1}^{n}{(|\tilde{X_k}|-\sqrt{2})^2>c(n)\rbrace}$?
My idea was to use the Likelihood-ratio test:
For one data I got
$$\frac{\mathrm{exp}\lbrace - \sqrt{2}|x|\rbrace \sqrt{2\pi}}{\sqrt{2}\mathrm{exp}\lbrace-x^2/2\rbrace}=\sqrt{\pi}\mathrm{exp}\lbrace \frac{1}{2}(x^2-2\sqrt{2}|x|)\rbrace=\sqrt{\pi}e^{-1}\mathrm{exp}\lbrace\frac{1}{2}(|x|-\sqrt{2})^2\rbrace$$
For all datas I got
$$\pi^{n/2}e^{-n}\mathrm{exp} \lbrace \frac{1}{2} \sum_{k=1}^{n}{(|x_k|-\sqrt{2})^2\rbrace}$$
So
$$K=\lbrace \pi^{n/2}e^{-n}\mathrm{exp}\lbrace \frac{1}{2}\sum_{k=1}^{n}{(|X_k|-\sqrt{2})^2 \rbrace > \tilde{c} \rbrace}=\lbrace \sum_{k=1}^{n}{(|X_k|-\sqrt{2})^2\rbrace>2 \mathrm{log}(\tilde{c}e^n\pi^{-n/2})\rbrace}$$
Now I don't know where to put $\mu$ to get $\tilde{X_k}$ instead of $X_k$.
Where do I have to change something?