In the Wikipedia page on the Neyman-Pearson Lemma, they discuss the example where there are $n$ iid Normal RVs with a known mean and a population variance that is the null hypothesis, and another that is the alternative hypothesis. They get the test statistic $\Sigma (x_i - \mu)^2$
They then jump to saying "we should reject $H_0$ if $\Sigma (x_i - \mu)^2$ is sufficiently large and that we can get the critical value by scaling the test-statistic to become a chi-square distribution, they never explain why.
Various sites online explain that basically converting the test statistic to become the sample variance over the population variance will turn the formula into a chi-square distribution. They do it as follows:
$\chi^2_{n-1} = \frac{(n-1)\frac{\Sigma (x_i - \mu)^2}{n-1}}{\sigma_0^2}$
$\chi^2_{n-1} = \frac{(n-1)s^2}{\sigma_0^2}$
But then I am confused as to how this would relate to a significance value $\alpha$. Like if it is $\alpha = 0.05$, how does this translate back into our original problem of "debunking/rejecting the null hypothesis"
Not looking for the answer, I just want some help explaining why the test statistic is the way it is.