Using neyman-pearson to reach a chi-square test for the variance of a population 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.
 A: Neyman-Pearson says reject for $Q = \sum_i (X_i - \mu)^2 > c,$ with $c$ chosen so that $P(Q > c) = \alpha$ under $H_0.$ Then the problem is to find the numerical value of $c.$
Under $H_0,$ we have that $Z_i = (X_i - \mu)/\sigma_0$ is standard normal
so that $Z_i^2\sim \mathsf{Chisq}(1).$ Thus $Q/\sigma_0^2 \sim\mathsf{Chisq}(n).$
We can use printed chi-squared tables or software to find the number $b$ with
$P(Q/\sigma_0^2 > b) = \alpha.$ What is $c$ in terms of $b?$

Addendum: Below we simulate a million samples, each with $n = 5$
observations from $\mathsf{Norm}(100, 15).$ 
In the histogram at left
we show that a histogram of simulated values of $(n-1)S^2/\sigma^2$
(called dist.1) fits $\mathsf{Chisq(4)}$ (solid curve), but not 
$\mathsf{Chisq(5)},$ as mentioned in a Comment. Informally, one says that one degree of freedom
has been 'lost' estimating unknown $\mu.$ 
By contrast, the histogram
at right illustrates that simulated values of $nQ/\sigma^2$ (called 'dist2`) do fit $\mathsf{Chisq}(5),$ as in the discussion above. 

R code for the simulation is shown below (but not the code to make the histograms).
set.seed(409)  # for reproducibility
m = 10^6;  n = 5;  mu = 100;  sg = 15
x = rnorm(m *n, mu, sg);  MAT = matrix(x, nrow=m)
s.sq = apply(MAT, 1, var)
dist.1 = (n-1)*s.sq/sg^2 col="brown", lwd=2, lty="dotted")
q = rowSums((MAT - mu)^2)
dist.2 = q/sg^2

