Finding size of a test I'm doing a hypothesis test question but I am stuck.
Here is the question.

Suppose $X_1,\ldots, X_n$ are iid $N(0,\sigma^2). $ Suppose the hypotheses being tested are 
  $$H_0=\sigma^2\geq \sigma_0^2 \quad\text{vs}\quad H_A:\sigma^2\lt\sigma^2_0.$$
  Show that the power function of the test based on the rejection criterion $\sum{x_i}^2\lt c$ is given by 
  $$Q(\sigma^2)=\Pr\left\{\chi_n^2\lt\frac{c}{\sigma^2}\right\}.$$

I was able to get that part out but I'm stuck on the following part:

Find $c$ such that the size of the test is 0.05.

How can I determine $c$ if I don't know $\sigma^2$?  Can anyone point me in the right direction?
 A: You already did the hard part.  Your solution must have involved the following reasoning.  Under the assumptions, the test statistic
$$T = X_1^2 + X_2^2 + \cdots + X_n^2$$
is distributed as $\sigma^2$ times a $\chi^2(n)$ variable.  The chance that such a variable is in the rejection region $\mathcal{R}=(-\infty, c)$ equals the chance that a $\chi^2(n)$ variable is less than $c/\sigma^2$, given by the $\chi^2(n)$ distribution function $F$ as
$$\Pr(T \in\mathcal{R}) =  F_{\chi^2(n)}\left(\frac{c}{\sigma^2}\right).$$
Now let's address the easy part about the test size.  Under the null hypothesis $H_0: \sigma^2 \ge \sigma_0^2,$ the chance of rejection is largest when $\sigma^2 = \sigma_0^2.$  (This is intuitively obvious, because larger values of $\sigma$ locate most of the probability on larger values, making it less and less likely to observe a small value of the test statistic.)  By definition of "size" you want this not to exceed $0.05$:
$$0.05 \ge F_{\chi^2(n)}\left(\frac{c}{\sigma_0^2}\right).$$
Since $F$ (when applied to positive numbers) is continuous and increasing, the unique solution is found by applying its inverse, aka "percentage point function" or "quantile function,"
$$\sigma_0^2\, F^{-1}_{\chi^2(n)}(0.05) = c.$$
You see, although you don't know $\sigma^2,$ you may compute as if you were in the worst case for the null hypothesis (where $\sigma^2$ is equal to $\sigma_0^2$), allowing you to produce a definite solution.

Let's return to that appeal to intuition.  I claimed that as $\sigma^2$ increases, the area to the left of a fixed value $c$ beneath the distribution function of $T$ grows smaller.  Consider two such density functions: one corresponding to a particular value of $\sigma^2,$ shown in red, and another corresponding to a larger value $\lambda \sigma^2,$ shown in gold:

The gold curve is the same as the red curve, but it has been stretched by a factor $\lambda \gt 1$ horizontally--and therefore had to be shrunk by the same factor vertically to preserve its unit area.  Because this process preserves corresponding areas everywhere, the area to the left of $c$ (shown in light red) beneath the first curve equals the area to the left of $c\lambda$ (shown in light gold) beneath the second curve.  Since $c\lt c\lambda$, the area under the second (gold) curve to the left of $c$ (where the two shaded regions overlap) must be smaller than the area to the left of $c\lambda,$ as claimed.
It may be worth remembering this argument because it applies to literally any distribution of non-negative numbers: it's not special to the $\chi^2$ distributions.
A: For the power function, we want the condition of rejecting the null hypothesis. 
Note that the sum of normal r.v's are distributed by $\chi_n^2$. Let's denote $X\sim \mathcal{N}(0,\sigma^2). $
$$
 Q(\sigma^2) = Pr\left(\sum_{i=1}^n X<\frac{c}{\sigma^2}\quad| \quad \sigma^2 <\sigma_0^2\right) $$
Note that $X_1 \sim \mathcal{N}(0,\sigma^2) \Rightarrow X_1 +X_2 \sim \mathcal{N}(0,2\sigma^2) \Rightarrow \sum X_i \sim \mathcal{N}(0,n\sigma^2).$
So, let's standardise this, by dividing the inequality by $\sqrt{n\sigma^2}$, so that:
\begin{align}Q(\sigma^2) &= Pr\left(\frac{\sum_{i=1}^n X}{\sigma\sqrt{n}}<\frac{c}{\sqrt{n\sigma^2}\sigma^2}\right) \\
&= Pr \left(Z<\frac{c}{\sigma^3\sqrt{n}}\right)
\end{align}
Therefore, you want to find $Pr\left(|Z| < \frac{c}{\sigma^3\sqrt{n}}\right)=\alpha = 0.05$. 
$$ \iff Pr(Z<\frac{c}{\sigma^3\sqrt{n}})-Pr(Z<-\frac{c}{\sigma^3\sqrt{n}}) = 0.05 \iff 2pr(Z<\frac{c}{\sigma^3\sqrt{n}})=0.05+1$$
This happens if $\frac{c}{\sigma^3\sqrt{n}}=\Phi^{-1}(2^{-1}\times 1.05)$, where $\Phi$ is the cdf of the standard normal. 
Conclude that $c=\sigma^3\sqrt{n}\Phi^{-1}(0.525)$.
