Confidence interval for $\sigma^2$ for linear regression Consider the linear regression $Y_i=\alpha_0+\beta_0 X_i+\epsilon_0$, where $i=1,2,...,n$, and $\epsilon \sim N(0, \sigma^2)$.
$(\hat{\alpha},\hat{\beta})$ estimate $(\alpha_0,\beta_0)$ and are found by minimizing $\sum_{i=1}^n (Y_i-\alpha-\beta X_i)^2.$
We get that $\hat{\alpha}=\bar{Y}-\hat{\beta} \bar{X}$ and $\hat{\beta}=\frac{\sum_{i=1}^n(X_i-\bar{X})Y_i}{\sum_{i=1}^n(X_i-\bar{X})^2}$.
Let $X_1,X_2,...,X_n$ be i.i.d $N(0,\sigma^2)$.
I have to show that $\sum_{i=1}^n X_i^2$ can be used to construct a pivot to find a confidence interval for $\sigma^2$, not necessarily the shortest confidence integral.
I know that a pivot has to have $N(0,1)$ distribution. 
Do I just have to show that  a form of $\sum_{i-1}^n X_i^2$ has $N(0,1)$ distribution using the fact that $X_i \sim N(0,\sigma^2)$?
I'm not sure how to continue, any help is appreciated.
 A: I think there is a typo in your formula, it should be
$$\sum_{i=1}^n X_i^2$$ not $\sum_{i-1}^n X_i^2$
Since $X_i$s are i.i.d $N(0,\sigma^2)$,
$$\frac{X_i}{\sigma}\sim N(0,1)$$
Therefore,
$$\frac{X_i^2}{\sigma^2}\sim \chi^2_{(1)}$$
Therefore,
$$\frac{X_1^2}{\sigma^2}+\frac{X_2^2}{\sigma^2}+...+\frac{X_n^2}{\sigma^2}\sim \chi^2_{(n)}$$
i.e $$\frac{\sum_{i=1}^n X_i^2}{\sigma^2}\sim \chi^2_{(n)}$$
Therefore, 
$$P(\chi^2_{(n)}(\frac{\alpha}{2})<\frac{\sum_{i=1}^n X_i^2}{\sigma^2}<\chi^2_{(n)}(1-\frac{\alpha}{2}))=1-\alpha$$
$$\Leftrightarrow P(\frac{1}{\chi^2_{(n)}(1-\frac{\alpha}{2})}<\frac{\sigma^2}{\sum_{i=1}^n X_i^2}<\frac{1}{\chi^2_{(n)}(\frac{\alpha}{2})})=1-\alpha$$
$$\Leftrightarrow P(\frac{\sum_{i=1}^n X_i^2}{\chi^2_{(n)}(1-\frac{\alpha}{2})}<\sigma^2<\frac{\sum_{i=1}^n X_i^2}{\chi^2_{(n)}(\frac{\alpha}{2})})=1-\alpha$$
Therefore, the $(1-\alpha)\%$ confidence intervale is 
$$(\frac{\sum_{i=1}^n X_i^2}{\chi^2_{(n)}(1-\frac{\alpha}{2})},\frac{\sum_{i=1}^n X_i^2}{\chi^2_{(n)}(\frac{\alpha}{2})})$$
Note $\chi^2_{(n)}(\frac{\alpha}{2})$ is just quantile function of the $\chi^2_{(n)}$ distribution.
