Estimating confidence Interval for unknown Variance, Normal distribution

Question

I've been stuck with this question for a while:

I've learnt how to use t-distribution to estimate CIs for an unknown variance, but I'm unsure how that applies to this situation.

Any help would be greatly appreciated.

Thank you!

Answer 1

$X_1,\ldots,X_n$ are i.i.d. $\mathcal N(\mu,\sigma^2)$ random variables and $\bar X=n^{-1}\sum_{i=1}^nX_i$. The distributions and confidence intervals are as follow.

(a) $$ \frac1{\sigma^2}\sum_{i=1}^n(X_i-\bar X)^2\sim\chi_{n-1}^2 $$ and $$ \Pr\biggl(\frac{\sum_{i=1}^n(X_i-\bar X)^2}{\chi_{n-1,\alpha/2}^2}\le\sigma^2\le\frac{\sum_{i=1}^n(X_i-\bar X)^2}{\chi_{n-1,1-\alpha/2}^2}\biggr)=1-\alpha. $$ (b) $$ \frac1{\sigma^2}\sum_{i=1}^n(X_i-\mu)^2\sim\chi_n^2 $$ and $$ \Pr\biggl(\frac{\sum_{i=1}^n(X_i-\mu)^2}{\chi_{n,\alpha/2}^2}\le\sigma^2\le\frac{\sum_{i=1}^n(X_i-\mu)^2}{\chi_{n,1-\alpha/2}^2}\biggr)=1-\alpha. $$

(c)

$$ \frac{n(\bar X-\mu)^2}{\sigma^2}\sim\chi_1^2 $$ and

$$ \Pr\biggl(\frac{n(\bar X-\mu)^2}{\chi_{1,\alpha/2}^2}\le\sigma^2\le\frac{n(\bar X-\mu)^2}{\chi_{1,1-\alpha/2}^2}\biggr)=1-\alpha. $$

I hope this helps.