Estimating confidence Interval for unknown Variance, Normal distribution 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!
 A: $X_1,\ldots,X_n$ 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.
A: I believe you have the denominators switched.  Also, say $c_1 = \chi^2_{1, 1-\alpha/2} $ and $c_2 =\chi^2_{1, \alpha/2 } $
then  for (c), the expected width of the interval is 
  $$ \mathbb{E}  \left[   n (\bar{X} - \mu )^2 ( \frac{1}{c_2}  - \frac{1}{c_1}) \right] =  
n ( \frac{1}{c_2}  - \frac{1}{c_1}) \times \mathbb{E}(\bar{X} - \mu )^2  =  
 n ( \frac{1}{c_2}  - \frac{1}{c_1}) \; \; \times \frac{\sigma^2}{n} = ( \frac{1}{c_2}  - \frac{1}{c_1}) \sigma^2 $$
since   $ \quad \quad   \frac{n}{\sigma^2} ( \bar{X} - \mu ) \sim \chi^2_1 $ 
and thus  $ \quad  \mathbb{E}(\bar{X} - \mu )^2 =  \frac{\sigma^2}{n}  $
strange the expected width does not depend on $n$...
