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I'm just unclear what $\sigma$ represents here. Am looking for a confidence interval where $\mu=\sigma$? or looking for the confidence interval where $\sigma=\sqrt{\sigma^2/n}$?? any hints on where to start this?

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up vote 1 down vote accepted

$\sigma$ is the standard deviation of your initial sample $X_1,\ldots,X_n$, and $S$ is the usual estimator of the standard deviation of a sample where also the mean is unknwon. What you're told is that $$ S\approx \mathcal{N}\left(\sigma,\frac{\sigma^2}{2n}\right) $$ where $\approx$ means approximately distributed as. So yes, the mean of $S$ is $\sigma$ where $\sigma$ is the standard deviation of $X_i$. Note that the above implies that $$ \frac{S-\sigma}{\sigma}=\frac{S}{\sigma}-1\approx\mathcal{N}\left(0,\frac{1}{2n}\right) $$ which enables you to arrive at that confidence interval.

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