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Supposed we have $n = 15$ independent samples $X_1, X_2, ..., X_n$ from distribution $N(\mu, \sigma)$. Sample mean $\bar{X} = 2.4$ and sample variance $\hat{\sigma^2} = 0.55$

What's the 95% confidence interval of $\mu$ and $\sigma$?

I understand the 95% confidence interval of $\mu$ -- I can estimate the standard error by $\hat{\sigma} / \sqrt{n}$, then I can obtain the CI. But how do I calculate the CI of $\sigma$?

P.S. As you can tell this is more of an homework-ish question. I am self-studying MIT 18.443 -- Statistics for Applications and this is on one of the practice exams. Unfortuantely they don't have solutions. In addition to answering this question, if someone can recommend another open course with solutions to homework/exams that would be appreciated too.

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  • $\begingroup$ @gung It doesn't quite look like a duplicate to me. There's a difference between obtaining a sampling variance (the thrust of the other question) and erecting a confidence interval. $\endgroup$ – whuber Mar 10 '15 at 20:28
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    $\begingroup$ These are likely to be useful: stats.stackexchange.com/q/76444/1934 and stats.stackexchange.com/q/7004/1934 $\endgroup$ – Wolfgang Mar 10 '15 at 20:32
  • $\begingroup$ @whuber, you're right, but I thought the answer already existed on the site. Wolfgang's links are more on-target here. $\endgroup$ – gung - Reinstate Monica Mar 10 '15 at 20:34
  • $\begingroup$ @gung Yes, those links just about do it. I suppose we only have to make the case that a CI for the SD can be derived in the obvious way from the CI for the variance. $\endgroup$ – whuber Mar 10 '15 at 20:47