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When we have an estimate of the variance $s^2$ obtained from $n$ observations, $x_1, x_2, \dots, x_n$, we can calculate a 95% confidence interval of the true variance

$$\left( \frac{(n-1)s^2}{\chi^2_{\alpha/2}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}} \right)$$

I want to get narrower confidence interval.

The best thing to do is to increase $n$, or we may use larger $\alpha$.

Are there any other available methods?

Any idea is appreciated.

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    $\begingroup$ while retaining the (unstated but implied) assumption of iid normality? $\endgroup$ – Glen_b Mar 30 '17 at 4:39
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    $\begingroup$ Do you have a particular problem in mind here? Information on the structure of your data may help in answering your question. For instance, a density plot would help. Otherwise, there is no other ways of getting a narrower confidence interval. $\endgroup$ – AaronDefazio Mar 30 '17 at 5:12
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    $\begingroup$ Also it seems that in your question you want $\alpha$ = 0.05. $\endgroup$ – Michael Chernick Mar 30 '17 at 6:13
  • $\begingroup$ @AaronDefazio This is very general question, so there is no exemplary data. $\endgroup$ – user67275 Apr 1 '17 at 6:51
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    $\begingroup$ Your question is based on a mistaken premise (made explicit in the comments) -- it is only the case that the formula you give is a $1-\alpha$ interval for the variance if the original distribution sampled from is iid normal. $\endgroup$ – Glen_b Apr 1 '17 at 8:35

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