0
$\begingroup$

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.

Improve this question
$\endgroup$
7
  • 3
    $\begingroup$ while retaining the (unstated but implied) assumption of iid normality? $\endgroup$
    – Glen_b
    Commented Mar 30, 2017 at 4:39
  • 2
    $\begingroup$ 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
    Commented Mar 30, 2017 at 5:12
  • 1
    $\begingroup$ Also it seems that in your question you want $\alpha$ = 0.05. $\endgroup$
    – Michael R. Chernick
    Commented Mar 30, 2017 at 6:13
  • $\begingroup$ @AaronDefazio This is very general question, so there is no exemplary data. $\endgroup$
    – user67275
    Commented Apr 1, 2017 at 6:51
  • 3
    $\begingroup$ 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
    Commented Apr 1, 2017 at 8:35

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.