I am sampling from a parameter with unknown distribution. I would like to calculate a 95% CI for the standard deviation of the sample.

@cardinal provides a nice general solution for calculating a CI in his [answer] to my previous question, Calculating required sample size, precision of variance estimate?. And @erik-p provides an estimate of the standard deviation of the variance of the sample.

However, in order to calculate the 95%CI for the sample variance, it seems that I need to know the distribution of the sample variance. Is it possible to calculate such an estimate without knowing the distribution from which the sample was taken?

A related question is Reference for $\mathrm{Var}[s^2]=\sigma^4 \left(\frac{2}{n-1} + \frac{\kappa}{n}\right)$?

  • $\begingroup$ @gung is that better? $\endgroup$
    – Abe
    Jun 6, 2012 at 14:54
  • $\begingroup$ That does help, thanks. I'm going to interpret the top sentence as 'I am trying to estimate a population parameter from a sample taken from a population where the distribution is unknown'. We don't usually speak of "sampling from a parameter". Sorry to be pedantic. $\endgroup$ Jun 6, 2012 at 15:30
  • $\begingroup$ @gung thanks. the context is that the parameter is a posterior predictive estimate of a model output. Each 'sample' takes a few hours for the model to run. $\endgroup$
    – Abe
    Jun 6, 2012 at 15:33

1 Answer 1


Without knowing the population distribution you cannot know the exact distribution of the sample variance. However each squared deviation from the mean has the same distribution and they are averaged and only weakly dependent. So for large n the sample variance is approximately normally distributed with mean σ$^2$ and variance as given above. So you can use the normal approximation to get an approximate 95% confidence interval. On the other hand if n is small and the CLT cannot be applied you can generate bootstrap confidence intervals. But keep in mind that in my paper with LaBudde I showed that for highly skewed distributions such as the lognormal the bootstrap intervals will severely undercover in the small sample size situation. But it will work fine for symmetric distributions and distributions that have mild skew.

  • 4
    $\begingroup$ In general, it helps if you provide a full reference or, preferably, a link when you mention a paper. $\endgroup$
    – MånsT
    Jun 6, 2012 at 16:25
  • $\begingroup$ I think I have referenced my paper previously on another question. It is Chernick, M. R. and LaBudde, R. A. (2010) Revisiting qualms about bootstrap confidene intervals. Am. J. Math. Manag. Sci. 29, 437-456. It can also be found discussed in our book Chernick, M. R. and LaBudde, R. A. (2011) An Introduction to Bootstrap Methods with Applications to R. pp. 90-94. Wiley, Hoboken. $\endgroup$ Jun 6, 2012 at 16:47
  • $\begingroup$ can you define "large n"? It is reasonable for me to get $n=500$ - does this justify the assumption of normality? Also, when you say "the variance as given above", do you mean $\sigma^4 \left(\frac{2}{n-1} + \frac{\kappa}{n}\right)$? If so, could you help clarify some confusion in the linked question as to whether the equation is correct? $\endgroup$
    – Abe
    Jun 6, 2012 at 16:50
  • $\begingroup$ @Abe I think others have addressed the formula with references. As far as what is large enough, that depends on the distribution. Some avergaes converge to normality very rapidly and others slowly. 500 is usually large enough. $\endgroup$ Jun 6, 2012 at 16:56

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.