0
$\begingroup$

I know how to estimate population standard deviation using a chi square distribution, but I don't know why it works. I'd like to have an intuition for the why.

I've tried googling around. I always find descriptions of the how, not the why. My guess as to what's going on:

  • The distribution of sample SDs is normally distributed (sorta makes sense, but I don't really understand why).
  • If they're normally distributed, when you square it, it becomes a chi square distribution with 1 degree of freedom (this makes sense to me). Since SD squared is Var, we have a chi square distribution of Vars.
  • From there, once we have a distribution, we can say that X% of the sample Vars are within a range (U, L).
  • There's some sort of adjustment, because sample Var is a biased estimator of population Var.
  • Where do multiple degrees of freedom come in to play?
$\endgroup$
3
  • $\begingroup$ "I know how to estimate population standard deviation using a chi square distribution" — Exactly what procedure are you referring to? Are you asking a question about Bessel's correction? $\endgroup$ Nov 25 '16 at 1:34
  • $\begingroup$ Squareroot((n-1)(s^2)/chi value) $\endgroup$ Nov 25 '16 at 2:40
  • $\begingroup$ See here and a bit lower down $\endgroup$
    – Glen_b
    Nov 25 '16 at 4:07
0
$\begingroup$

No I don't believe your rationale is correct. If X has a normal distribution then its sample variance is proportional to a chi square distribution with n-1 degrees of freedom where n is the sample size. So the chi-square distribution can be used to draw inferences about the population variance and hence also the standard deviation. But if X is not normally distributed there is no reason to use the chi-square distribution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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