Re: Compute the confidence interval of the standard deviation

Thanks for the reply, comments inline.

Herman Rubin wrote:
In article <1166122000.459047.152020@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Dani Camps <danicamps81@xxxxxxxxx> wrote:
Hi all,

I would need some help to compute the 95% confidence interval of the
standard deviation, in principle nothing can be said about the
underlaying population. Any hint?

If you are willing to assume that the fourth moment exists
and the tails are not too bad, the central limit theorem
applied to the standard deviation should work reasonably
well. Without that, the sample size needed for a good
interval will be immense.

Most statistics textbooks use the chi-squared distribution
for this purpose, but this can be bad if the fourth moment
is not that of the normal distribution.


If I understand correctly, you mean that I can assume that S^2=1/(N-1)
sum (X-mean(X))^2 should have a distribution that tends to be normal
because of the central limit theorem. But from here on I am a bit lost,
how the mean of the normal distribution that the former expression
tends to is related with the variance of my population ? is it exactly
the same ? And how the deviation of that normal distribution is related
with the deviation I am trying to estimate ?

I don't know about the chi-squared distribution, could you please give
a bit more details or point e to a good reference to know how to
continue?

Thanks

Daniel

--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@xxxxxxxxxxxxxxx Phone: (765)494-6054 FAX: (765)494-0558

.

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