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Topic: Which sample variance should I choose?
Replies: 7   Last Post: Sep 5, 2011 3:50 AM

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Paige Miller

Posts: 361
Registered: 12/7/04
Re: Which sample variance should I choose?
Posted: Aug 31, 2011 10:52 AM
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On Aug 31, 1:06 am, Steven D'Aprano <steve
+comp.lang.pyt...@pearwood.info> wrote:
> The population variance is given by:
>
> ?^2 = ?(x - µ)^2 / n
>
> with µ = population mean, the summation being over all the x in the
> population.
>
> (for brevity, I haven't attempted to show subscript-i on the x).
>
> If you don't have the entire population, you can estimate the variance with
> the sample variance:
>
> s^2 = ?(x - m)^2 / n  (Eq. 1)
>
> where m = sample mean (usually written as x bar), and the sum is over all
> the x in the sample. A second estimator is:
>
> s^2 = ?(x - m)^2 / (n-1)  (Eq. 2)
>
> which some people prefer because it is unbiased (that is, the average of all
> the possible sample variances equals the true population variance if you
> use the (n-1) version).
>
> See alsohttp://mathworld.wolfram.com/SampleVariance.html
>
> I have a set of data with an (allegedly) known population mean µ but an
> unknown ?^2. I wish to estimate ?^2. Under what circumstances should I
> prefer Eq.1 over Eq.2?
>
> Or should I ignore the sample mean altogether, and plug the known population
> mean into one of the two equations? I.e.:
>
> s^2 = ?(x - µ)^2 / n      (Eq. 3)
> s^2 = ?(x - µ)^2 / (n-1)  (Eq. 4)
>
> Under what circumstances should I prefer each of these four estimators of
> ?^2 and what are the pros and cons of each?



There is no answer until you tell us what you are planning to use the
variance for.

--
Paige Miller
paige\dot\miller \at\ kodak\dot\com



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