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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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Richard Ulrich

Posts: 2,961
Registered: 12/13/04
Re: Which sample variance should I choose?
Posted: Aug 31, 2011 2:32 PM
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On Wed, 31 Aug 2011 06:40:22 -0700 (PDT), leoldv <>

>On Aug 31, 1:06 am, Steven D'Aprano <steve
>> 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 also
>> 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?
>> Thanks in advance,
>> --
>> Steven

>If you know mu, the population mean, divide by n.

That's a pretty complete answer.

But if you are cynical about the pop. mean being the pop. mean,
then I suppose you could offer two answers, one using the
observed mean. If you want tests using ANOVA, you use an
unbiased estimate of the variance.

- To a question posted later: No, I don't see any place or
justification for using the population mean in a computation
with (n-1) as denominator.

Rich Ulrich

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