```Date: Sep 1, 2011 4:30 AM
Author: Steven D'Aprano
Subject: Re: Which sample variance should I choose?

On Thu, 1 Sep 2011 04:34 am Paul wrote:> If you know the population mean, then>  s2_1 = ?(x - µ)^2 / n> is unbiased.> > If you don't know the population mean, then>  s2_2 = ?(x - m)^2 / (n-1)> is unbiased, while>  s2_3 = ?(x - m)^2 / n> is biased but nevertheless more accurate than s2_2.Thanks Paul, that's exactly the sort of thing I'm looking for.Not that I don't believe you :) but if you also have a reference (especiallyone that's online) that would be really helpful.Thanks to everyone who answered.Steven.> None of these distinctions matters if n is reasonably large> > On Aug 31, 1:02 pm, Steven D'Aprano <steve> +comp.lang.pyt...@pearwood.info> wrote:>> Paige Miller wrote:>> > On Aug 31, 1:06 am, Steven D'Aprano <steve>> > +comp.lang.pyt...@pearwood.info> wrote:>> [...]>> >> 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.>>>> That's exactly what I'm trying to find out. Under which circumstances>> should I prefer one method over the others?>>>> I don't actually have a *specific* usage in mind, other than answering>> the question "what's the sample variance of this data?" But I would like>> to understand why somebody might choose one version or another.>>>> E.g.>>>> the unbiased sample variance (divide by n-1) has the advantage that, on>> average, it will equal the population variance (provided certain>> assumptions hold, such as sampling with replacement);>>>> but the unbiased sample variance also has a larger spread, so although it>> is the most accurate on average, there's a chance that it will be much>> further off. The biased sample variance (divide by n) is less accurate>> but more precise (the results are clustered more closely together, so the>> chances of getting a result that is *way* off is much reduced);>>>> etc. Or at least, this is what I *think* is the case.>>>> I'm not even sure that it is mathematically valid to substitute µ into>> the sample variance formulae instead of the sample mean. I can't see why>> it wouldn't be, but I'm not sure.>>>> For reference, here's the suggested sample variance formulae again:>>>> s^2 = ?(x - m)^2 / n      (Eq. 1) Biased, using sample mean>> s^2 = ?(x - m)^2 / (n-1)  (Eq. 2) Unbiased, using sample mean>> s^2 = ?(x - µ)^2 / n      (Eq. 3) Biased, using population mean>> s^2 = ?(x - µ)^2 / (n-1)  (Eq. 4) Unbiased, using population mean>>>> where the sums are over each x in the sample.>>>> -->> Steven-- Steven
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