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The unbaised variance of a population from a single sample ($n\ll N$), $s^2=\sum_i(x_i-\bar{x})/(n_x-1)$. ($n_x$ being the sample size.) However, the standard error of the mean: $SE=s/\sqrt{n_m}$, not $s/\sqrt{n_m-1}$. But the $n_m$ samples that go into the $SE$ are merely samples of the whole powerset population of all possible samples. In fact, it's commonly the case that the number of samples is much less than the size of any given sample, that is: $n_m \ll n_x$, so shouldn't an unbiased estimator of $SE$ also be $s/\sqrt{n_m-1}$, or would that be "double-unbiasing", or somesuch?

jackisquizzical
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  • Is either flavor of standard error an unbiased estimator? Taking the square root of unbiased $S^2$ results in an $S$ that, amazingly, is a biased estimator of standard deviation. – Dave Feb 26 '23 at 21:24
  • The standard error is calculated from a single sample as well. – Michael M Feb 26 '23 at 21:34
  • @MichaelM We usually calculate statistics based on only a single sample, so it is not clear what your gripe is. Could you please elaborate? – Dave Feb 26 '23 at 21:38
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    I think my comment can be translated as: "reading the question over and over again and still don't understand it"... – Michael M Feb 26 '23 at 22:11
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    Note that the standard error of the sample mean does not involve calculating a "sample mean of the sample mean"; we lose a degree of freedom when constructing an unbiased estimate of the variance due to having to estimate the sample mean, but we don't when re-using that estimate to calculate the standard error of the sample mean. – jbowman Feb 26 '23 at 22:53
  • @Dave To be somewhat clearer, I should have said: "...But the $_$ SAMPLE MEANS that go into the ...". – jackisquizzical Feb 27 '23 at 00:18
  • @jbowman Ah. Right, so we lost 1df in computing the mean for the SSE, but don't have to do it again in order to compute the SE. Checking my understanding: If we were to use multiple sample means, then we could compute an unbiased SE using $n-1$, but that's just computing a sample $s$ of a sample of sample means, which is a different computation. – jackisquizzical Feb 27 '23 at 00:24
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    Exactly so. If you did that, though, you'd be better off combining all the data into one big sample, calculating one standard deviation for the sample, and using that as the basis for your SE calculation. – jbowman Feb 27 '23 at 00:26
  • I don't know why this was marked as duplicate. It's not at all a duplicate of the one that it was linked to, which is only about the sample sd, neither about se nor samples of sample means. (I don't know how to cast a vote to reopen, or at least to remove the dup marker.) – jackisquizzical Feb 28 '23 at 15:24

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