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In this resource, towards the bottom, the authors write:

The next step is to estimate the standard error of the mean. If we knew the population variance, we could use the following formula, $sigma_M = \frac{\sigma}{\sqrt{N}}$. Instead we compute an estimate of the standard error (sM): $s_M = \frac{s}{\sqrt{N}}$.

I'm confused here. We don't know the population variance, thus we must estimate the sampling distribution variance, which follows the formula $s_M = \frac{s}{\sqrt{N-1}}$.

Did the authors make an error?

Or is the formula, $s_M = \frac{s}{\sqrt{N-1}}$, only applicable when we are trying to estimate the sampling distribution variance rather than the standard error of the sampling distribution mean?

Edit: Adding a second source,

If the variance in a sample is used to estimate the variance in a population, then the previous formula underestimates the variance and the following formula should be used, $s^2= \frac{\Sigma{(X-M)^2}}{N-1}$

So it seems that my proposal in italics is correct. The Bessel's correction (N-1) is only relevant when trying to estimate the population variance given a sample; it is not relevant when estimating the standard error of anything, such as standard error of the sampling distribution mean.

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  • $\begingroup$ You have confused formulas for the standard deviation with formulas for the standard error. The fraction $1/\sqrt N$ comes from computing the variance of the mean of $N$ iid sample values. It has nothing to do with estimation, bias, Bessel's correction, etc. $\endgroup$
    – whuber ♦
    May 31 at 13:56
  • $\begingroup$ @whuber, when is Bessel's correction applicable? $\endgroup$
    – jbuddy_13
    May 31 at 14:05
  • $\begingroup$ It doesn't apply to computing standard errors from standard deviations. If you're still curious about Bessel's correction, see our thread about it. $\endgroup$
    – whuber ♦
    May 31 at 14:29

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