# Standard error of sampling distribution mean vs standard deviation of the sampling distribution?

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?

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, $$s^2= \frac{\Sigma{(X-M)^2}}{N-1}$$
• 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.