Estimating the population variance [duplicate]

Question

I'm trying to understand the emphasized phrase in the following passage:

The usual method of determining the probability that the mean of the population lies within a given distance of the mean of the sample is to assume a normal distribution about the mean of the sample with a standard deviation equal to $s/\sqrt{n}$, where $s$ is the standard deviation of the sample, and to use the tables of the probability integral.

If I understand the phrase in question correctly, the author claims that "the usual method" uses $s/\sqrt{n}$ as an estimator for the population's standard deviation, or equivalently, that it uses $s^2/n$ as an estimator for the population's variance, where

$$s^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \overline{x})^2$$

and $\overline{x}$ is the sample mean:

$$\overline{x} = \frac{1}{n}\sum_{i=1}^n x_i$$

If so, I find this confusing. I thought that $s^2$, as defined above, not $s^2/n$, is the usual estimator of the population variance.

Am I misunderstanding something?

Answer 1

The distinction is between the standard deviation of a random variable and the standard deviation of the sample average of that random variable. Note that the sample average is itself a random variable, it varies based on sampling variation. For reference, the standard deviation of a sample statistic (such as an average) is often called its standard error.

In other words imagine you have a random variable $X$, maybe it is the height of a randomly selected individual. Now you take a sample of size $N$ and compute the sample's average height. If you repeat this sampling process a million times, each repetition using a sample of size $N$ to compute a sample average height, then you will have a million sample averages. These averages are a random variable with a probability distribution (side note: the Central Limit Theorem tells us it will be essentially Normal no matter what distribution $X$ is drawn from). The standard deviation of the distribution of sample averages is called its standard error and given by the formula you cite.