The t-distribution

Often $\sigma$ is an unknown value, much like $\mu$. It can be shown that replacing $\sigma$ with $s$ results in a $t$-distribution, so we often use that.

The t-distribution in terms of sample standard deviation

To show this, note the alternative form for ${\chi }^2$:

$$\begin{array}{r}{\chi }^2\sim \sum z^2\\ {\chi }^2\sim n\frac{s^2}{{\sigma }^2}\\ ⟹s^2\sim n{\sigma }^2{\chi }^2\end{array}$$

Applying that to the typical standardization formula:

$$\begin{array}{rl}\frac{x-{\mu }_0}{s}& =\frac{x-{\mu }_0}{s}\\ & =\frac{x-\mu }{\sqrt{n{\sigma }^2{\chi }^2}}\\ & =\frac{x-\mu }{\sigma \sqrt{n{\chi }^2}}\\ & =\frac{\left(\frac{x-\mu }{\sigma }\right)}{\sqrt{n{\chi }^2}}\\ & =\frac{z}{\sqrt{n{\chi }^2}}\end{array}$$

This right hant side is known as the t-distribution $t\sim \frac{z}{\sqrt{n{\chi }^2}}$, this can be verified with R:

N = 3000
n = 5
qqplot(
  rt(N, n - 1),
  rnorm(N) / sqrt(n * rchisq(N, n - 1))
)

Sampling Distributions

Recall from Biased Estimators: Variance of a Sample Mean that:

$${\sigma }_{\bar{x}}=\frac{\sigma }{\sqrt{n}}$$

Likewise:

$$s_{\bar{x}}=\frac{s}{\sqrt{n}}$$