When σ2 is known

If \(\{X_1, X_2,\dots, X_n\}\) is a random sample from a \(\NormalDistn(\mu, \sigma^2)\) distribution and \(\sigma^2\) is known, then

\[ \overline{X} \;\;\sim\;\; \NormalDistn(\mu,\;\frac{\sigma^2}{n}) \]

and

\[ \frac{\overline{X} - \mu}{\diagfrac{\sigma}{\sqrt{n}}} \;\;\sim\;\; \NormalDistn(0,\;1) \]

is a pivot for the parameter \(\mu\). From this, we can find a 95% confidence interval for \(\mu\)

\[ \overline{X} -1.96 \frac{\sigma}{\sqrt{n}} \;\;\lt\;\; \mu \;\;\lt\;\; \overline{X} +1.96 \frac{\sigma}{\sqrt{n}} \]

Unknown σ2

When \(\sigma^2\) is an unknown parameter, the quantity \(\displaystyle \frac{\overline{X} - \mu}{\diagfrac{\sigma}{\sqrt{n}}}\) cannot be used as a pivot since it involves both \(\mu\) and \(\sigma^2\).

Pivot for μ when σ² is unknown

If \(\overline{X}\) and \(S^2\) are the mean and variance of a random sample of size \(n\) from a \(\NormalDistn(\mu, \sigma^2)\) distribution,

\[ \frac{\overline{X} - \mu}{\diagfrac{S}{\sqrt{n}}} \;\;\sim\;\; \TDistn(n-1 \text{ df}) \]

is a pivot for the parameter \(\mu\).

(Proved in full version)