Normal-theory confidence interval for the mean

Proof

Assume $X_1,\dots,X_n$ are i.i.d. $\N(\mu,\sigma^2)$ and define \begin{equation*} \bar X = \frac{1}{n}\sum_{i=1}^n X_i, \qquad s^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2. \end{equation*}

Since $X_i\sim\N(\mu,\sigma^2)$, \begin{equation*} \bar X \sim \N\left(\mu,\frac{\sigma^2}{n}\right). \end{equation*}

Hence \begin{equation*} Z = \frac{\bar X - \mu}{\sigma/\sqrt{n}}\sim\N(0,1). \end{equation*}

For normal samples, \begin{equation*} \frac{(n-1)s^2}{\sigma^2}\sim\chi^2_{n-1}, \end{equation*} and $\bar X$ and $s^2$ are independent.

Let \begin{equation*} U=\frac{(n-1)s^2}{\sigma^2}. \end{equation*}

Consider \begin{equation*} T = \frac{Z}{\sqrt{U/(n-1)}} = \frac{\bar X - \mu}{s/\sqrt{n}}. \end{equation*}

By definition of the Student $t$ distribution, \begin{equation*} T \sim t_{n-1}. \end{equation*}

Let $t_{1-\alpha/2,\,n-1}$ denote the $1-\alpha/2$ quantile of $t_{n-1}$. Then \begin{equation*} P\!\left(- t_{1-\alpha/2,\,n-1} \le \frac{\bar X - \mu}{s/\sqrt{n}} \le t_{1-\alpha/2,\,n-1} \right) = 1-\alpha. \end{equation*}

Rearranging, \begin{equation*} P\!\left(\bar X - t_{1-\alpha/2,\,n-1}\frac{s}{\sqrt{n}} \le \mu \le \bar X + t_{1-\alpha/2,\,n-1}\frac{s}{\sqrt{n}} \right) = 1-\alpha. \end{equation*}

Therefore, a two-sided $(1-\alpha)$ confidence interval for $\mu$ is \begin{equation*} \bar X \pm t_{1-\alpha/2,\,n-1}\frac{s}{\sqrt{n}}. \end{equation*}