If \(X_1, X_2, \ldots, X_n\) are normally distributed with mean \(\mu\) and variance \(\sigma^2\), then:

\(T=\dfrac{\bar{X}-\mu}{S/\sqrt{n}}\)

follows a \(T\) distribution with \(n-1\) degrees of freedom.

Proof

The proof is as simple as recalling a few distributional results from our work in Stat 414. Recall the definition of a \(T\) random variable, namely if \(Z\sim N(0,1)\) and \(U\sim \chi^2_{(r)}\) are independent, then:

\(T=\dfrac{Z}{\sqrt{U/r}}\)

follows the \(T\) distribution with \(r\) degrees of freedom. Furthermore, recall that if \(X_1, X_2, \ldots, X_n\) are normally distributed with mean \(\mu\) and variance \(\sigma^2\), then:

1. \(Z=\dfrac{\bar{X}-\mu}{\sigma/\sqrt{n}}\sim N(0,1)\)

2. \(\dfrac{(n-1)S^2}{\sigma^2}\sim \chi^2_{n-1}\)

3. \(\bar{X}\) and \(S^2\) are independent

Now, we just have to put all that we've remembered together:

\(T=\dfrac{ \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} }{\sqrt{\frac{\frac{(n-1)s^2}{\sigma^2}}{n-1}}}=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}\left(\frac{\sigma}{s}\right)=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}\sim t_{n-1}\)

The first equality simply defines a \(T\) random variable using the first, second, and third bullet point above. The second equality comes from canceling out the \(n-1\) terms in the denominator. The third equality comes from canceling out the \(\sigma\) terms, leaving us with:

\(T=\dfrac{\bar{X}-\mu}{S/\sqrt{n}}\)

following a \(T\) distribution with \(n-1\) degrees of freedom, as was to be proved!

If \(X_1, X_2, \ldots, X_n\) are normally distributed random variables with mean \(\mu\) and variance \(\sigma^2\), then a \((1-\alpha)100\%\) confidence interval for the population mean \(\mu\) is:

\(\bar{x}\pm t_{\alpha/2,n-1}\left(\dfrac{s}{\sqrt{n}}\right)\)

This interval is often referred to as the "\(t\)-interval for the mean."

Proof

The proof is very similar to that for the \(Z\)-interval for the mean. We start by drawing a picture of a \(T\)-distribution with \(n-1\) degrees of freedom:

From the diagram, we can see that the following probability statement is true:

\(P[-t_{\alpha/2,n-1}\leq T \leq t_{\alpha/2,n-1}]=1-\alpha \)

Then, simply replacing \(T\), we get:

\(P\left[-t_{\alpha/2,n-1}\leq \dfrac{\bar{X}-\mu}{s/\sqrt{n}} \leq t_{\alpha/2,n-1}\right]=1-\alpha \)

Let's again focus only on the inequality inside the brackets for a bit. Because we manipulate each of the three sides of the inequality equally, each of the following statements are equivalent:

\begin{array}{rccl} -t_{\alpha/2,n-1} & \leq & \dfrac{\bar{X}-\mu}{s/\sqrt{n}} & \leq & t_{\alpha/2,n-1}\\ -t_{\alpha/2,n-1}\left(\dfrac{s}{\sqrt{n}}\right) & \leq & \bar{X}-\mu & \leq & +t_{\alpha/2,n-1}\left(\dfrac{s}{\sqrt{n}}\right)\\ -\bar{X}-t_{\alpha/2,n-1}\left(\dfrac{s}{\sqrt{n}}\right) & \leq & -\mu & \leq & -\bar{X}+t_{\alpha/2,n-1}\left(\dfrac{s}{\sqrt{n}}\right)\\ \bar{X}-t_{\alpha/2,n-1}\left(\dfrac{s}{\sqrt{n}}\right) & \leq & \mu &\leq & \bar{X}+t_{\alpha/2,n-1}\left(\dfrac{s}{\sqrt{n}}\right) \end{array}

That is, we have shown that a \((1-\alpha)100\%\) confidence interval for the mean \(\mu\) is:

\(\left[\bar{X}-t_{\alpha/2,n-1}\left(\dfrac{s}{\sqrt{n}}\right),\bar{X}+t_{\alpha/2,n-1}\left(\dfrac{s}{\sqrt{n}}\right)\right]\)

as was to be proved.