Intervals Based on a Normal Population Distribution: The T-Distribution

Posted by Beetle B. on Sun 16 July 2017

Say you take a sample where \(n\) is not large. Then the CLT doesn’t apply. We must then know/assume a distribution.

Suppose we know it is normal with known \(\mu,\sigma\), both unknown.

The random variable \(T=\frac{\bar{X}-\mu}{S/\sqrt{n}}\) has a pdf called the t-distribution with \(n-1\) degrees of freedom.

The numerator and the denominator are independent random variables (not proven).

Note that now that \(n\) is small, we cannot use \(S\approx\sigma\).

Properties of t-Distributions

The degrees of freedom are denoted by \(\nu\). Let \(t_{\nu}\) be the pdf:

  1. Each \(t_{\nu}\) curve is bell shaped and centered at 0.
  2. Each \(t_{\nu}\) curve is more spread out than the normal curve.
  3. As \(\nu\) increases, the spread decreases.
  4. As \(\nu\rightarrow\infty,t_{\nu}\rightarrow\) normal curve.

Notation: Let \(t_{\alpha,\nu}\) be the number such that the area of the curve to the right is \(\alpha\). It is called the t-critical value.

Then:

\begin{equation*} P\left(-t_{\alpha/2,n-1}<T<t_{\alpha/2,n-1}\right)=1-\alpha \end{equation*}

And the \(100(1-\alpha)\) CI for \(\mu\) is \(\bar{x}\pm t_{\alpha/2,n-1}s/\sqrt{n}\)

If you want a one sided CI, it is \(\bar{x}\pm t_{\alpha,n-1}s/\sqrt{n}\)

Figuring out the needed \(n\) is hard.