Hypothesis Tests for One Mean: Unknown Population Variance

When testing a hypothesis about a population mean $\mu$, if the population variance $\sigma^2$ is unknown -- which is the typical situation in practice -- we cannot use a z-test. Instead, we replace $\sigma$ with the sample standard deviation $s$. Assuming the underlying population is approximately normal, the resulting standardized statistic follows a Student's t-distribution with $n-1$ degrees of freedom.

The one-sample t-statistic is

$$t = \dfrac{\bar{x} - \mu_0}{s/\sqrt{n}}, \qquad \text{df} = n - 1.$$

The procedure mirrors the z-test, except that the critical value comes from the t-distribution. For a right-tailed test at significance level $\alpha$:

1. State $H_0: \mu = \mu_0$ versus $H_1: \mu > \mu_0$.
2. Compute $t = \dfrac{\bar{x} - \mu_0}{s/\sqrt{n}}$.
3. Look up $t_{\alpha,\, n-1}$, the upper $\alpha$ critical value with $n-1$ degrees of freedom.
4. Reject $H_0$ if $t > t_{\alpha,\, n-1}$; otherwise, fail to reject.

For example, suppose a sample of $n=10$ observations yields $\bar{x}=52$ and $s=6$, and we test $H_0: \mu = 50$ versus $H_1: \mu > 50$ at $\alpha = 0.05$. Then

$$t = \dfrac{52 - 50}{6/\sqrt{10}} = \dfrac{2}{1.897} \approx 1.054.$$

With df $= 9$, the critical value is $t_{0.05,\,9} = 1.833$. Since $1.054 < 1.833$, we fail to reject $H_0$.