4.1 Tests about one mean and one variance

Tests about one mean and one variance

In the previous paragraphs it was assumed that we were sampling from a normal distribution and the variance was known. The null hypothesis was generally of the form ${\text{H}}_{\text{0}}\text{: }\mu \text{= }{\mu }_{\text{0}}$ .

There are essentially tree possibilities for the alternative hypothesis, namely that $\mu$ has increased,

• ${\text{H}}_{\text{1}}\text{: }\mu >{\mu }_{\text{0}}$ ; $\mu$ has decreased,
• ${\text{H}}_{\text{1}}\text{: }\mu <{\mu }_{\text{0}}$ ; $\mu$ has changed, but it is not known if it has increased or decreased, which leads to a two-sided alternative hypothesis
• $H_1;\mu \ne {\mu }_0$ .

To test $H_0;\mu ={\mu }_0$ against one of these tree alternative hypotheses, a random sample is taken from the distribution, and an observed sample mean, $\overline{x}$ , that is close to ${\mu }_0$ supports ${\text{H}}_{\text{0}}$ . The closeness of $\overline{x}$ to ${\mu }_0$ is measured in term of standard deviations of $\overline{X}$ , $\sigma /\sqrt{n}$ which is sometimes called the standard error of the mean . Thus the statistic could be defined by

$$Z=\frac{\overline{X}-{\mu }_0}{\sqrt{\sigma 2}/n}=\frac{\overline{X}-{\mu }_0}{\sigma /\sqrt{n}},$$ and the critical regions, at a significance level $\alpha$ , for the tree respective alternative hypotheses would be:

• $z\ge z_{\alpha }$
• $z\le z_{\alpha }$
• $\left|z\right|=z_{\alpha /2}$

In terms of $\overline{x}$ these tree critical regions become

• $\overline{x}\ge {\mu }_0+z_{\alpha }\sigma /\sqrt{n},$
• $\overline{x}\le {\mu }_0-z_{\alpha }\sigma /\sqrt{n},$
• $\left|\overline{x}-{\mu }_0\right|\ge z_{\alpha }\sigma /\sqrt{n}$

These tests and critical regions are summarized in TABLE 1 . The underlying assumption is that the distribution is $N\left(\mu ,{\sigma }^2\right)$ and ${\sigma }^2$ is known. Thus far we have assumed that the variance ${\sigma }^2$ was known. We now take a more realistic position and assume that the variance is unknown. Suppose our null hypothesis is $H_0;\mu ={\mu }_0$ and the two-sided alternative hypothesis is $H_1;\mu \ne {\mu }_0$ . If a random sample $X_1,X_2,\mathrm{...},X_n$ is taken from a normal distribution $N\left(\mu ,{\sigma }^2\right)$ ,let recall that a confidence interval for $\mu$ was based on $$T=\frac{\overline{X}-\mu }{\sqrt{S^2/n}}=\frac{\overline{X}-\mu }{S/\sqrt{n}}.$$

This suggests that T might be a good statistic to use for the test $H_0;\mu ={\mu }_0$ with $\mu$ replaced by ${\mu }_0$ . In addition, it is the natural statistic to use if we replace ${\sigma }^2/n$ by its unbiased estimator $S^2/n$ in $\left(\overline{X}-{\mu }_0\right)/\sqrt{{\sigma }^2/n}$ in a proper equation. If $\mu ={\mu }_0$ we know that T has a t distribution with n -1 degrees of freedom. Thus, with $\mu ={\mu }_0$ ,

$$P\left[\left|T\right|\ge t_{\alpha /2}\left(n-1\right)\right]=P\left[\frac{\left|\overline{X}-{\mu }_0\right|}{S/\sqrt{n}}\ge t_{\alpha /2}\left(n-1\right)\right]=\alpha .$$

Accordingly, if $\overline{x}$ and s are the sample mean and the sample standard deviation, the rule that rejects $H_0;\mu ={\mu }_0$ if and only if $$\left|t\right|=\frac{\left|\overline{x}-{\mu }_0\right|}{s/\sqrt{n}}\ge t_{\alpha /2}\left(n-1\right).$$

Provides the test of the hypothesis with significance level $\alpha$ . It should be noted that this rule is equivalent to rejecting $H_0;\mu ={\mu }_0$ if ${\mu }_0$ is not in the open $100\left(1-\alpha \right)\%$ confidence interval $$\left(\overline{x}-t_{\alpha /2}\left(n-1\right)s/\sqrt{n},\overline{x}+t_{\alpha /2}\left(n-1\right)s/\sqrt{n}\right).$$

Table 2 summarizes tests of hypotheses for a single mean, along with the three possible alternative hypotheses, when the underlying distribution is $N\left(\mu ,{\sigma }^2\right)$ , ${\sigma }^2$ is unknown, $t=\left(\overline{x}-{\mu }_0\right)/\left(s/\sqrt{n}\right)$ and $n\le 31$ . If n >31, use table 1 for approximate tests with $\sigma$ replaced by s .

In this example the use of the t -statistic with a one-sided alternative hypothesis will be illustrated.