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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,
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:
In terms of $\overline{x}$ these tree critical regions become
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 $$T=\frac{\overline{X}-\mu}{\sqrt{{S}^{2}/n}}=\frac{\overline{X}-\mu}{S/\sqrt{n}}.$$
${H}_{0}$ | ${H}_{1}$ | Critical Region |
---|---|---|
$\mu ={\mu}_{0}$ | $\mu >{\mu}_{0}$ | $z\ge {z}_{\alpha}$ or $\overline{x}\ge {\mu}_{0}+{z}_{\alpha}\sigma /\sqrt{n}$ |
$\mu ={\mu}_{0}$ | $\mu <{\mu}_{0}$ | $z\le -{z}_{\alpha}$ or $\overline{x}\le {\mu}_{0}-{z}_{\alpha}\sigma /\sqrt{n}$ |
$\mu ={\mu}_{0}$ | $\mu \ne {\mu}_{0}$ | $\left|z\right|\ge {z}_{\alpha /2}$ or $\left|\overline{x}-{\mu}_{0}\right|\ge {z}_{\alpha /2}\sigma /\sqrt{n}$ |
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 .
${H}_{0}$ | ${H}_{1}$ | Critical Region |
---|---|---|
$\mu ={\mu}_{0}$ | $\mu >{\mu}_{0}$ | $t\ge {t}_{\alpha}\left(n-1\right)$ or $\overline{x}\ge {\mu}_{0}+{t}_{\alpha}\left(n-1\right)s/\sqrt{n}$ |
$\mu ={\mu}_{0}$ | $\mu <{\mu}_{0}$ | $t\le -{t}_{\alpha}\left(n-1\right)$ or $\overline{x}\le {\mu}_{0}-{t}_{\alpha}\left(n-1\right)s/\sqrt{n}$ |
$\mu ={\mu}_{0}$ | $\mu \ne {\mu}_{0}$ | $\left|t\right|\ge {t}_{\alpha /2}\left(n-1\right)$ or $\left|\overline{x}-{\mu}_{0}\right|\ge {t}_{\alpha /2}\left(n-1\right)s/\sqrt{n}$ |
Let X (in millimeters) equal the growth in 15 days of a tumor induced in a mouse. Assume that the distribution of X is $N\left(\mu ,{\sigma}^{2}\right)$ . We shall test the null hypothesis ${H}_{0}:\mu ={\mu}_{0}=4.0$ millimeters against the two-sided alternative hypothesis is ${H}_{1}:\mu \ne 4.0$ . If we use n =9 observations and a significance level of $\alpha $ =0.10, the critical region is $$\left|t\right|=\frac{\left|\overline{x}-4.0\right|}{s/\sqrt{9}}\ge {t}_{\alpha /2}\left(8\right)={t}_{0.05}\left(8\right)=\mathrm{1.860.}$$
If we are given that n =9, $\overline{x}$ =4.3, and s =1.2, we see that $$t=\frac{4.3-4.0}{1.2/\sqrt{9}}=\frac{0.3}{0.4}=\mathrm{0.75.}$$
Thus $\left|t\right|=\left|0.75\right|<1.860$ and we accept (do not reject) ${H}_{0}:\mu =4.0$ at the $\alpha $ =10% significance level. See Figure 1 .
In this example the use of the t -statistic with a one-sided alternative hypothesis will be illustrated.
In attempting to control the strength of the wastes discharged into a nearby river, a paper firm has taken a number of measures. Members of the firm believe that they have reduced the oxygen-consuming power of their wastes from a previous mean $\mu $ of 500. They plan to test ${H}_{0}:\mu =500$ against ${H}_{1}:\mu <500$ , using readings taken on n =25 consecutive days. If these 25 values can be treated as a random sample, then the critical region, for a significance level of $\alpha $ =0.01, is $$t=\frac{\overline{x}-500}{s/\sqrt{25}}\le -{t}_{0.01}\left(24\right)=-\mathrm{2.492.}$$
The observed values of the sample mean and sample standard deviation were $\overline{x}$ =308.8 and s =115.15. Since $$t=\frac{308.8-500}{115.15/\sqrt{25}}=-8.30<-2.492,$$ we clearly reject the null hypothesis and accept ${H}_{1}:\mu <500$ . It should be noted, however, that although an improvement has been made, there still might exist the question of whether the improvement is adequate. The 95% confidence interval $308.8\pm 2.064\left(115.15/5\right)$ or $\left[\text{261}\text{.27,356}\text{.33}\right]$ for $\mu $ might the company answer that question.
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