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In general, without changing the sample size or the type of the test of the hypothesis, a decrease in $\alpha$ causes an increase in $\beta$ , and a decrease in $\beta$ causes an increase in $\alpha$ . Both probabilities $\alpha$ and $\beta$ of the two types of errors can be decreased only by increasing the sample size or, in some way, constructing a better test of the hypothesis.

## Example

If n =100 and we desire a test with significance level $\alpha$ =0.05, then $\alpha =P\left(\overline{X}\ge c;\mu =60\right)=0.05$ means, since $\overline{X}$ is $\text{N(}\mu \text{,100/100=1)}$ ,

$P\left(\frac{\overline{X}-60}{1}\ge \frac{c-60}{1};\mu =60\right)=0.05$ and $c-60=1.645$ . Thus c =61.645. The power function is

$K\left(\mu \right)=P\left(\overline{X}\ge 61.645;\mu \right)=P\left(\frac{\overline{X}-\mu }{1}\ge \frac{61.645-\mu }{1};\mu \right)=1-\Phi \left(61.645-\mu \right).$

In particular, this means that $\beta$ at $\mu$ =65 is $=1-K\left(\mu \right)=\Phi \left(61.645-65\right)=\Phi \left(-3.355\right)\approx 0;$ so, with n =100, both $\alpha$ and $\beta$ have decreased from their respective original values of 0.1587 and 0.0668 when n =25. Rather than guess at the value of n , an ideal power function determines the sample size. Let us use a critical region of the form $\overline{x}\ge c$ . Further, suppose that we want $\alpha$ =0.025 and, when $\mu$ =65, $\beta$ =0.05. Thus, since $\overline{X}$ is $\text{N(}\mu \text{,100/n)}$ ,

$0.025=P\left(\overline{X}\ge c;\mu =60\right)=1-\Phi \left(\frac{c-60}{10/\sqrt{n}}\right)$ and $0.05=1-P\left(\overline{X}\ge c;\mu =65\right)=\Phi \left(\frac{c-65}{10/\sqrt{n}}\right).$

That is, $\frac{c-60}{10/\sqrt{n}}=1.96$ and $\frac{c-65}{10/\sqrt{n}}=-1.645$ .

Solving these equations simultaneously for c and $10/\sqrt{n}$ , we obtain $c=60+1.96\frac{5}{3.605}=62.718;$ $\frac{10}{\sqrt{n}}=\frac{5}{3.605}.$

Thus, $\sqrt{n}=7.21$ and $n=51.98$ . Since n must be an integer, we would use n =52 and obtain $\alpha$ =0.025 and $\beta$ =0.05, approximately.

For a number of years there has been another value associated with a statistical test, and most statistical computer programs automatically print this out; it is called the probability value or, for brevity, p -value . The p -value associated with a test is the probability that we obtain the observed value of the test statistic or a value that is more extreme in the direction of the alternative hypothesis, calculated when ${\text{H}}_{\text{0}}$ is true. Rather than select the critical region ahead of time, the p -value of a test can be reported and the reader then makes a decision.

Say we are testing against with a sample mean $\overline{X}$ based on n =52 observations. Suppose that we obtain the observed sample mean of $\overline{x}=62.75$ . If we compute the probability of obtaining an $\overline{x}$ of that value of 62.75 or greater when $\mu$ =60, then we obtain the p -value associated with $\overline{x}=62.75$ . That is,

$\begin{array}{l}p-value=P\left(\overline{X}\ge 62.75;\mu =60\right)=P\left(\frac{\overline{X}-60}{10/\sqrt{52}}\ge \frac{62.75-60}{10/\sqrt{52}};\mu =60\right)\\ =1-\Phi \left(\frac{62.75-60}{10/\sqrt{52}}\right)=1-\Phi \left(1.983\right)=0.0237.\end{array}$

If this p -value is small, we tend to reject the hypothesis . For example, rejection of if the p -value is less than or equal to 0.025 is exactly the same as rejection if $\overline{x}=62.718$ .That is, $\overline{x}=62.718$ has a p -value of 0.025. To help keep the definition of p -value in mind, we note that it can be thought of as that tail-end probability , under ${\text{H}}_{\text{0}}$ , of the distribution of the statistic, here $\overline{X}$ , beyond the observed value of the statistic. See Figure 1 for the p -value associated with $\overline{x}=62.75.$

Suppose that in the past, a golfer’s scores have been (approximately) normally distributed with mean $\mu$ =90 and ${\sigma }^{2}$ =9. After taking some lessons, the golfer has reason to believe that the mean $\mu$ has decreased. (We assume that ${\sigma }^{2}$ is still about 9.) To test the null hypothesis against the alternative hypothesis , the golfer plays 16 games, computing the sample mean $\overline{x}$ .If $\overline{x}$ is small, say $\overline{x}\le c$ , then ${H}_{0}$ is rejected and ${H}_{1}$ accepted; that is, it seems as if the mean $\mu$ has actually decreased after the lessons. If c =88.5, then the power function of the test is

$K\left(\mu \right)=P\left(\overline{X}\le 88.5;\mu \right)=P\left(\frac{\overline{X}-\mu }{3/4}\le \frac{88.5-\mu }{3/4};\mu \right)=\Phi \left(\frac{88.5-\mu }{3/4}\right).$

Because 9/16 is the variance of $\overline{X}$ . In particular, $\alpha =K\left(90\right)=\Phi \left(-2\right)=1-0.9772=0.0228.$

If, in fact, the true mean is equal to $\mu$ =88 after the lessons, the power is $K\left(88\right)=\Phi \left(2/3\right)=0.7475$ . If $\mu$ =87, then $K\left(87\right)=\Phi \left(2\right)=0.9772$ . An observed sample mean of $\overline{x}=88.25$ has a

$p-value=P\left(\overline{X}\le 88.25;\mu =90\right)=\Phi \left(\frac{88.25-90}{3/4}\right)=\Phi \left(-\frac{7}{3}\right)=0.0098,$

and this would lead to a rejection at $\alpha$ =0.0228 (or even $\alpha$ =0.01).

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