1. Calculate the probability of a Type II error for a $Z$ test for $\mu$.
2. Compute the power of a $Z$ test for $\mu$ and construct a power curve.

• Step 1 Recall that $Z_{\text{crit}}$ divides the critical region from the noncritical region. Let ${\overline{x}}_{\text{crit}}$ be the value of the sample mean $\overline{x}$ associated with $Z_{\text{crit}}$. The following table shows how to calculate ${\overline{x}}_{\text{crit}}$ for the three forms of the hypothesis test.

  Form of test		Value of ${\overline{x}}_{\text{crit}}$
  Right-tailed	$H_0:\mu ={\mu }_0\hspace{1em}vs.\hspace{1em}{H}_a:\mu >{\mu }_0$	${\overline{x}}_{\text{crit}}={\mu }_0+Z_{\text{crit}}·\frac{\sigma }{\sqrt{n}}$
  Left-tailed	$H_0:\mu ={\mu }_0\hspace{1em}vs.\hspace{1em}{H}_a:\mu <{\mu }_0$	${\overline{x}}_{\text{crit}}={\mu }_0-Z_{\text{crit}}·\frac{\sigma }{\sqrt{n}}$
  Two-tailed	$\begin{array}{ccc}{H}_0:\mu ={\mu }_0& vs.& H_a:\mu \ne {\mu }_0\end{array}$	${\overline{x}}_{\text{crit},\text{ lower}}={\mu }_0-Z_{\text{crit}}·\frac{\sigma }{\sqrt{n}}$
  		${\overline{x}}_{\text{crit},\text{ upper}}={\mu }_0+Z_{\text{crit}}·\frac{\sigma }{\sqrt{n}}$

  Here, ${\mu }_0$ is the hypothesized value of the population mean, σ is the population standard deviation, and $n$ is the sample size.

• Step 2 Let ${\mu }_a$ represent a particular value for the population mean $\mu$ chosen from the values indicated in the alternative hypothesis $H_a$. Draw a normal curve centered at ${\mu }_a$, with the value or values of ${\overline{x}}_{\text{crit}}$ from Step 1 indicated (see Example 34).

• Step 3 Calculate $\beta$ for the particular ${\mu }_a$ chosen using the following table.

$\begin{array}{lll}{H}_0:\mu =11\hfill & \text{versus}\hfill & H_a:\mu >11\hfill \end{array}$

1. State what a Type II error would be in this case.
2. Let ${\mu }_a=13$. That is, suppose the population mean debit card usage is actually 13 times per month. Calculate $\beta$, the probability of making a Type II error when ${\mu }_a=13$.

1. We make a Type II error when we do not reject $H_0$ when $H_0$ is false. In this case, a Type II error would be to conclude that the population mean debit card usage was 11 times per month when in actuality it was more than 11 times per month.
2. We follow the steps for calculating $\beta$.

   • Step 1 We have a right-tailed test, so that

     ${\overline{x}}_{\text{crit}}={\mu }_0+z_{\text{crit}}·\frac{\sigma }{\sqrt{n}}=11+2.33·\frac{3}{\sqrt{36}}=12.165$

   • Step 2 Figure 49 shows the normal curve centered at ${\mu }_a=13$, with ${\overline{x}}_{\text{crit}}=12.165$ labeled.

     FIGURE 49 $\beta$ probability of Type II error.

     

   • Step 3 The right-tailed test tells us that $\beta$ equals the area under the normal curve drawn in Step 2 to the left of ${\overline{x}}_{\text{crit}}=12.165$. This is the shaded area in Figure 49. Area represents probability, so we have

     $\beta =p\left(\overline{x}<12.165\right)\text{when }{\mu }_{\mathrm{a}}=13$

     Standardizing with ${\mu }_a=13,\sigma =3$, and $n=36$:

     $\begin{array}{c}\beta =P\left(\overline{x}<12.165\right)\\ =P\left(Z<\frac{12.165-13}{3/\sqrt{36}}\right)\\ =P\left(Z<-1.67\right)=0.0475\end{array}$

     This is a Case 1 problem from Table 8 of Chapter 6 on page 355.

     Thus, $\beta =0.0475$. This represents the probability of making a Type II error, that is, of not rejecting the hypothesis that the population mean debit card usage is 11 times per month when in actuality it is 13 times per month.

$\text{power }=1-\beta$

$\text{power }=1-\beta =1-0\text{.0475}=0\text{.9525}$

1. Calculate the power of the hypothesis test from Example 34 for the following values of ${\mu }_a:11.0,\hspace{0.17em}11.5,\hspace{0.17em}12.0,\hspace{0.17em}12.165,\hspace{0.17em}12.5,\hspace{0.17em}13.5$.
2. Construct the power curve by graphing the values for the power of the test on the vertical axis against the values of ${\mu }_a$ on the horizontal axis.

1. We have ${\overline{x}}_{\text{crit}}=12.165,\hspace{0.17em}\sigma =3,$, and $n=36$. The calculations are provided in the following table.

   ${\mu }_a$	Probability of Type II error: $\beta$	Power of the test: $1-\beta$
   11.0	$P\left(Z<\frac{12.165-11}{3/\sqrt{36}}\right)=P\left(Z<2.33\right)=0.9901$	$1-0.9901=0.0099$
   11.5	$P\left(Z<\frac{12.165-11.5}{3/\sqrt{36}}\right)=P\left(Z<1.33\right)=0.9082$	$1-0.9082=0.0918$
   12.0	$P\left(Z<\frac{12.165-12}{3/\sqrt{36}}\right)=P\left(Z<0.33\right)=0.6293$	$1-0.6293=0.3707$
   12.165	$P\left(Z<\frac{12.165-12.165}{3/\sqrt{36}}\right)=P\left(Z<0.00\right)=0.5$	$1-0.5=0.5$
   12.5	$P\left(Z<\frac{12.165-12.5}{3/\sqrt{36}}\right)=P\left(Z<-0.67\right)=0.2514$	$1-0.2514=0.7486$
   13.5	$P\left(Z<\frac{12.165-13.5}{3/\sqrt{36}}\right)=P\left(Z<-2.67\right)=0.0038$	$1-0.0038=0.9962$

2. Figure 51 represents a power curve, because it plots the values for the power of the test on the vertical axis against the values of ${\mu }_a$ on the horizontal axis. Note that, as ${\mu }_a$ moves farther away from the hypothesized mean ${\mu }_0=11$, the power of the test increases. This is because it is more likely that the null hypothesis will be correctly rejected as the actual value of the mean ${\mu }_a$ gets farther away from the hypothesized value ${\mu }_0$.

   

   FIGURE 51 Power curve.