Chapter 8, Problem 1CTC

Critical Thinking Challenges

The power of a test (1 − β) can be calculated when a specific value of the mean is hypothesized in the alternative hypothesis; for example, let H0: μ = 50 and let H1: μ = 52. To find the power of a test, it is necessary to find the value of β. This can be done by the following steps:

Step 1 For a specific value of α find the corresponding value of $\bar{X}$, using $z=\frac{\bar{X}-\mu }{\sigma /\sqrt{n}}V$ where μ is the hypothesized value given in H0. Use a right-tailed test.

Step 2 Using the value of $\bar{X}$ found in step 1 and the value of μ in the alternative hypothesis, find the area corresponding to z in the formula . $z=\frac{\bar{X}-\mu }{\sigma /\sqrt{n}}V$

Step 3 Subtract this area from 0.5000. This is the value of β.

Step 4 Subtract the value of β from 1. This will give you the power of a test. See Figure 8–41.

1. Find the power of a test, using the hypotheses given previously and α = 0.05, σ = 3, and n = 30.

2. Select several other values for μ in H1 and compute the power of the test. Generalize the results.

FIGURE 8–41 Relationship Among α, β, and the Power of a Test

To determine

To obtain: The power of a test, using the given hypotheses and $\alpha =0.05$, $\sigma =3$ and $n=30$.

Answer to Problem 1CTC

The power of test is 0.5222.

Explanation of Solution

Given info:

The hypotheses are $H_0:\mu =50$ and $H_1:\mu =52$.

Calculation:

For z score:

Here, the level of significance is 0.05.

The level of significance is 0.05 represents the area of 0.05 to the right of z.

The z score is obtained as follows:

$\begin{array}{c}P\left(\text{Area to the left}\right)=1-P\left(\text{Area to the right}\right)\\ =1-0.05\\ =0.95\end{array}$

Use Table E: The Standard Normal Distribution to find z score.

Procedure:

• Locate an approximate area of 0.95 in the body of the Table E.
• Move left until the first column and note the values as 1.6.
• Move upward until the top row is reached and note the values as 0.04 and 0.05.

$\begin{array}{c}z=\frac{1.64+1.65}{2}\\ =\frac{3.29}{2}\\ =1.645\end{array}$

Thus, the z score is 1.645.

For $\overline{X}$:

The formula for finding $\overline{X}$ is,

$\overline{X}=\mu +z\left(\frac{\sigma }{\sqrt{n}}\right)$

Substitute 50 for $\mu$, 1.645 for z, 3 for $\sigma$ and 30 for n

$\begin{array}{c}\overline{X}=50+1.645\left(\frac{3}{\sqrt{30}}\right)\\ =50+1.645\times 0.5477\\ =50+0.9010\\ =50.901\end{array}$

For z using alternative hypothesis mean 52:

The formula for finding the z score is,

$z=\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}$

Substitute 50.901 for $\overline{X}$, 52 for $\mu$, 3 for $\sigma$ and 30 for n

$\begin{array}{c}z=\frac{50.901-52}{\frac{3}{\sqrt{30}}}\\ =\frac{-1.099}{0.5477}\\ =-2.01\end{array}$

The z score value of –2.01 represents the area to the left of –2.01.

Use Table E: The Standard Normal Distribution to find the area.

Procedure:

• Locate –2.0 in the left column of the table.
• Obtain the value in the corresponding row below 0.01.

That is, $P\left(z<-2.01\right)=0.0222$

Thus, the area to the left of –2.01 is 0.0222.

For $\beta$:

The value of $\beta$ is obtained by subtract the value of 0.0222 from 0.5000.

$\begin{array}{c}\beta =0.5000-0.0222\\ =0.4778\end{array}$

Power of a test:

The formula for finding the power of test is,

$\text{Power of test}=1-\beta$

Substitute 0.4778 for $\beta$

$\begin{array}{c}\text{Power of test}=1-0.4778\\ =0.5222\end{array}$

Thus, the power of test is 0.5222.

To determine

To compute: The power of a test for different mean values in alternative hypothesis.

Answer to Problem 1CTC

The power of a test for different mean values in alternative hypothesis is 0.5001.

Explanation of Solution

Given info:

The null hypothesis is $H_0:\mu =50$.

Calculation:

Consider alternative hypothesis $H_1:\mu =53$ and $H_1:\mu =54$

From part (a), the value of $\overline{X}$ is 50.901.

For $H_1:\mu =53$:

For z using alternative hypothesis mean 53:

The formula for finding the z score is,

$z=\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}$

Substitute 50.901 for $\overline{X}$, 53 for $\mu$, 3 for $\sigma$ and 30 for n

$\begin{array}{c}z=\frac{50.901-53}{\frac{3}{\sqrt{30}}}\\ =\frac{-2.099}{0.5477}\\ =-3.83\end{array}$

The z score value of –3.83 represents the area to the left of –3.83.

From the Table E, the area of the z score is 0.0001.

That is, $P\left(z<-3.83\right)=0.0001$

Thus, the area to the left of –3.83 is 0.0001.

For $\beta$:

The value of $\beta$ is obtained by subtract the value of 0.0001 from 0.5000.

$\begin{array}{c}\beta =0.5000-0.0001\\ =0.4999\end{array}$

Power of a test:

The formula for finding the power of test is,

$\text{Power of test}=1-\beta$

Substitute 0.4999 for $\beta$

$\begin{array}{c}\text{Power of test}=1-0.4999\\ =0.5001\end{array}$

Thus, the power of test is 0.5001.

For $H_1:\mu =54$:

For z using alternative hypothesis mean 54:

The formula for finding the z score is,

$z=\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}$

Substitute 50.901 for $\overline{X}$, 54 for $\mu$, 3 for $\sigma$ and 30 for n

$\begin{array}{c}z=\frac{50.901-54}{\frac{3}{\sqrt{30}}}\\ =-\frac{3.099}{0.5477}\\ =-5.66\end{array}$

The z score value of 3.47 represents the area to the left of 3.47.

From the Table E, the area of the z score is 0.0001.

That is, $P\left(z<-5.66\right)=0.0001$

Thus, the area to the left of –5.66 is 0.0001.

For $\beta$:

The value of $\beta$ is obtained by subtract the value of 0.9997 from 0.5000.

$\begin{array}{c}\beta =0.5000-0.0001\\ =0.4999\end{array}$

Power of a test:

The formula for finding the power of test is,

$\text{Power of test}=1-\beta$

Substitute 0.4999 for $\beta$

$\begin{array}{c}\text{Power of test}=1-0.4999\\ =0.5001\end{array}$

Thus, the power of test is 0.5001.