Chapter 6.13, Problem 5E

A tire company claims that the lifetimes of its tires average 50,000 miles. The standard deviation of tire lifetimes is known to be 5000 miles. You sample 100 tires and will test the hypothesis that the mean tire lifetime is at least 50,000 miles against the alternative that it is less. Assume, in fact, that the true mean lifetime is 49,500 miles.

1. a. State the null and alternate hypotheses. Which hypothesis is true?
2. b. It is decided to reject H₀ if the sample mean is less than 49,400. Find the level and power of this test.
3. c. If the test is made at the 5% level, what is the power?
4. d. At what level should the test be conducted so that the power is 0.80?
5. e. You are given the opportunity to sample more tires. How many tires should be sampled in total so that the power is 0.80 if the test is made at the 5% level?

To determine

State the null and alternate hypotheses and also identify which hypothesis is true.

Explanation of Solution

Given info:

A tire company claims that the lifetimes of its tires average 50,000 miles. The standard deviation of tire lifetimes is 5,000 miles.

Assume that the true mean lifetime is 49,500 miles.

Justification:

State the null and alternate hypotheses.

Null hypothesis:

$H_0:\mu \ge 50,000$

That is, the mean tire lifetime is at least 50,000 miles.

Alternative hypothesis:

$H_1:\mu <50,000$

That is, the mean tire lifetime is less than 50,000 miles.

The alternative hypothesis, $H_1$ is true because the true mean lifetime is 49,500 miles.

To determine

Find the level and power of the test.

Answer to Problem 5E

The level is 0.1151.

The power is 0.4207.

Explanation of Solution

Calculation:

Type-1 error: Rejecting the null hypothesis $\left(H_0\right)$ when it is true. It is denoted by α.

$\begin{array}{c}\alpha \text{ = P( type I error) }\\ \text{= P( rejecting }{H}_0\text{when }{H}_0\text{is true)}\end{array}$

Under $H_0$, the sample mean is approximately normally distributed with mean and standard deviation is 50,000 and 500 $\left(=\frac{5000}{\sqrt{100}}\right)$, respectively.

The level is calculated as follows:

$\begin{array}{c}\text{Level}=P\left(\overline{X}\le 49,400\right)\\ =P\left(z\le \frac{49,400-50,000}{500}\right)\\ =P\left(z\le \frac{-600}{500}\right)\\ =P\left(z\le -1.2\right)\end{array}$

Software Procedure:

Step-by-step procedure to obtain the $P\left(z\le -1.2\right)$ using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Left Tail for the region of the curve to shade.
• Enter the data value as -1.2.
• Click OK.

Output using the MINITAB software is given below:

From the output, $P\left(z\le -1.2\right)=0.1151$.

Therefore, the level is 0.1151.

Power:

The probability of rejecting a null hypothesis when it is false is called the power of the test.

$\begin{array}{c}\text{Power}=P\left(\overline{X}\le 49,400\right)\\ =P\left(z\le \frac{49,400-49,500}{500}\right)\\ =P\left(z\le \frac{-100}{500}\right)\\ =P\left(z\le -0.2\right)\end{array}$

Software Procedure:

Step-by-step procedure to obtain the $P\left(z\le -0.2\right)$ using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Left Tail for the region of the curve to shade.
• Enter the data value as -0.2.
• Click OK.

Output using the MINITAB software is given below:

From the output, $P\left(z\le -0.2\right)=0.4207$.

Thus, the power is 0.4207.

To determine

Find the power of the test.

Answer to Problem 5E

The power is 0.2578.

Explanation of Solution

Calculation:

The 5% rejection region is $\overline{X}\le x_5$ because the alternative hypothesis is of the form $\mu <{\mu }_0$.

5^th percentile:

Software Procedure:

Step-by-step procedure to obtain the 5^th percentile using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose Probability Value and Left Tail for the region of the curve to shade.
• Enter the Probability value as 0.5.
• Click OK.

Output using the MINITAB software is given below:

From the output, the z-score corresponding to the 5^th percentile is –1.645.

The value of $x_5$ is calculated as follows:

$\begin{array}{c}{x}_5=50,000-1.645\left(500\right)\\ =50,000-\text{822}\text{.5}\\ =\text{49,177}\text{.5}\end{array}$

Therefore, the rejection region is, $\overline{X}\le 49,177.5$.

Power:

$\begin{array}{c}\text{Power}=P\left(\overline{X}\le 49,177.5\right)\\ =P\left(z\le \frac{49,177.5-49,500}{500}\right)\\ =P\left(z\le \frac{-322.5}{500}\right)\\ =P\left(z\le -0.65\right)\end{array}$

Software Procedure:

Step-by-step procedure to obtain the $P\left(z\le -0.65\right)$ using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Left Tail for the region of the curve to shade.
• Enter the data value as -0.65.
• Click OK.

Output using the MINITAB software is given below:

From the output, $P\left(z\le -0.65\right)=0.2578$.

Thus, the power is 0.2578.

To determine

Find the level.

Answer to Problem 5E

The power is 0.4364.

Explanation of Solution

Calculation:

The rejection region is $\overline{X}\le x_0$. That is $P\left(\overline{X}\le x_0\right)=0.80$.

80^th percentile:

Software Procedure:

Step-by-step procedure to obtain the 80^th percentile using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose Probability Value and Left Tail for the region of the curve to shade.
• Enter the Probability value as 0.80.
• Click OK.

Output using the MINITAB software is given below:

From the output, the z-score corresponding to the 80^th percentile is 0.84.

The value of $x_0$ is calculated as follows:

$\begin{array}{c}{x}_0=49,500+0.84\left(500\right)\\ =49,500+\text{420}\\ =\text{49,920}\end{array}$

Therefore, the rejection region is, $\overline{X}\le 49,920$.

Power:

$\begin{array}{c}\text{Power}=P\left(\overline{X}\le 49,920\right)\\ =P\left(z\le \frac{49,920-50,000}{500}\right)\\ =P\left(z\le \frac{-80}{500}\right)\\ =P\left(z\le -0.16\right)\end{array}$

Software Procedure:

Step-by-step procedure to obtain the $P\left(z\le -0.16\right)$ using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Left Tail for the region of the curve to shade.
• Enter the data value as -0.16.
• Click OK.

Output using the MINITAB software is given below:

From the output, $P\left(z\le -0.16\right)=0.4364$.

Thus, the power is 0.4364.

To determine

Find the sample size if the test is made at the 5% level.

Answer to Problem 5E

About 618 tires should be sampled in total so that the power is 0.80 if the test is made at the 5% level.

Explanation of Solution

Calculation:

The 5% rejection region is $\overline{X}\le x_5$ because the alternative hypothesis is of the form $\mu <{\mu }_0$.

From part c., the z-score corresponding to the 5^th percentile is –1.645.

The value of $x_5$ is calculated as follows:

$x_5=50,000-1.645\left(\frac{5000}{\sqrt{n}}\right)$

For power 0.80, the rejection region is $\overline{X}\le x_0$. That is $P\left(\overline{X}\le x_0\right)=0.80$.

From part d., the z-score corresponding to the 80^th percentile is 0.84.

The value of $x_0$ is calculated as follows:

$x_0=49,500+0.84\left(\frac{5000}{\sqrt{n}}\right)$

Therefore,

$\begin{array}{c}50,000-1.645\left(\frac{5000}{\sqrt{n}}\right)=49,500+0.84\left(\frac{5000}{\sqrt{n}}\right)\\ 0.84\left(\frac{5000}{\sqrt{n}}\right)+1.645\left(\frac{5000}{\sqrt{n}}\right)=50,000-49,500\\ \left(\frac{\text{4,200}}{\sqrt{n}}\right)+\left(\frac{\text{8225}}{\sqrt{n}}\right)=500\\ \frac{\text{4,200+8225}}{\sqrt{n}}=500\end{array}$

                    $\begin{array}{c}\frac{\text{12,425}}{\sqrt{n}}=500\\ \sqrt{n}=\frac{\text{12,425}}{500}\\ \text{=617}\text{.5225}\\ \approx \text{618}\end{array}$

Thus, the required sample size is 618.