Chapter 13, Problem 13.2.2RE

(a)

To determine

To state: The test hypotheses.

To identify : The claim for the given situation.

Answer to Problem 13.2.2RE

The testing hypotheses are given below:

Null hypothesis $H_0$:

$H_0:\text{Median}=40,000\text{miles}$

Alternative hypothesis $H_1$:

$H_1:\text{Median}\ne 40,000\text{miles}$

The claim is that, the median lifetime of a certain brand of truck tires is 40,000 miles.

Explanation of Solution

Given info:

The claim given by a tier manufacturer is the median lifetime of a certain brand of truck tires is 40,000 miles. To examine this a random sample of 30 tires are tested and among them 12 tires are above 40,000 miles. The level of significance $\alpha =0.05$.

Justification:

Null hypothesis states that the median lifetime of a certain brand of truck tires is 40,000 miles.

Alternative hypothesis states that the median lifetime of a certain brand of truck tires is not equal to 40,000 miles.

Here, the claim is to test whether the median lifetime of a certain brand of truck tires is 40,000 miles.

Hence, the claim is that the median lifetime of a certain brand of truck tires is 40,000 miles.

(b)

To determine

To find: The critical value.

Answer to Problem 13.2.2RE

The critical value is $\pm 1.96$.

Explanation of Solution

Calculation:

Critical value:

Here, the test is two tailed test.

For the level of significance $\alpha =0.05$,

$\begin{array}{c}\text{Two}\text{tailed}\text{area}=\frac{\alpha }{2}\\ =\frac{0.05}{2}\\ =0.025\end{array}$

Hence, the cumulative area to the left is,

$\begin{array}{c}\text{Area}\text{to}\text{the}\text{left}=1-\text{Area}\text{to}\text{the}\text{right}\\ =1-0.025\\ =0.975\end{array}$

From Table E of the standard normal distribution from Appendix A, the critical value is $\pm 1.96$.

Thus, the two tailed critical value of z for the level of significance $\alpha =0.05$ is $\left(z_{\frac{\alpha }{2}}\right)=\pm 1.96$

(c)

To determine

To find: The test statistic value.

Answer to Problem 13.2.2RE

The test statistic is -0.9129.

Explanation of Solution

Calculation:

Test statistic:

If $n\le 25,$ the test statistic for sign test is $x$.

If $n>25$, the test statistic for sign test is $z=\frac{\left(x+0.5\right)-0.5n}{\frac{\sqrt{n}}{2}}$

Where $x$ is the smaller number of $(+)$& $(-)$ and $n$ is the sample size.

Here, $n=30$ which is greater than 25. That is $30>25$.

And the number of observation greater than the median is 12.

That is, the value of $X=12$

The test statistics is derived as follows:

$\begin{array}{c}Z=\frac{\left(X+0.5\right)-0.5n}{\frac{\sqrt{n}}{2}}\\ =\frac{\left(12+0.5\right)-0.5\left(30\right)}{\frac{\sqrt{30}}{2}}\\ =\frac{\left(12.5\right)-15}{2.7386}\\ =-0.9129\end{array}$

Thus the test statistic is –0.9129.

(d)

To determine

Whether the null hypothesis is rejected for the given situation.

Answer to Problem 13.2.2RE

The null hypothesis $H_0$ should not be rejected.

Explanation of Solution

Decision rule for a two tailed test:

• If $z<-z_{\frac{\alpha }{2}}\left(=-1.96\right)$, then reject the null hypothesis
• If $z>z_{\frac{\alpha }{2}}\left(=+1.96\right)$, then reject the null hypothesis

Conclusion:

The value of tests statistic is -0.9129 and the critical value is $\pm 1.96$.

Here, the test statistic value is not less than the critical value -1.96 .

That is, $-0.9129\left(=z\right)\text{>}-1.96\left(=-z_{\frac{\alpha }{2}}\right)$

And $-0.9129\left(=z\right)\text{<}1.96\left(=z_{\frac{\alpha }{2}}\right)$

By the rejection rule “failed to reject the null hypotheses”.

Thus, the null hypothesis is not rejected.

(e)

To determine

To summarize: The results.

Answer to Problem 13.2.2RE

The conclusion is that, there is no enough evidence to reject the claim that the median lifetime of a certain brand of truck tires is not equal to 40,000 miles.

Explanation of Solution

From part (d), the null hypothesis is not rejected.

Hence, there is no enough evidence to reject the claim that the median lifetime of a certain brand of truck tires is not equal to 40,000 miles.