Chapter 9.1, Problem 5E

For Exercises 5 through 16, perform each of the following steps.

a. State the hypotheses and identify the claim.

b. Find the critical value(s).

c. Compute the test value.

d. Make the decision.

e. Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified.

5. Recreational Time A researcher wishes to see if there is a difference between the mean number of hours per week that a family with no children participates in recreational activities and a family with children participates in recreational activities. She selects two random samples and the data are shown. At α = 0.10, is there a difference between the means?

To determine

To identify: The claim and state $H_0$ and $H_1$.

Answer to Problem 5E

The claim is that “there is a difference between the mean number of hours per week that a family with no children participates in recreational activities and a family with children participates in recreational activities”.

Null hypothesis: $H_0:{\mu }_1={\mu }_2$

Alternative hypothesis: $H_1:{\mu }_1\ne {\mu }_2$ (claim).

Explanation of Solution

Given info:

For no children, ${\overline{X}}_1=8.6$, ${\sigma }_1=2.1$ and $n_1=36$

For children, ${\overline{X}}_2=10.6$, ${\sigma }_2=2.7$ and $n_2=36$

Justification:

Here, the claim is that there is a difference between the mean number of hours per week that a family with no children participates in recreational activities and a family with children participates in recreational activities. This can be written as ${\mu }_1\ne {\mu }_2$. The complement of the claim is ${\mu }_1={\mu }_2$. In the given experiment, the alternative hypothesis indicates the claim.

To determine

To find: The critical value and rejection region for a two-tailed test with $\alpha =0.10$.

Answer to Problem 5E

The critical value at $\alpha =0.10$ is ±1.65.

The rejection region for a two-tailed test with $\alpha =0.10$ is, $z<-z_0\left(=-1.65\right)$, $z>+z_0\left(=+1.65\right)$.

Explanation of Solution

Calculation:

Here, the test is two tailed test.

From the Standard Normal Table, the critical value at $\alpha =0.10$ is ±1.65.

The rejection region for a two-tailed test with $\alpha =0.10$ is, $z<-z_0\left(=-1.65\right)$, $z>+z_0\left(=+1.65\right)$.

To determine

To find: The standardized test statistic z.

Answer to Problem 5E

The standardized test statistic z is –3.51.

Explanation of Solution

Calculation:

$\begin{array}{c}z=\frac{\left({\overline{X}}_1-{\overline{X}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_1}}}\\ =\frac{8.6-10.6}{\sqrt{\frac{{\left(2.1\right)}^2}{36}+\frac{{\left(2.7\right)}^2}{36}}}\\ =\frac{-2}{0.57}\\ =-3.51\end{array}$

To determine

To decide: Whether to reject or fail to reject the null hypothesis at a level of significance of $\alpha =0.10$.

Answer to Problem 5E

The decision is “reject the null hypothesis”.

Explanation of Solution

Calculation:

Decision rule:

• If $z<z_0\left(=-1.65\right)$, then reject the null hypothesis.
• If $z>z_0\left(=+1.65\right)$, then reject the null hypothesis.

Here, the test statistic is less than the critical value. That is, $z\left(=-3.51\right)<z_0\left(=-1.65\right)$.

Thus, the decision is “reject the null hypothesis”.

To determine

To interpret: The decision in the context of the original claim.

Answer to Problem 5E

The conclusion is that, there is enough evidence to support the claim that there is a difference between the mean number of hours per week that a family with no children participates in recreational activities and a family with children participates in recreational activities.

Explanation of Solution

From part (d), the null hypothesis is rejected. Thus, there is enough evidence to support the claim that there is a difference between the mean number of hours per week that a family with no children participates in recreational activities and a family with children participates in recreational activities at 10% level of significance.