Chapter 10.3, Problem 17E

A survey published in the American Journal of Sports Medicine² reported the number of meters (m) per week swum by two groups of swimmers—those who competed exclusively in breaststroke and those who competed in the individual medley (which includes breaststroke). The number of meters per week practicing the breaststroke was recorded for each swimmer, and the summary statistics are given below. Is there sufficient evidence to indicate that the average number of meters per week spent practicing breaststroke is greater for exclusive breaststrokers than it is for those swimming individual medley?

1. a State the null and alternative hypotheses
2. b What is the appropriate rejection region for an α = .01 level test?
3. c Calculate the observed value of the appropriate test statistic.
4. d What is your conclusion?
5. e What is a practical reason for the conclusion you reached in part (d)?

To determine

Provide the appropriate alternative and null hypotheses.

Answer to Problem 17E

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

Alternative hypothesis: $H_a:{\mu }_1>{\mu }_2$

Explanation of Solution

The aim is to check whether the average distance for breaststroke is greater than that of individual medley.

Therefore, the test hypotheses are given below:

Denote ${\mu }_1$ as the average distance for breaststroke and ${\mu }_2$ as the average distance for individual medley.

Null hypothesis:

 $H_0:{\mu }_1={\mu }_2$.

That is, the average distance for breaststroke is same as that of individual medley.

Alternative hypothesis:

$H_a:{\mu }_1>{\mu }_2$

That is, the average distance for breaststroke is greater than that of individual medley.

To determine

Provide the appropriate rejection region for an $\alpha =0.1$ level test.

Answer to Problem 17E

The rejection region under the level of significance of 0.01 is $\underset{\_}{z>2.33}$.

Explanation of Solution

Rejection region:

Here, the alternative is right-tailed test. In context, the level of the test $\alpha$ is 0.01.

If  $z>z_{\alpha }$ or , reject the null hypothesis.

Critical value:

From “Appendix 3, Table 4 Normal curve areas”, the critical value for the level of significance of 0.01 is 2.33.

Thus, the rejection region under the level of significance of 0.01 is $z>2.33$.

To determine

Find the observed value of the appropriate test statistic.

Answer to Problem 17E

The observed value of the appropriate test statistic is 4.76.

Explanation of Solution

Test statistic:

For the large sample n, the sample means of ${\overline{Y}}_1\text{and}{\overline{Y}}_2$ are a point estimate of ${\mu }_1\text{ and }{\mu }_2$. Thus, the test statistic is given below:

$Z=\frac{\left({\overline{Y}}_1-{\overline{Y}}_2\right)-D_0}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}$

Where $D_0={\mu }_1-{\mu }_2$ is the difference, $n_1\text{and }{n}_2$ are the sample sizes of two population and ${\sigma }_1^2\text{and}{\sigma }_2^2$ are the population variances, respectively.

For the large samples $(n_i>30)$, the population variance is not known. The sample variances provide a good estimate of the population.

Substitute ${\overline{Y}}_1$ as 9,017, ${\overline{Y}}_2$ as 5,853, $D_0$ as 0, $s_1$ as 7,162, $s_2$ as 1,961, $n_1$ as 130, and $n_2$ as 80.

$\begin{array}{c}Z=\frac{\left({\overline{Y}}_1-{\overline{Y}}_2\right)-D_0}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{9,017-5,853}{\sqrt{\frac{7,{162}^2}{130}+\frac{1,{961}^2}{80}}}\\ =\frac{3,164}{665.312}\\ =4.76\end{array}$

Therefore, the observed value of the appropriate test statistic is 4.76.

To determine

Provide the conclusion.

Answer to Problem 17E

The null hypothesis is rejected.

Explanation of Solution

From Part (c), the test statistic is 4.76.

Decision rule:

• If $z>2.33$, reject the null hypothesis.
• Otherwise, fail to reject the null hypothesis.

Conclusion:

Here, the test statistic Z falls in the rejection region.

Therefore, by the decision rule, reject the null hypothesis.

Therefore, there is evidence to conclude that the average distance for breaststroke is greater than that of individual medley.

To determine

Provide the practical conclusion for the conclusion obtained in Part (d).

Explanation of Solution

The practical reason for the conclusion test is that the swimmers who compete in individual medley have to practice other forms of swimming and should only concentrate on breaststroke, whereas the swimmers who also compete exclusively in breaststroke can solely concentrate on breaststroke.