Chapter 10.9, Problem 82E

Exercises 8.83 and 10.73 presented some data collected in a 1993 study by Susan Beckham and her colleagues. In this study, measurements of anterior compartment pressure (in millimeters of mercury) were taken for ten healthy runners and ten healthy cyclists. The researchers also obtained pressure measurements for the runners and cyclists at maximal O₂ consumption. The data summary is given in the accompanying table.

1. a Is there sufficient evidence to support a claim that the variability of compartment pressure differs for runners and cyclists who are resting? Use α = .05.
2. b i What can be said about the attained significance level using a table in the appendix?

ii Applet Exercise What can be said about the attained significance level using the appropriate applet?

c Is there sufficient evidence to support a claim that the variability in compartment pressure between runners and cyclists differs at maximal O₂ consumption? Use α = .05.

d i What can be said about the attained significance level using a table in the appendix?

ii Applet Exercise What can be said about the attained significance level using the appropriate applet?

To determine

State whether there is sufficient evidence to claim that the variability of compartment pressure for runners who are resting is different from the variability of compartment pressure for cyclists who are resting using $\alpha =0.05$.

Answer to Problem 82E

There is no evidence that the variability of compartment pressure for runners who are resting is different from the variability of compartment pressure for cyclists who are resting.

Explanation of Solution

The test hypotheses are given as follows:

Denote ${\sigma }_1^2$ as the population variance compartment pressure for runners who are resting and ${\sigma }_2^2$ as the population variance compartment pressure for runners who are cyclists.

Null hypothesis:

$H_0:{\sigma }_1^2={\sigma }_2^2$

That is, the variability of compartment pressure for runners who are resting is the same as the variability of compartment pressure for cyclists who are resting.

Alternative hypothesis:

$H_a:{\sigma }_1^2\ne {\sigma }_2^2$

That is, the variability of compartment pressure for runners who are resting is different from the variability of compartment pressure for cyclists who are resting.

Test statistic:

$F=\frac{S_1^2}{S_2^2}$

Where $S_1^2,S_2^2$ are the sample variances.

The test statistic is obtained as given below:

Substitute $S_1$ as 3.92 and $S_2$ as 3.98.

$\begin{array}{c}F=\frac{{3.92}^2}{{3.98}^2}\\ =\frac{15.3664}{15.8404}\\ =0.97\end{array}$

The numerator degrees of freedom, ${\nu }_1$ is obtained as given below:

$\begin{array}{c}{\nu }_1=n_1-1\\ =10-1\\ =9\end{array}$

The denominator degrees of freedom, ${\nu }_2$ is obtained as given below:

$\begin{array}{c}{\nu }_2=n_2-1\\ =10-1\\ =9\end{array}$

Rejection region:

In context, the level of the test, $\alpha$ is 0.05.

Here, the alternative is a two-tailed test. The level of the test is $0.025\left(=\frac{0.05}{2}\right)$. Therefore, the rejection region will be $F>F_{\alpha }$.

The critical value for $F_{0.025}$ is obtained as given below:

Step-by-step procedure to obtain the critical value using Applet:

• Select F-Ratio Probabilities and Quantiles.
• In df1, enter 9.
• In df2, enter 9.
• In probability:, enter 0.025.

The output obtained is as follows:

Therefore, the rejection region for the two-tailed test is $F>4.026$.

Decision rule:

• If $F>4.026$, reject the null hypothesis.
• Otherwise fail to reject the null hypothesis.

Conclusion:

Here, the test statistic, F does not falls in the rejection region.

By the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence to conclude that the variability of compartment pressure for runners who are resting is different from the variability of compartment pressure for cyclists who are resting at $\alpha =0.05$.

To determine

Obtain the attained ­p-value using a table in the appendix.

Answer to Problem 82E

The attained ­p-value is $0.2<p\text{-value}<1$.

Explanation of Solution

Here, the probability of the test statistic $\left[P\left(F<0.97\right)\right]$ for a two-tailed test is obtained as given below:

From the “Appendix 3, Table 7 Percentage Points of the F Distributions”, the test statistic value of 0.97 lies between 0.1 and 0.5 quantiles of F distribution with (9, 9) degrees of freedom.

As $F=0.97$ lies between 0.1 and 0.5 critical values,  the p-value for the two-tailed test will be as follows:

$\begin{array}{l}0.1<2\times p\text{-value}<0.5\\ 0.2<p\text{-value}<1\end{array}$

To determine

Obtain the exact p-value using the Applet F-Ratio Probabilities and Quantiles.

Answer to Problem 82E

The exact p-value is 0.964.

Explanation of Solution

From Part (a), the test statistic is 0.97.

p-value:

The probability of the test statistic $\left[P\left(F<0.97\right)\right]$ for a two-tailed test is obtained as given below:

$2\times \left[P\left(F<0.97\right)\right]=2\left(1-P\left(F>0.97\right)\right)$

Step-by-step procedure to obtain the probability using Applet:

• Select F-Ratio Probabilities and Quantiles.
• In df1, enter 9.
• In df2, enter 9.
• In x:, enter 0.97.

Output obtained is as follows:

From the output, it is clear that the probability of the test statistic $\left[P\left(F>0.97\right)\right]$ is 0.518.

$\begin{array}{c}2\times \left[P\left(F<0.97\right)\right]=2\left(1-P\left(F>0.97\right)\right)\\ =2\times \left[1-0.518\right]\\ =2\times 0.482\\ =0.964\end{array}$

Therefore, the probability of the test statistic is 0.964.

To determine

State whether there is sufficient evidence to claim that the variability of compartment pressure for runners at maximal $O_2$ consumption is different from the variability of compartment pressure for cyclists at maximal $O_2$ consumption using $\alpha =0.05$.

Answer to Problem 82E

There is evidence that the variability of compartment pressure for runners at maximal $O_2$ consumption is different from the variability of compartment pressure for cyclists at maximal $O_2$ consumption.

Explanation of Solution

The test hypotheses are given as follows:

Denote ${\sigma }_1^2$ as the population variance compartment pressure for runners at maximal $O_2$ consumption and ${\sigma }_2^2$ as the population variance compartment pressure for runners at maximal $O_2$ consumption.

Null hypothesis:

$H_0:{\sigma }_1^2={\sigma }_2^2$

That is, the variability of compartment pressure for runners at maximal $O_2$ consumption is different from the variability of compartment pressure for cyclists at maximal $O_2$ consumption.

Alternative hypothesis:

$H_a:{\sigma }_1^2\ne {\sigma }_2^2$

That is, the variability of compartment pressure for runners at maximal $O_2$ consumption is different from the variability of compartment pressure for cyclists at maximal $O_2$ consumption.

Test statistic:

$F=\frac{S_1^2}{S_2^2}$

Where $S_1^2,S_2^2$ are the sample variances.

The test statistic is obtained as given below:

Substitute $S_1$ as 16.9 and $S_2$ as 4.67.

$\begin{array}{c}F=\frac{{16.9}^2}{{4.67}^2}\\ =\frac{285.61}{21.8089}\\ =13.096\end{array}$

The numerator degrees of freedom, ${\nu }_1$ is obtained as given below:

$\begin{array}{c}{\nu }_1=n_1-1\\ =10-1\\ =9\end{array}$

The denominator degrees of freedom, ${\nu }_2$ is obtained as given below:

$\begin{array}{c}{\nu }_2=n_2-1\\ =10-1\\ =9\end{array}$

From Part (a), the rejection region for the two-tailed test is $F>4.026$.

Decision rule:

• If $F>4.026$, reject the null hypothesis.
• Otherwise fail to reject the null hypothesis.

Conclusion:

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

By the decision rule, reject the null hypothesis.

Therefore, there is evidence to conclude that the variability of compartment pressure for runners at maximal $O_2$ consumption is different from the variability of compartment pressure for cyclists at maximal $O_2$ consumption at $\alpha =0.05$.

To determine

Obtain the attained ­p-value using a table in the appendix.

Answer to Problem 82E

The attained ­p-value is $p\text{-value}<0.01$.

Explanation of Solution

Here, the probability of the test statistic $\left[P\left(F>13.096\right)\right]$ for a two-tailed test is obtained as given below:

From the “Appendix 3, Table 7 Percentage Points of the F Distributions”, the test statistic value of 13.096 lies above 0.005 quantiles of F distribution with (9, 9) degrees of freedom.

As $F=13.096$ lies greater than 0.005 critical values,  the p-value for the two-tailed test will be as follows:

$\begin{array}{l}2\times p\text{-value}<0.005\\ p\text{-value}<0.01\end{array}$

To determine

Obtain the exact p-value using the Applet F-Ratio Probabilities and Quantiles.

Answer to Problem 82E

The exact p-value is 0.

Explanation of Solution

From Part (a), the test statistic is 13.096.

p-value:

The probability of the test statistic $\left[P\left(F>13.096\right)\right]$ for a two-tailed test is obtained as given below:

Step-by-step procedure to obtain probability using Applet:

• Select F-Ratio Probabilities and Quantiles.
• In df1, enter 9.
• In df2, enter 9.
• In x:, enter 13.096.

Output obtained is as follows:

From the output, it is clear that the value of x must be in the range [0.01, 10.11]. Therefore, the probability greater than 13.096 will be assumed to be 0.

Therefore, the probability of the test statistic is 0.