Chapter 16, Problem 42DE

Refer to the Baseball 2012 data, which report information on the 2012 Major League Baseball season.

1. a. Rank the teams by the number of wins and their total team salary. Compute the coefficient of rank correlation between the two variables. At the .01 significance level, can you conclude that it is greater than zero?
2. b. Assume that the distributions of team salaries for the American League and National League do not follow the normal distribution. Conduct a test of hypothesis to see whether there is a difference in the two distributions.
3. c. Rank the 30 teams by attendance and by team salary. Determine the coefficient of rank correlation between these two variables. At the .05 significance level, is it reasonable to conclude the ranks of these two variables are related?

To determine

Find the rank correlation coefficient between the number of Wins and total team salary.

Conduct a hypothesis test to see whether the rank correlation is greater than zero.

Answer to Problem 42DE

The rank correlation coefficient between the number of Wins and team salary is 0.105.

Explanation of Solution

Step-by-step procedure to obtain the rank correlation using Minitab software:

• Select Stat > Basic Statistics > Correlation.
• In Variables, select Wins, and team salary from the box on the left.
• Click OK.

Output obtained using Minitab is as follows:

Thus, the rank correlation coefficient between the number of Wins and team salary is 0.105.

Denote the population correlation as $\rho$.

The null and alternative hypotheses are stated as follows:

Null hypothesis:

$H_0:\rho =0$

That is, the correlation between the ranks of the number of wins and team salary.

Alternative hypothesis:

$H_1:\rho >0$

That is, there is positive correlation in the ranks of the number of wins and team salary.

Test statistic:

The test statistic is as follows:

$t=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}$

Here, sample size is 30 and correlation coefficient is 0.105.

The test statistic is computed as follows:

$\begin{array}{c}t=\frac{0.105\sqrt{30-2}}{\sqrt{1-{\left(0.105\right)}^2}}\\ =\frac{0.105\times \sqrt{28}}{\sqrt{1-0.011025}}\\ =\frac{0.5556}{0.9945}\\ \approx 0.559\end{array}$

Thus, the test statistic value is 0.559.

Degrees of freedom:

$\begin{array}{c}df=n-2\\ =30-2\\ =28\end{array}$

The level of significance is 0.01.

Step-by-step software procedure to obtain the critical value using EXCEL software is as follows:

• Open an EXCEL file.
• In cell A1, enter the formula “=T.INV (0.01, 12)”.

Output using the EXCEL is given as follows:

From the EXCEL output, the critical value for the right tailed test is 2.467$\left(=t_{\alpha }\right)$ .

Decision rule:

Reject the null hypothesis H₀ if, $\left|t\right|\text{-calculated}>\left|t_{\alpha }\right|$.

Otherwise fail to reject H₀.

Conclusion:

The value of test statistic is 0.559 and critical value is 2.467.

Here, $\left|t\right|\text{-calculated}\left(=0.559\right)<\left|t_{\alpha }\right|\left(=2.467\right)$.

By the rejection rule, fail to reject the null hypothesis at the 0.01 significance level.

Thus, there is no sufficient evidence to conclude that there is a positive correlation in the ranks of the number of wins and team salary.

To determine

Conduct a test of hypothesis to see whether there is a difference in the distribution of team salaries for the American League and National League.

Answer to Problem 42DE

There is no difference in the distributions of the team salaries for the American League and National League.

Explanation of Solution

The Mann-Whitney test is suitable for the comparison of two inpendent groups.

The null and alternative hypotheses are stated as follows:

Null hypothesis: There is no difference in the distributions of the team salaries for the American League and National League.

Alternative hypothesis: There is a difference in the distributions of the team salaries for the American League and National League.

Step-by-step procedure to obtain the test statistic using Excel MegaStat software is as follows:

• Choose MegaStat > Nonparametric Tests > Wilcoxon – Mann/Whitney Test.
• In Group 1, enter the column of American League.
• In Group 2, enter the second column of National League.
• Select on Continuity correction and Select the Alternative.
• Click OK

Output obtained using Excel Mega stat software is as follows:

From the above output, the value of the test statistic is 0.478 and the p-value is 0.6326.

Conclusion:

Consider the significance level is 0.05.

Here, the p-value is greater than the significance level 0.05.

By the rejection rule, fail to reject the null hypothesis at the 0.05 significance level.

Therefore, there is no difference in the distributions of the team salaries for the American League and National League.

To determine

Find the coefficient of rank correlation between attendance and team salary.

Check whether it is reasonable to conclude the ranks of these two variables are related.

Answer to Problem 42DE

The coefficient of rank correlation between attendance and team salary is 0.805.

Explanation of Solution

Step-by-step procedure to obtain the rank correlation using Minitab software:

• Select Stat > Basic Statistics > Correlation.
• In Variables, select Attendance, and team salary from the box on the left.
• Click OK.

Output obtained using Minitab is as follows:

From the above output, the coefficient of rank correlation between attendance and team salary is 0.805.

Denote the population correlation as $\rho$.

The null and alternative hypotheses are stated below:

Null hypothesis:

$H_0:\rho =0$

That is, there is no correlation between attendance and team salary.

Alternative hypothesis:

$H_1:\rho \ne 0$

That is, there is a correlation between attendance and team salary.

Here, sample size is 30 and correlation coefficient is 0.805.

The test statistic is as follows:

$\begin{array}{c}t=\frac{0.805\sqrt{30-2}}{\sqrt{1-{\left(0.805\right)}^2}}\\ =\frac{4.25966}{0.593275}\\ \approx 7.18\end{array}$

Thus, the test statistic value is 7.18.

Degrees of freedom:

$\begin{array}{c}df=n-2\\ =30-2\\ =28\end{array}$

The level of significance is 0.05.

Step-by-step software procedure to obtain the critical value using EXCEL software is as follows:

• Open an EXCEL file.
• In cell A1, enter the formula “=T.INV.2T (0.05, 28)”.

Output using the EXCEL is given as follows:

From the EXCEL output, the critical value is 2.048$\left(=t_{\alpha }\right)$ .

Conclusion:

The value of test statistic is 7.18 and critical value is 2.048.

Here, $\left|t\right|\text{-calculated}\left(=7.18\right)>\left|t_{\alpha }\right|\left(=2.048\right)$.

By the rejection rule, reject the null hypothesis at the 0.05 significance level.

Therefore, it is reasonable to conclude that the ranks of these two variables are related.