Chapter 16, Problem 42DA

a.

To determine

Obtain the correlation between the rankings of the number of wins and total team salary.

State whether it can be concluded that there is a positive association between the rankings of number of wins and total team salary.

Answer to Problem 42DA

The correlation between the rankings of the number of wins and total team salary is 0.727.

The conclusion is that there is evidence that there is a positive association between the number of wins and total team salary.

Explanation of Solution

Spearman’s coefficient of rank correlation:

$r_s=1-\frac{6{\displaystyle \sum d^2}}{n\left(n^2-1\right)}$

Here, d is the difference between ranks of each pair.

n is the number of paired observations.

Step-by-step procedure to obtain correlation using MINITAB is given below:

• Select Stat > Basic Statistics > Correlation.
• In Variables, select Wins, and Team salary from the box on the left.
• In drop down box, select Spearman Rho.
• Click OK.

Output obtained using MINITAB is given below:

The rank correlation value of 0.165 reveals that there is a slight positive correlation between the rankings of the number of wins and total team salary.

The test hypothesis is given as follows:

Null hypothesis:

$H_0:$ The rank correlation in the population is zero.

Alternative hypothesis:

$H_1:$ There is a positive association between the rankings of the number of wins and total team salary.

If the sample size is greater than 10, then the sampling distribution of $r_s$ follows the t distribution with n–2 df.

Hypothesis test for rank correlation:

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

Degrees of freedom:

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

Decision rule:

• If $t>t_{0.01}$, reject the null hypothesis.
• Otherwise, fail to reject the null hypothesis.

In this context, the critical value $t_{0.01}(t_{\alpha })$ for the right-tailed test is obtained as 2.467 using the EXCEL formula, “=T.INV (0.99,28)”.

The test statistic will be obtained as follows:

Substitute $r_s$ as 0.165, n as 30.

$\begin{array}{c}t=0.165\sqrt{\frac{30-2}{1-{\left(0.165\right)}^2}}\\ =0.165\sqrt{\frac{28}{0.9728}}\\ =0.165\left(5.365\right)\\ =0.885\end{array}$

Conclusion:

Here, the test statistic is less than the critical value.

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

Therefore, there is no evidence to support the claim that there is a positive association between the rankings of the number of wins and total team salary.

b.

To determine

State whether there is a difference between two populations.

Answer to Problem 42DA

There is no difference between two populations.

Explanation of Solution

In this context, 1^st population has 15 observations and 2^nd population has 15 observations and the assumed significance level is 0.05.

The test hypothesis is given as follows:

Null hypothesis:

$H_0:$ The two populations are same.

Alternative hypothesis:

$H_1:$ The two populations are not same.

Decision rule:

• If $z<-z_{0.025}\text{ or }z>z_{0.025}$, reject the null hypothesis.
• Otherwise, fail to reject the null hypothesis.

In this context, the critical value for $z_{0.975}$ is obtained as 1.96 using the EXCEL formula “=NORM.S.INV (0.975)”.

The test statistic will be obtained under the following two categories:

$z=\frac{W-\frac{n_1\left(n_1+n{2}_{}+1\right)}{2}}{\sqrt{\frac{n_1{n}_2\left(n_1+n{2}_{}+1\right)}{12}}}$.

Here, $n_1$ is the number of observations in the first population,

$n_2$ is the number of observations in the second population, and

W is the sum of ranks from the first population.

The table represents the ranks for the two populations:

American	Rank	National	Rank
118.9	18	65.8	1
168.7	26	89.6	8
110.7	13	117.2	16
87.7	7	117.7	17
172.8	27	98.3	9
69.1	2	230.4	30
112.9	14	84.6	5
146.4	24	98.7	10
108.3	12	100.1	11
213.5	29	133	22
80.8	4	85.9	6
123.2	20	126.6	21
74.8	3	166.5	25
144.8	23	120.3	19
116.4	15	174.5	28

The sum of the ranks from the first population is as follows:

$\begin{array}{c}W=18+26+\mathrm{...}+15\\ =237\end{array}$

The test statistic will be obtained as given below:

Substitute W as 237, n₁ as 15, and n₂ as 15.

$\begin{array}{c}z=\frac{237-\frac{15\left(15+15+1\right)}{2}}{\sqrt{\frac{\left(15\right)\left(15\right)\left(15+15+1\right)}{12}}}\\ =\frac{237-232.5}{24.109}\\ =-0.187\end{array}$

Conclusion:

Here, the test statistic is greater than the critical value $-z_{0.025}=-1.96$.

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

Therefore, there is no evidence to support the claim that the two populations are different.

c.

To determine

Obtain the correlation between the rankings of the attendance and total team salary.

State whether it can be concluded that there is a positive association between the rankings of the attendance and total team salary.

Answer to Problem 42DA

The correlation between the rankings of the attendance and total team salary is 0.727.

The conclusion is that there is evidence that there is a association between the number of attendance and total team salary.

Explanation of Solution

Step-by-step procedure to obtain correlation using MINITAB is given below:

• Select Stat > Basic Statistics > Correlation.
• In Variables, select Attendance, and Team salary from the box on the left.
• In drop down box, select Spearman Rho.
• Click OK.

Output obtained using MINITAB is given below:

The rank correlation value of 0.681 reveals that there is a positive correlation between rankings of the number of attendance and total team salary.

The test hypothesis is given as follows:

Null hypothesis:

$H_0:$ The rank correlation in the population is zero.

Alternative hypothesis:

$H_1:$ There is an association between the rankings of the number of attendance and total team salary.

If the sample size is greater than 10, then the sampling distribution of $r_s$ follows the t distribution with n–2 df.

Hypothesis test for rank correlation:

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

Degrees of freedom:

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

Decision rule:

• If $t>t_{0.05}$, reject the null hypothesis.
• Otherwise, fail to reject the null hypothesis.

In this context, the critical value $t_{0.05}(t_{\alpha })$ for the two-tailed test is obtained as 2.048 using the EXCEL formula, “=T.INV (0.95,28)”.

The test statistic will be obtained as follows:

Substitute $r_s$ as 0.681, n as 30.

$\begin{array}{c}t=0.681\sqrt{\frac{30-2}{1-{\left(0.681\right)}^2}}\\ =0.681\sqrt{\frac{28}{0.4638}}\\ =0.681\left(7.77\right)\\ =5.29\end{array}$

Conclusion:

Here, the test statistic is greater than the critical value.

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

Therefore, there is enough evidence to support the claim that there is an association between the rankings of the number of attendance and total team salary.