Chapter 16, Problem 38CE

a.

To determine

Obtain the degree of association between the citizens ranked with sports and citizens ranked with world events.

Answer to Problem 38CE

The degree of association between the citizens ranked with sports and citizens ranked with world events is 0.486.

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.

The table represents the difference between ranks of each pair:

(1) Citizen	(2) Sports	(3) Rank 1	(4) World Events	(5) Rank 2	(6) Difference $=\left\{\left(3\right)-\left(5\right)\right\}$	(7) $d^2$
1	47	5	49	4	$1\left(=5-4\right)$	1
2	12	1	10	1	$0\left(=1-1\right)$	0
3	62	10	76	11	$-1\left(=10-11\right)$	1
4	81	11	92	14	$-3\left(=11-14\right)$	9
5	90	14	86	12.5	$1.5\left(=14-12.5\right)$	2.25
6	35	3	42	3	$0\left(=3-3\right)$	0
7	61	9	61	7	$2\left(=9-7\right)$	4
8	87	12.5	75	9.5	$3\left(=12.5-9.5\right)$	9
9	59	7	86	12.5	$-5.5\left(=7-12.5\right)$	30.25
10	40	4	61	7	$-3\left(=4-7\right)$	9
11	87	12.5	18	2	$10.5\left(=12.5-2\right)$	110.25
12	16	2	75	9.5	$-7.5\left(=2-9.5\right)$	56.25
13	50	6	51	5	$1\left(=6-5\right)$	1
14	60	8	61	7	$1\left(=8-7\right)$	1
						${\displaystyle \sum d^2}=234$

In this context, the number of paired observation, n  is 14.

The Spearman’s coefficient of rank correlation obtained is given below:

Substitute the corresponding values to get the rank correlation.

$\begin{array}{c}{r}_s=1-\frac{6\left(234\right)}{14\left({14}^2-1\right)}\\ =1-\frac{1,404}{2,730}\\ =1-0.5143\\ =0.486\end{array}$

Thus, the Spearman’s coefficient of rank correlation is 0.486.

b.

To determine

State whether the rank correlation between the sports and world events “knowledge” scores is greater than zero.

Answer to Problem 38CE

The rank correlation between the sports and world events “knowledge” scores is greater than zero.

Explanation of Solution

The test hypothesis is given as follows:

Null hypothesis:

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

Alternative hypothesis:

$H_1:$ The rank correlation between the sports and world events “knowledge” scores is greater than zero.

In this context, the number of paired observations is 14.

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:

$\begin{array}{c}n-2=14-2\\ =12\end{array}$

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 right-tailed test is obtained as 1.782 using the EXCEL formula, “=T.INV (0.95,12)”.

From Part (a), the rank correlation, $r_s$ is 0.486.

The test statistic will be obtained as follows:

Substitute $r_s$ as 0.486, n as 14.

$\begin{array}{c}t=0.486\sqrt{\frac{14-2}{1-{\left(0.486\right)}^2}}\\ =0.486\sqrt{\frac{12}{0.764}}\\ =0.486\left(3.964\right)\\ =1.926\end{array}$

Conclusion:

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

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

Therefore, there is evidence to support the claim that the rank correlation between the sports and world events “knowledge” scores is greater than zero.