Chapter 11.3, Problem 8P

(i)

To determine

Construct the ranks for a Spearman rank correlation test.

Explanation of Solution

Calculation:

Let variable x denotes the abuse, and y denotes the runaways.

Procedure for assigned rank to data values:

• First combine both the samples.
• Arrange the data values in ascending order.
• Rank each of the data value in sequential order.

The rank for each of tied data is computed as,

• First assign the sequential position ranks for all the values that are same.
• Take the mean of all the position ranks that are assigned for same data values.
• Assign the mean rank as the rank position for all the tied data.
• The assigned mean rank is used in calculating the test statistic.

The ranks are,

City	x	y	x rank	y rank
1	49	382	7	7
2	74	510	4	3
3	87	581	3	2
4	10	163	15	14
5	26	210	12	11.5
6	119	791	1	1
7	35	275	10	10
8	13	153	14	15
9	89	491	2	4
10	45	351	8	8
11	53	402	6	6
12	22	209	13	13
13	65	410	5	5
14	38	312	9	9
15	29	210	11	11.5

(ii)

To determine

(a)

Find the level of significance.

State the null and alternative hypothesis.

(b)

Find the value of the sample test statistic.

(c)

Find the P-value of the sample test statistic.

(d)

Mention the conclusion of the test.

(e)

Interpret the conclusion in the context of the application.

Answer to Problem 8P

(a)

The level of significance is 0.01.

(b)

The value of the sample test statistic is 0.985.

(c)

The P-value is $P\text{-value}<0.001$.

(d)

The null hypothesis is rejected.

Explanation of Solution

Calculation:

(a)

From the given information the value of $\alpha$ is 0.01, and to test the claim that there is a monotone-increasing relationship between the ranks of incidents of abuse and number of runaway children.

Hence, the level of significance is 0.01.

Let ${\rho }_s$ denotes the Spearman rank correlation coefficient.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:{\rho }_s=0$

That is, the relation between rank in abuse and rank in runaways is no monotone.

Alternative hypothesis:

$H_1:{\rho }_s>0$

That is, the relation between rank in abuse and rank in runaways is monotone-increasing.

(b)

Test statistic:

The Spearman rank correlation coefficient is,

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

In the formula d is the difference between x and y, and n is the sample size.

The ranks are,

City	x rank	y rank	$d=x-y$	$d^2$
1	7	7	0	0
2	4	3	1	1
3	3	2	1	1
4	15	14	1	1
5	12	11.5	0.5	0.25
6	1	1	0	0
7	10	10	0	0
8	14	15	–1	1
9	2	4	–2	4
10	8	8	0	0
11	6	6	0	0
12	13	13	0	0
13	5	5	0	0
14	9	9	0	0
15	11	11.5	–0.5	0.25
				${\displaystyle \sum d^2}=8.5$

Test statistic:

Substitute ${\displaystyle \sum d^2}$ as 8.5, and n as 15 in the test statistic formula

$\begin{array}{c}{r}_s=1-\frac{6\left(8.5\right)}{15\left({15}^2-1\right)}\\ =1-\frac{51}{3,360}\\ =1-0.015\\ =0.985\end{array}$

Hence, the value of the sample test statistic is 0.985.

(c)

The sample size is 15, and the test is one tailed.

Use the Tables 9: Critical Values for Spearman Rank Correlation:

• Locate the value 15 in column n.
• Locate the value 0.985 in corresponding row for the value 15.
• The value 0.985 is less than 0.750 in the table.
• The one tailed area corresponding to 0.750 is less than 0.001.

The P value is $P<0.001$.

Hence, the P-value is $P\text{-value}<0.001$.

(d)

Rejection rule:

• If the P-value is less than or equal to $\alpha$, then reject the null hypothesis and the test is statistically significant. That is, $P\text{-value}\le \alpha$.

Conclusion:

The P-value is $P<0.001$ and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, $\left(<0.001\right)\left(=P\text{-value}\right)<0.01\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is rejected.

Hence, the data is statistically significant at level 0.01.

(e)

From part (d), the null hypothesis is rejected. This shows that, there is sufficient evidence that the relation between rank in abuse and rank in runaways is monotone-increasing at level of significance 0.01.