Chapter 15.3, Problem 17E

Exercise 17 demonstrates that the results of the signed-rank test may be misleading when the assumption of symmetry is violated.

Asymmetric differences: Consider the following paired samples and their differences.

1. Verify that the median of the sample differences is equal to 0.
2. Compute the test statistic for the signed-rank test.
3. Do you reject the null hypothesis that the median is equal to 0? Use the $\alpha =0.05$ level of significance.

To determine

To verify:The median of the sample difference is $0$ .

Explanation of Solution

Given information:

The level of significance is $0.05$ and the data of samples is shown in table-1 and difference table is shown in Table-2 below.

Sample 1
68	59	66	67	64	50 60 55	74	59	66	55	68
58	70	71	54	68	65 82 34	89	97	73	78
Sample 2
80	70	76	76	72	57 66 60	78	62	68	56	68
45	56	56	38	51	47 63 14	68	75	50	54

  Table-1

Differences
-12	-11	-10	-9	-8	-7	-6	-5	-4	-3	-2	-1	0
13	14	15	16	17	18	19	20	21	22	23	24

  Table-2

Calculation:

From Table-2, the sample size is $25$ , which is odd.

The median for odd data is,

  $\text{median}={\left(\frac{n+1}{2}\right)}^{th}\text{term}$

Substitute the values in above formula.

  $\begin{array}{c}\text{median}={\left(\frac{25+1}{2}\right)}^{th}\text{term}\\ ={\left(13\right)}^{th}\text{term}\end{array}$

Since, the thirteenth term of table-2 is zero.

Thus, the median of difference data is $0$ .

To determine

To find:The test statistics.

Answer to Problem 17E

The test statistics is $78$ .

Explanation of Solution

Given information:

The level of significance is $0.05$ and the data of samples is shown in table-1 and difference table is shown in Table-2 below.

Sample 1
68	59	66	67	64	50 60 55	74	59	66	55	68
58	70	71	54	68	65 82 34	89	97	73	78
Sample 2
80	70	76	76	72	57 66 60	78	62	68	56	68
45	56	56	38	51	47 63 14	68	75	50	54

  Table-1

Differences
-12	-11	-10	-9	-8	-7	-6	-5	-4	-3	-2	-1	0
13	14	15	16	17	18	19	20	21	22	23	24

  Table-2

Calculation:

The MINITAB output is shown below.

  

  Figure-1

From Figure-1, it is clear that the test statistics is $78$ .

Therefore, the test statistics is $78$ .

To determine

To find:Whether to reject the null hypothesis.

Answer to Problem 17E

The null hypothesis is rejected.

Explanation of Solution

Given information:

The level of significance is $0.05$ and the data of samples is shown in table-1 and difference table is shown in Table-2 below.

Sample 1
68	59	66	67	64	50 60 55	74	59	66	55	68
58	70	71	54	68	65 82 34	89	97	73	78
Sample 2
80	70	76	76	72	57 66 60	78	62	68	56	68
45	56	56	38	51	47 63 14	68	75	50	54

  Table-1

Differences
-12	-11	-10	-9	-8	-7	-6	-5	-4	-3	-2	-1	0
13	14	15	16	17	18	19	20	21	22	23	24

  Table-2

Calculation:

The null hypothesis is, the median is equal to zero.

The alternative hypothesis is, the median is not equal to zero.

Since, the test statistics is $78$ .

For sample size $25$ and level of significance of $0.05$ the critical value is $89$ .

Since, the test statistics is less than the critical value.

Thus, hypothesis $H_0$ is rejected.

Therefore, the null hypothesis is rejected.