Question

Using the Wilcoxon Signed-Ranks Test. In Exercises 5–8, refer to the sample data for the given exercises in Section 13-2 on page 611. Use the Wilcoxon signed-ranks test to test the claim that the matched pairs have differences that come from a population with a median equal to zero. Use a 0.05 significance level.

Exercise 5 “Speed Dating: Attributes”

Short answer

There is not enough evidence to conclude that the matched pairs of ratings have differences that do not come from a population with a median equal to 0.

Step-by-step solution

Given information

Referring to Exercise 5 in section 13-2, the data of female attribute ratings and male ratings is given as follows:

Rating of Male by Female	29	38	36	37	30	34	35	23	43
Rating of Female by Male	36	34	34	33	31	17	31	30	42

Define Wilcoxon signed-rank test

The Wilcoxon signed-rank test is used to test the difference between the values of two related samples.

It is a non-parametric test.

Identify the null hypothesis and the alternative hypothesis

The null hypothesis is as follows:

The matched pairs of ratings have differences that come from a population with a median of 0.

The alternative hypothesis is as follows:

The matched pairs of ratings have differences that do not come from a population with a median of 0.

Determine the signed-ranks

The signed-ranks can be calculated in the following steps:

• Compute the differences by subtracting the value inthe second sample from the corresponding value of the first sample.

• The following table shows the differences along with their signs:

Rating of Male by Female	29	38	36	37	30	34	35	23	43
Rating of Female by Male	36	34	34	33	31	17	31	30	42
Difference	-7	+4	+2	+4	-1	+17	+4	-7	+1

• Compute the ranks of absolute differences by sorting themfrom smallest to largest.

• Assign the smallest observation therank of 1, and increase the ranks until the largest observation.

• If any of the observations are repeated, assign the mean value of the ranks to all those observations.

• The following table shows the ranks:

Rating of Male by Female	29	38	36	37	30	34	35	23	43
Rating of Female by Male	36	34	34	33	31	17	31	30	42
Difference	–7	+4	+2	+4	–1	+17	+4	–7	+1
Ranks of |d|	7.5	5	3	5	1.5	9	5	7.5	1.5

• Assign the sign to the ranks according to the sign of the difference.

• The following table shows the sign of the ranks:

Rating of Male by Female	29	38	36	37	30	34	35	23	43
Rating of Female by Male	36	34	34	33	31	17	31	30	42
Difference	–7	+4	+2	+4	–1	+17	+4	–7	+1
Ranks of |d|	7.5	5	3	5	1.5	9	5	7.5	1.5
Signed-Ranks	–7.5	+5	+3	+5	–1.5	+9	+5	–7.5	+1.5

Determine the sum of the ranks

Compute the sum of the positive ranks as shown below:

\(\begin{array}{c}Su{m_{positive}} = 5 + 3 + 5 + 9 + 5 + 1.5\\ = 28.5\end{array}\)

Compute the sum of the negative ranks and then calculate its absolute values.

\(\begin{array}{c}\left| {Su{m_{negative}}} \right| = \left( { - 7.5} \right) + \left( { - 1.5} \right) + \left( { - 7.5} \right)\\ = \left| { - 16.5} \right|\\ = 16.5\end{array}\)

Calculate the test statistic and the critical value

Consider the smaller of the sums as the test statistic.

Here, the smaller sum is 16.5.Thus, T is equal to 16.5.

The critical value for n= 9 and\(\alpha \)= 0.05 for a two-tailed test is equal to 6.

As the test statistic value is greater than the critical value, the null hypothesis fails to reject.

Determine the conclusion of the test

There is not enough evidence to conclude that the matched pairs of ratings have differences that do not come from a population with a median of 0.