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 6 “Speed Dating: Attractiveness”

Wilcoxon signed-rank test is the non-parametric counterpart of the t-test.

In reference to exercise 6 in section 13-2, the data of female attractiveness and male attractiveness ratings is given as shown below:

The Wilcoxon signed-rank test is the distribution-free test that can be used to analyze the difference between two samples that are matched on certain characteristics.

The null hypothesis is as follows:

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

The alternative hypothesis is as follows:

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

The signed ranks can be obtained as:

• Compute the ranks of absolute differences by sorting themfrom smallest to largest.
• Assign the smallest observation the rank of 1, and increase the ranks until the largest observation.
• Discard the values with a difference of 0.
• If any of the observations are repeated, assign the mean value of the ranks to all those observations. The following table shows the ranks:

Compute the sum of the positive ranks as shown below:

\(\begin{array}{c}Su{m_{positive}} = 2 + 2 + 2 + 8 + 5\\ = 19\end{array}\)

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

\(\begin{array}{c}\left| {Su{m_{negative}}} \right| = \left| {\left( { - 5} \right) + \left( { - 5} \right) + \left( { - 9} \right) + \left( { - 7} \right)} \right|\\ = \left| { - 26} \right|\\ = 26\end{array}\)

Consider the smaller sum as the test statistic.

Here, the smaller sum is 19.

Thus, T is equal to 19.

The critical value for n=12 and\(\alpha \)= 0.05 for a two-tailed test is equal to 14.

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

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