Using Nonparametric Tests. In Exercises 1–10, use a 0.05 significance level with the indicated test. If no particular test is specified, use the appropriate nonparametric test from this chapter.

Speed Dating Repeat the preceding exercise using the Wilcoxon signed-ranks test.

A sample is given showing the male attractiveness ratings by females.

The Wilcoxon signed-ranks test is used to test the claim that the sample comes from a population with a median equal to 5.

The null hypothesis is as follows:

The given sample comes from a population with a median equal to 5.

The alternative hypothesis is as follows:

The given sample comes from a population with a median not equal to 5.

First subtract all values from 5.

Compute the ranks of all the absolute difference values by assigning a rank of 1 to the lowest value, 2 to the next lowest value and so on until the greatest value.

If some values are equal, assign the mean of the ranks to all those values.

For example, if 2 observations have the same value and occupy the positions of 8 and 9, then the mean of 8 and 9 equal to 8.5 is assigned to both the observations.

While ranking, exclude all differences with a value of 0.

Write the sign of the ranks corresponding to the sign of the difference.

The following table shows the signed-ranks:

Compute the sum of all the positive ranks as shown:

\(\begin{array}{c}Su{m_{positive}} = 7 + 7 + 7\\ = 21\end{array}\)

Compute the absolute value of the sum of all the negative ranks:

\(\begin{array}{c}\left| {Su{m_{neg}}} \right| = \left| {\left( { - 15} \right) + \left( { - 15} \right) + .... + \left( { - 7} \right)} \right|\\ = \left| { - 210} \right|\\ = 210\end{array}\)

Let T be the smaller of the two sums.

Thus, T=21.

Here,n be the number of pared set of values for the which the differences is not equal to 0.

There are 21 values for which the difference is not equal to 0. So, the value of n is equal to 21.

Since n is less than 30, the test statistic value is equal to T.

The critical value of T for n=21 and\(\alpha = 0.05\)for the two-tailed test is equal to 59.

Since the test statistic is less than the critical value, the null hypothesis is rejected.

There is enough evidence to conclude that the given sample does not come from a population with a median equal to 5.