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Using the Wilcoxon Signed-Ranks Test. In Exercises 5鈥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 鈥淪peed Dating: Attractiveness鈥

Short Answer

Expert verified

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.

Step by step solution

01

Given information

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:

Rating of Male by Female

4

8

7

7

6

8

6

4

2

5

9.5

7

Rating of Female by Male

6

8

7

9

5

7

5

4

6

8

6

5

02

Define Wilcoxon signed-rank test

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.

03

Identify the statistical hypotheses

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.

04

Calculate the signed ranks

The signed ranks can be obtained as:

  • Compute the differences by subtracting the value in the second sample from the corresponding value in the first sample. The following table shows the differences along with their signs:

Rating of Male by Female

4

8

7

7

6

8

6

4

2

5

9.5

7

Rating of Female by Male

6

8

7

9

5

7

5

4

6

8

6

5

Sign of Difference

鈥2

0

0

鈥2

+1

+1

+1

0

鈥4

鈥3

+3.5

+2

  • 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:

Rating of Male by Female

4

8

7

7

6

8

6

4

2

5

9.5

7

Rating of Female by Male

6

8

7

9

5

7

5

4

6

8

6

5

Difference

鈥2

0

0

鈥2

+1

+1

+1

0

鈥4

鈥3

+3.5

+2

Rank of |d|

5

\( \times \)

\( \times \)

5

2

2

2

\( \times \)

9

7

8

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

4

8

7

7

6

8

6

4

2

5

9.5

7

Rating of Female by Male

6

8

7

9

5

7

5

4

6

8

6

5

Difference

鈥2

0

0

鈥2

+1

+1

+1

0

鈥4

鈥3

+3.5

+2

Rank of |d|

5

\( \times \)

\( \times \)

5

2

2

2

\( \times \)

9

7

8

5

Signed-Ranks

鈥5

\( \times \)

\( \times \)

鈥5

+2

+2

+2

\( \times \)

鈥9

鈥7

+8

+5

05

Calculate the sum of 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.

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

06

Calculate the test statistic and the degrees of freedom

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.

07

Draw a conclusion

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.

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Sign Test vs. Wilcoxon Signed-Ranks Test Using the data in Exercise 1, we can test for no difference between body temperatures at 8 AM and 12 AM by using the sign test or the Wilcoxon signed-ranks test. In what sense does the Wilcoxon signed-ranks test incorporate and use more information than the sign test?

Wilcoxon Signed-Ranks Test for Body Temperatures The table below lists body temperatures of seven subjects at 8 AM and at 12 AM (from Data Set 3 鈥淏ody Temperatures in Appendix B). The data are matched pairs because each pair of temperatures is measured from the same person. Assume that we plan to use the Wilcoxon signed-ranks test to test the claim of no difference between body temperatures at 8 AM and 12 AM.

a. What requirements must be satisfied for this test?

b. Is there any requirement that the samples must be from populations having a normal distribution or any other specific distribution?

c. In what sense is this sign test a 鈥渄istribution-free test鈥?

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