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Using Nonparametric Tests. In Exercises 1鈥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.

Presidents, Popes, Monarchs Listed below are numbers of years that U.S. presidents, popes, and British monarchs lived after their inauguration, election, or coronation, respectively. Assume that the data are samples randomly selected from larger populations. Test the claim that the three samples are from populations with the same median.

Presidents

10

29

26

28

15

23

17

25

0

20

4

1

24

16

12


4

10

17

16

0

7

24

12

4

18

21

11

2

9

36


12

28

3

16

9

25

23

32








Popes

2

9

21

3

6

10

18

11

6

25

23

6

2

15

32


25

11

8

17

19

5

15

0

26







Monarchs

17

6

13

12

13

33

59

10

7

63

9

25

36

15


Short Answer

Expert verified

There is not enough evidence to warrant rejection of the claim that the three samples come from the populations with the same median.

Step by step solution

01

Given information

Three samples given are showing the number of years the US presidents, popes, and British monarchs lived after their inauguration, election or coronation respectively.

02

Appropriate test

As the number of samples is three and the equality of medians of the populations of the samples need to be tested, the Kruskal Wallis test is required to be conducted.

03

Identify the hypothesis

The null hypothesis for testing the equality of medians is as follows:

Thethree samples come from populations with the same median.

The alternative hypothesis is as follows:

Thethree samples do not come from populations with the same median.

04

Assign ranks

Combine the three samples and write the A for presidents, B for pope, and C for monarchs as the sample name.

Denote a rank of 1 to the smallest observation, 2 to the next smallest observation until all the observations are assigned ranks.

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

The following table shows the ranks of all the values:

Values

Ranks

Sample

Values

Ranks

Sample

10

26.5

A

2

6

B

29

69

A

9

22.5

B

26

65.5

A

21

53.5

B

28

67.5

A

3

8.5

B

15

39.5

A

6

15.5

B

23

56

A

10

26.5

B

17

46.5

A

18

49.5

B

25

62

A

11

30

B

0

2

A

6

15.5

B

20

52

A

25

62

B

4

11

A

23

56

B

1

4

A

6

15.5

B

24

58.5

A

2

6

B

16

43

A

15

39.5

B

12

33.5

A

32

70.5

B

4

11

A

25

62

B

10

26.5

A

11

30

B

17

46.5

A

8

20

B

16

43

A

17

46.5

B

0

2

A

19

51

B

7

18.5

A

5

13

B

24

58.5

A

15

39.5

B

12

33.5

A

0

2

B

4

11

A

26

65.5

B

18

49.5

A

17

46.5

C

21

53.5

A

6

15.5

C

11

30

A

13

36.5

C

2

6

A

12

33.5

C

9

22.5

A

13

36.5

C

36

73.5

A

33

72

C

12

33.5

A

59

75

C

28

67.5

A

10

26.5

C

3

8.5

A

7

18.5

C

16

43

A

63

76

C

9

22.5

A

9

22.5

C

25

62

A

25

62

C

23

56

A

36

73.5

C

32

70.5

A

15

39.5

C

05

Test Statistic

Let\({n_1}\)denote the sample size corresponding to presidents鈥 ages.

Let\({n_2}\)denote the sample size corresponding to popes鈥 ages.

Let\({n_3}\)denote the sample size corresponding to monarchs鈥 ages.

Thus,

\(\begin{array}{l}{n_1} = 38\\{n_2} = 24\\{n_3} = 14\end{array}\)

The value of N is equal to

\(\begin{array}{c}N = 38 + 24 + 14\\ = 76\end{array}\)

The sum of the ranks corresponding to presidents is computed below:

\(\begin{array}{c}{R_1} = 26.5 + 69 + ... + 70.5\\ = 1485.5\end{array}\)

The sum of the ranks corresponding to popes is computed below:

\(\begin{array}{c}{R_2} = 6 + 22.5 + .... + 65.5\\ = 806.5\end{array}\)

The sum of the ranks corresponding to popes is computed below:

\(\begin{array}{c}{R_3} = 46.5 + 15.5 + .... + 39.5\\ = 634\end{array}\)

Thus, the value of the test statistic is computed as follows:

\(\begin{array}{c}H = \frac{{12}}{{N\left( {N + 1} \right)}}\left( {\frac{{{R_1}^2}}{{{n_1}}} + \frac{{{R_2}^2}}{{{n_2}}} + \frac{{{R_3}^2}}{{{n_3}}}} \right) - 3\left( {N + 1} \right)\\ = \frac{{12}}{{76\left( {76 + 1} \right)}}\left( {\frac{{{{1485.5}^2}}}{{38}} + \frac{{{{806.5}^2}}}{{24}} + \frac{{{{634}^2}}}{{14}}} \right) - 3(76 + 1)\\ = 2.5288\end{array}\)

Thus, the value of H comes out to be equal to 2.5288.

06

Conclusion

Let k be the number of samples.

Thus, k=3.

The degrees of freedom are computed as follows:

\(\begin{array}{c}df = k - 1\\ = 3 - 1\\ = 2\end{array}\)

The critical value of\({\chi ^2}\)with 2 degrees of freedom at\(\alpha = 0.05\)is equal to 5.9915.

Since the value of H is less than the critical value, the decision is fail to reject the null hypothesis.

There is not enough evidence to warrant rejection of the claim that the three samples come from the populations with the same median.

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