Exercise Wicoxon Matched Pairs Sign-Ranked Test

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EXERCISE 4.4 (Small Sample)

Piggot et.al* paired 10 psychotic and 10 normal children on the age and gender. They then

compared subjects for different in respiratory sinus arrhythmia under conditions of spontaneous

and 5-, 10- and 15- second interval breathing. They recorded cardiac rate and respiratory changes

simultaneously. Table 4.8 shows the differences in duration of the cardiac respiratory phase

following the beginning of aspiratory (psychotics compared to the controls for the respiration).

Do these data provide sufficient evidence to indicate a difference between psychotic and normal

children? Let What is the P-value for this test?

*Source: Leonard R. Piggot, Albert F.Ax, Jaccqueline L.Bamford and Joanne M.Fetzer, “Respiration Sinus

Arrhythmia in Psychotic Children, “Psychophysiology, 10(1973), 401-414; copyright 1973, The society for psycho

physiological Research; reprinted with permission of the publisher.

Duration of cardiac acceleration, seconds, timed respiration means for 15-second interval

breathing Table 4.8

Pair 1 2 3 4 5 6 7 8 9 10

Psychotic (X) 1.74 1.44 2.12 1.80 2.00 2.70 1.96 1.46 1.82 1.40

Control (Y) 2.46 1.88 2.38 1.94 2.14 1.60 1.96 1.82 1.80 1.84



Solution:

1)

( claim)

2) Test statistics

Psychotic (X) Control (Y) Rank of Signed Rank of

+ve -ve

1.74 2.46 0.72 8 +8.0

1.44 1.88 0.44 6.6 +6.5

2.12 2.38 0.26 4 +4.0

1.80 1.94 0.14 2.5 +2.52.00 2.14 0.14 2.5 +2.5

2.70 1.60 -1.10 9 -9.0

1.96 1.96 0 Omit - -

1.46 1.82 0.36 5 +5.0

1.82 1.80 -0.02 1 -1.0

1.40 1.84 0.44 6.5 +6.5

3)Since is less than , the test statistic is

, n = 9,

From table A.3,

P-value, 2(0.0820) = 0.164

Since P-value, 0.164 > 0.05

4) Do not reject

5) Do not enough evidence to support the claim that provide sufficient evidence to indicate a

difference between psychotic and normal children



Exercise 4.16 (Large Sample)



Heiman et al* randomly placed one member of each of thirty two matched pairs of students

reading below grade level in a supplementary reading program. Subjects were matched nearly as

possible on the discrepancy between their reading level and their current grade level. The

supplementary program consisted of a point system to reward attention to and identification of

letter and word combinations. Table 4.25 derived from author’s results, shows the differences

between the scores made by the subjects on a reading test after and before the program. Use the

procedure based on the Wilcoxon test to construct a 95% confidence interval for the median

difference.

Source: Julia R. Heiman, Mark J. Fisher, and Alan O. Ross, “ A Supplementary Behavioral Program to Improve

Deficient reading Performance,” J. Abnormal Child Psychol. 1(1973), 390-399; published by Plenum Publishing

Corporation New York.

Difference in reading test score made by thirty two matched pairs of subject, one member

of which was assigned to an experimental program and the other to a control group (before

score subtracted from the score)

Pair 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

Experimental (X) 0.2 1.

3

0.6 0.4 0.3 0.7 1.0 1.1 1.3 1.2 0.2 0.4 0.7 0.8 0.1 1.0

Control (Y) 1.2 0.

2

0.8 1.1 1.1 1.6 0.4 1.1

6

0.5 0.5 0.8 0.9 1.1 1.1 0.3 0.9

Pair 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Experimental (X) 0.5 1.0 0.6 0.

1

1.

3

0.

1

1.

0

0.

2

0.

4

0.

6

0.

7

0.

9

1.

2

0.

8

0.

5

1.4

Control (Y) 0.8 1.1 -0.1 0.

3

0.

2

1.

5

1.

3

0.

6

1.

2

0.

3

0.

8

1.

0

0.

5

1.

6

0.

1

0.5



Solution:

1)

2) Test statistics

Experimental (X) Control (Y) Rank of Signed Rank of

+ve -ve

0.5 0.8 0.3 7.5 +7.5

1.0 1.1 0.1 2.5 +2.50.6 -0.1 -0.8 22.5 -22.5

0.1 0.3 0.2 4.5 +4.5

1.3 0.2 -1.1 30.5 -30.5

0.1 1.5 1.4 32 +32

1.0 1.3 0.3 7.5 +7.5

0.2 0.6 0.4 11 +11

0.4 1.2 0.5 13.5 +13.5

0.6 0.3 0.6 16 +16

0.7 0.8 1.0 28 +28

0.9 1.0 1.0 28 +28

1.2 0.5 -0.7 19 -190.8 1.6 0.8 22.5 +22.5

0.5 0.1 -0.4 11 -11

1.4 0.5 -0.9 25.5 -25.5

0.2 1.2 1.0 28 +28

1.3 0.2 -1.1 30.5 -30.5

0.6 0.8 0.3 7.5 +7.5

0.4 1.1 0.7 19 +19

0.3 1.1 0.8 22.5 +22.5

0.7 1.6 0.9 25.5 +25.5

1.0 0.4 -0.6 16 -161.1 1.16 0.06 1 +1

1.3 0.5 -0.8 22.5 -22.5

1.2 0.5 -0.7 19 -19

0.2 0.8 0.6 16 +16

0.4 0.9 0.5 13.5 +13.5

0.7 1.1 0.4 11 +11

0.8 1.1 0.3 7.5 +7.5



0.1 0.3 0.2 4.5 +4.5

1.0 0.9 0.1 25 +25

3)

Since less than , the test statistic is

=

= - 0.0236

P-value, P( Z< -0.0236)

= 2(0.49059)

= 0.98118

Since P-value = 0.98118 > 0.05

4) Do not reject.

5) Do not enough evidence to support the claim that the differences between the scores

made by the subjects on a reading test after and before the program.



EXERCISE 4.5 (Small Sample)

Bhatia et al.* reported the data shown in table 4.9. Can we conclude on the basis of these data

that treatment lowers the stroke index in patients of this type? Let α = 0.05

Stroke index (ml/beat/ ) in pre – and post-treatment studies of coronary circulation in

chronic severe anaemia

Case 1 2 3 4 5 6 7 8



Before treatment (X) 109 57 53 57 68 72 51 65

After treatment: (Y) 56 44 55 40 62 46 49 41*Source: M.L.Bhatia,S.C.Manchanda, and Sujoy B.Roy, “coronary Haemodynamic Studies in Chronic Severe

Anaemia,”Br.Herat.J.,31 (1969),365-374. Reprinted by permission of the authors and the editor.

Solution:

1) Hypotheses

=

= ( claim)

2) Test statistic

The calculation of the test statistic is shown in the table below:

Before

treatment (X)

After

treatment ( Y)

Rank of Signed rank of

109 56 -53 8 -8

57 44 -13 4 -4

53 55 +2 1.5 + 1.5

57 40 -17 5 -5

68 62 -6 3 -3

72 46 -26 7 -751 49 -2 1.5 -1.5

65 41 -24 6 -6

Since the test statistic is T = 1.5

3) n = 8 , α = 0.05

since 1.5 between 1 and 2

From the table A.3

p – value = 0.0078 and 0.0117

therefore p – value < 0.05



4) Reject

5) Enough evidence to support the claim that treatment lower stroke index in patient.

EXAMPLE (Large Sample)

Thirty two healthy high school students participated in an encounter-group experience in which

half the participants were handicapped persons about the same age. Before and after the group

experience, the 32 handicapped students took a test designed to measure their understanding of

handicapped persons. Table 4.28 show results of the test. Can one conclude from the data that

such an experience increases student’s understanding (as indicated by a higher score) of

handicapped persons? Let α = 0.05, and determine the p –value.



Score made by 32 high school students on a test to measure understanding of handicapped

person before and after an encounter – group experience.

Student 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Before (X) 55 63 54 61 63 60 59 67 55 64 68 57 81 84 90 97

After (Y) 60 68 69 64 67 61 63 62 58 62 70 56 89 88 94 96

Student 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

Before (X) 80 72 64 55 54 60 70 35 58 70 80 55 49 42 53 60

After (Y) 80 65 74 70 52 42 40 20 42 52 71 51 30 41 52 53

Solution:

1) Hypotheses

( claim)

2) Test statistic:

Student Before ( X) After (Y) Rank of Signed rank of

1 55 60 +5 17 +17



2 63 68 +5 17 +17

3 54 69 +15 2 +25

4 61 64 +3 9.5 +9.5

5 63 67 +4 13 +13

6 60 61 +1 3 +3

7 59 63 +4 13 +138 67 62 -5 17 -17

9 55 58 +3 9.5 + 9.5

10 64 62 -2 7 -7

11 68 70 +2 7 +7

12 57 56 -1 3 -3

13 81 89 +8 21 +21

14 84 88 +4 13 +13

15 90 94 +4 13 +13

16 97 96 -1 3 -3

17 80 80 0 - Omit

18 72 65 -7 19.5 -19.519 64 74 +10 23 +23

20 55 70 +15 25 +25

21 54 52 -2 7 -7

22 60 42 -18 28.5 -28.5

23 70 40 -30 31 -31

24 35 20 -15 25 -25

25 58 42 -16 27 -27

26 70 52 -18 28.5 -28.5

27 80 71 -9 22 -22

28 55 51 -4 13 -13

29 49 30 -19 30 -30

30 42 41 -1 3 -3

31 53 52 -1 3 -3

32 60 53 -7 19.5 -19.5

3) The test statistic is T= 287

So,

When n = 31 ,T=287 , α = 0.05



=

= 0.764265

So, p – value = p (z > 0.764265)

p- value is 0.22235>0.05

4) Do not reject

5) Not enough evidence to support the claim that such an experience increases student’s

understanding (as indicated by a higher score) of handicapped persons.

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Exercise Wicoxon Matched Pairs Sign-Ranked Test