Mann Whitney Example

8/3/2019 Mann Whitney Example

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Learning Outcome

-Student should be able to make

inferences about the differencebetween two location by using Mann

- Whitney test


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Mann-Whitney test was proposed by Henry

Berthold Mann and Donald Ransom Whitney.

The test is sometimes also referred to as the

Wilcoxon Rank Sum Test.


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a) The data consist of a random sample ofobservation X1,X2,.,Xn1 from population 1

with unknown median Mx and another

random sample of observations Y1,Y2,.,Yn2

from population 2 with unknown median

My.

b) The two sample are independent

c) The variable observed is a continuous

random variable


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d) The measurement scale employed isat least ordinal

e) The distribution functions of the two

populations differ only with respect to

location, if they differ at all.


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Case A(two-sided)

Case B(one-sided)

Case C(one-sided)

H0 : Mx = My H0 : Mx My H0 : Mx My

H1 : Mx My H1 : Mx < My H1 : Mx > My


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1) Combine two samples.

2) Rank all sample observations from

smallest to largest.3) Sum the ranks observations from

population 1(that is, the Xs).


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Test statistic is :

Where S is the sum of the rank assigned to

the sample observations from population 1


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CaseCase HypothesesHypotheses Reject HReject H00 forfor Reject HReject H00 ifif

A

(two-sided)

H0 : Mx = My

H1 : Mx My

Either sufficiently

small or sufficiently

large values of T

or

Where,

B

(one-sided)

H0 : Mx My

H1 : Mx < My

Sufficiently small

values of T

C

(one-sided)

H0 : Mx My

H1 : Mx > My

Sufficiently large

values of T Where,


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We wish to see whether we canconclude on the basis of these data

that the two represented populations

are different with respect to their

medians. Let = 0.05


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Improved subjects Unimproved subjects

Subject Score(X) Subject Score(Y)

1 11.9 1 6.62 11.7 2 5.8

3 9.5 3 5.4

4 9.4 4 5.1

5 8.7 5 5.0

6 8.2 6 4.37 7.7 7 3.9

8 7.4 8 3.3

9 7.4 9 2.4

10 7.1 10 1.7

11 6.912 6.8

13 6.3

14 5.0

15 4.2

16 4.1

17 2.2


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1) Hypotheses

H0: Mx = My

H1: Mx My (claim)

2) Test statistic

- Rank all sample observations from smallest to

largest.- Sum the ranks observations from population 1

(that is, the Xs)


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X Score Rank Y Score Rank

2.2 2 1.7 1

4.1 6 2.4 3

4.2 7 3.3 45.0 9.5 3.9 5

6.3 14 4.3 8

6.8 16 5.0 9.5

6.9 17 5.1 11

7.1 18 5.4 127.4 19.5 5.8 13

7.4 19.5 6.6 15

7.7 21

8.2 22

8.7 239.4 24

9.5 25

11.7 26

11.9 27

Total 296.5


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By using formula:

Where S = Sum of rank in sample X = 296.5

n1 = Sample size of X = 17


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3) DecisionTable A.7shows that

for


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Thus, by equation:

Where

We reject H0


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4) Conclusion

There is enough evidence to support the

claim that the two population of parameters

are different.


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For present example, we consultTable A.7forn1=17, n2=10 and we find that the computed of

our test statistic, 143.5 between

(17)(10) 26 = 144

and

(17)(10) 35 = 135

Consequently, for this test

2(0.005) > P > 2(0.001) or

0.010 > P > 0.002


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Only consider if n1 or n2 > 20

Using Z test,


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West conducted an experiment with adult aphasic

subjects, in which each was required to respond to one 62

commands. Five subjects received an experimental

treatment program, and five controls received

conventional speech therapy. Table 3.8 shows thepercentage of correct responses of each subject in the two

groups following treatment. Do these data provide

sufficient evidence to indicate that the experimental

treatment improves the proportion of correct responses?

Let = 0.05.


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Experimental (x) Experimental (Y)

73 50

42 23

90 68

58 40

62 45

Table 3.8 : Percentage of correct responses to 62

commands by aphasic subject in two treatment

groups.


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Table 3.9 shows the tidal volume of 37 adultssuffering from atrial septal defect. In 26 of these,

pulmonary hypertension was absent, and in 11 it was

present. The data were reported by Ressl et al. do

these provide sufficient evidence to indicate a lower

tidal volume in subjects without pulmonary

hypertension? Let = 0.05.


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PulmonaryHypertension

absent

Case 1 2 3 4 5 6 7 8 9 10 11 12 13

(x) 652 556 618 500 500 526 511 538 440 547 605 500 437

Case 14 15 16 17 18 19 20 21 22 23 24 25 26

(X) 481 572 589 605 436 724 515 552 722 778 677 680 428

Pulmonaryhypertension

absent

Case 1 2 3 4 5 6 7 8 9 10 11

(Y) 876 556 493 348 530 780 569 546 766 819 710

Table 3.9: Tidal volume, in millimeters, in two groups of subjects.


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To study the effects of prolonged inhalation of cadmium,

Princi and Greever exposed 10 dogs to cadmium oxide,

while 10 dogs serving as controls were not exposed to this

substance. At the end of the experiment, they determined

the levels of hemoglobin of the 20 dogs, shown in Table

3.45. Let =0.05 and use the Mann-Whitney test to

determine if one may conclude that, on the average,

inhalation of cadmium causes a reduction in hemoglobin

levels in dogs.


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Score (X) Score (Y)

14.6 15.5

15.8 17.9

16.4 15.5

14.6 16.7

14.9 17.6

14.3 16.8

14.7 16.7

17.2 16.8

16.8 17.2

16.1 18.0

Table 3.45 : Hemoglobin determinations, grams, in twenty dogs


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EXERCISE ANSWER

EXERCISE 3.3 T = 20

H0 cannot be rejected

EXERCISE 3.4 T = 102.5

H0 cannot be rejected

EXERCISE 3.27 T = 17.5

Reject H0


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Link you tube :

http://www.youtube.com/watch?v=ZR_ml-GOL7U&featu
http://www.youtube.com/watch?v=ZR_ml-GOL7U&feature=youtu.behttp://www.youtube.com/watch?v=ZR_ml-GOL7U&feature=youtu.be

Mann Whitney Example

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