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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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