mann-whitney U test, wilcoxan rank sum test, krushalwallis test and friedman test

1. Mann – Whitney U test

2. • It is a non-parametric statistical method that
compares two groups that are independent of
sample data.
• It is used to test the null hypothesis that the
two samples have similar median or whether
observations in one sample are likely to have
larger values than those in other sample
• The parametric equivalent of Mann-Whitney
U test is t- test of unrelated sample

3. Assumption
• The two samples are random
• Two samples are independent of each other
• Measurement is of ordinal type thus
observations are arranged in ranks

4. Steps to perform
• The null hypothesis and alternative hypothesis
are identified.
• The significance level [alpha] related with null
hypothesis is stated. Usually alpha is set at 5%
and therefore, the confidence level is 95 %
• All of the observations are arranged in terms
of magnitude.

5. • The Ra denotes the sum of the ranks in group
a
• The Rb denotes the sum of ranks in group b
• U statistics is determined by
Verify Ua + Ub = nanb

6. • Evaluate U = min [ Ua,Ub]
• The obtained value is smaller of the two
statistics
• Using table of critics evaluate the possibility of
obtaining value of U or lower
• The critical value is compared with the
obtained value.
• The results are then interpreted to draw
conclusion.

7. Perform the Mann-Whitney U test
Treatment A Treatment B
3 9
4 7
2 5
6 10
2 6
5 8

8. Why Mann-Whitney U test
• Student t test I s preferred for this data but
• Data are not normal
• Sample size is small

9. Obsevations Arranged in order
1 2
2 2
3 3
4 4
5 5
6 5
7 6
8 6
9 7
10 8
11 9
12 10

10. There is no difference between the
rank of each treatment
Rank observation
1.5 2
1.5 2
3 3
4 4
5.5 5
5.5 5
7.5 6
7.5 6
9 7
10 8
11 9
12 10

11. TA Rank a Tb Rank b
3 3 9 11
4 4 7 9
2 1.5 5 5.5
6 7.5 10 12
2 1.5 6 7.5
5 5.5 8 10
Sum of Ra 23 Sum of Rb 55

12. Cross check

13. • Ua= 23- 6[6+1]/2
• = 23- 42/2
• = 23-21
• = 2
• Ub = 55- 6[ 6+1]/2
• = 55-21
• = 34

14. We have to choose lowest value
hence U= 2
• Use u table= critical value
• N1=6 n2=6
• U critics from table = 5
• We should get the calculated value as equal to
or greater than table value.
• Here we got lesser value than table value
hence null hypothesis is rejected.

15. Wilcoxon Rank sum test
• It is non-parametric dependent samples t test
that can be performed on ranked or ordinal
data.
• Mann-Whitney Wilcoxon test
• It is used to test null hypothesis
• It is used to assess whether the distribution of
observations obtained between two separate
groups on a dependent variable are
systematically different from one another.

16. • It is used to evaluate the populations that are
equally distributed or not
• A population is set of similar items or data
obtained from experiment
• Rank basically two types of rank given Ra large
and Rb small.

17. It can be used in the place of
• One sample t test
• Paired t test
• For ordered categorical data where a
numerical scale is in appropriate but where it
is possible to rank the observations

18. General way to perform test
• State the null hypothesis Ho and the
alternative hypothesis H1
• Define alpha level
• Define decision rule
• Calculate Z statistics
• Calculate results
• Make conclusion

19. For paired data
• State the null hypothesis
• Calculate each paired difference
• Rank di ignoring signs [ assign rank 1 to the
smallest , rank 2 to the next etc ]
• Designate each rank along with its sign. Based
on the sign of di

20. • Calculate W+ the sum of the ranks of positive
di and W- the sum of the ranks of the negative
di.
• [W+] + [W-] = n [n+1]2

21. Problem
Group A p1 Group B p2
41 66
56 43
64 72
42 62
50 55
70 80
44 74
57 75
63 77
78
N1=9 N2=10

22. Group s = P1 + P2 Group Rank
41 A 1
42 A 2
43 B 3
44 A 4
50 A 5
55 B 6
56 A 7
57 A 8
62 B 9
63
64
A
A
10
11
66 B 12
70 A 13
72 B 14

23. Group s = P1 + P2 Group Rank
74 B 15
75 B 16
77 B 17
78 B 18
80 B 19

24. Group A Rank sum
Group s = P1 + P2 Group Rank
41 A 1
42 A 2
44 A 4
50 A 5
56 A 7
57 A 8
63
64
A
A
10
11
70 A 13
SUM OF Rank a 61

25. Group B Rank sum
Group s = P1 + P2 Group Rank
43 B 3
55 B 6
62 B 9
66 B 12
72 B 14
74 B 15
75 B 16
77 B 17
78 B 18
80 B 19
Sum of Group B 129

26. Small Rank sum is chosen: 61
• μr=n1 [n1+ n2 + 1] /2
• μr= 9 [ 9+ 10+1 ] /2
• μr= 9 [20]/2 = 180/2 = 90
• σr =

29. Krushal –Wallis H-test
• H test
• Non parametric statistical procedure used for
comparing more than two independent
sample
• Parametric equivalent to this test is one way
ANOVA
• H test is for non-normally distributed data.

30. Krushal –Wallis H-test
• It is a generalization of the Mann- Whitney
test which is a test for determining whether
the two samples selected are taken from the
same population.
• The p values in both the Krushal –Wallis and
the Mann-Whitney tests are equal
• It is used for samples to evaluate their degree
of association.

31. Description of sample
• 3 independently drawn sample.
• Data in each sample should be more than 5
• Both distribution and population have same
shape
• Data must be ranked
• Samples must be independent
• K independent sample k> 3 or K=3

32. Characteristics
• Test statistics is applied when data is not normally
distributed
• Test uses k samples of data.
• Test can be used for one nominal and one ranked
variable
• Significance level is denoted with α
• Data is ranked and df is n-1
• The rank of each sample is calculated
• Average rank is applied in case if there is tie

33. Problem
• Null hypothesis
• K independent sample drawn from population
which are identically distributed.
• Alternative hypothesis
• K independent sample drawn from population
which are not identically distributed.

34. Notation
Sampl1 obseravtion
1 Xxx Xxx Xxx Xxx Xxx
2 Xxx Xxx Xxx Xxx Xxx Xxx xxx
3 Xxx Xxx Xxx xxx Xxx Xxx Xxx Xxx
K Xxx Xxx Xxx Xxx Xxx Xxx xxx
Observation more than five
K =3 or K>3

35. Procedure
• Define null H0 and alternative H1 hypothesis.
• Rank the sample observations in the
combined series.
• Compute Ti sum of ranks

36. • Apply chi square variate with K-1 degree of
freedom
• K = number of sample
• Conclusion
• Take the table value from Chi 2 [k-1][α]
• If calculated H value > Chi 2 [k-1][α]
• We reject H0

37. Use krushal wallis H test at 5 % level of
significance if three methods are
equally effective
Method
1
99 64 101 85 79 88 97 95 90 100
Method
2
83 102 125 61 91 96 94 89 93 75
Method
3
89 98 56 105 87 90 87 101 76 89

38. Step I
• Null hypotheis
• H0 : μ a = μ b = μc
• Three methods are equally effective
• Alternative hypothesis H1= at least two of the
μ are different
• Three methods are not equally effective
• n1+n2+n3=30

39. Step II
Met
hod
1
99 64 101 85 79 88 97 95 90 100 T1
Rank 24 3 26.5 8 6 11 22 20 15.5 25 161
Met
hod
2
83 102 125 61 91 96 94 89 93 75 T2
Rank 7 28 30 2 17 21 19 13 18 4 159
Met
hod
3
89 98 56 105 87 90 87 101 76 89 T3
Rank 13 23 1 29 9.5 15.5 9.5 26.5 5 13 145
n1 10 n2 10 n3 10

40. Compute statistics H

41. Compute statistics H

42. • Df = K-1 =3-1=2
• Table value = 5.99
• Calculated value is 0.196
• Since the calculated H value
• 0.196 < 5.99
• We fail to reject H0
• All the teaching methods are equal

43. Friedman test
• It is a non parametric test developed and
implemented by Milton Friedman.
• It is used for finding differences in treatments
across multiple attempts by comparing three
or more dependent samples.
• It is an alternative to ANOVA when the
assumption of normality is not met

44. Friedman test
• The test is calculated using ranks of data
instead of unprocessed data
• It is used to test for differences between
groups when the dependent variable being
measured is ordinal.
• It can also be used for continuous data that
has marked as deviations from normality with
repeated measures.

45. • It is a repeated measures of ANOVA that can
be performed on the ordinal data.

46. Descriptions and requirements
• Dependent variable should be measured at the
ordinal or continuous level
• Data comes from a single group measured on at
least three different occasions.
• Random sampling method must be used
• All of the pairs are independent.
• Observations are ranked within blocks with no
ties
• Samples need not be normally distributed.

47. Problem ordinal data is given .is there
a difference between weeks 1,2,3
using alpha as 0.05
week1 Week 2 Week 3
27 20 34
2 8 31
4 14 3
18 36 23
7 21 30
9 22 6

48. Steps
• Define null and alternative hypothesis.
• State alpha
• Calculate degree of freedom
• State decision rule
• Calculate the statistic
• State result
• conclusion

49. Step 1
• Null hypothesis
• H0 there is no difference between three
conditions.
• Alternative hypotheis
• H1 there is a difference between three
conditions

50. Step 2
• State alpha
• 0.05

51. Step3
• Degree of freedom’
• df=K-1
• K = no of groups
• =3-1=2
• df=2

52. Step 4
• Decision rule – chi square table
Chi square is greater than 5.991 than you can
reject the null hypothesis

53. Step 5Rank the value
Week 1 Week 2 Week 3
2 1 3
1 2 3
2 3 1
1 3 2
1 2 3
2 3 1
R=9 R=14 R=13

54. Step 6 Calculation of statistics

55. Step 7.state result
• If chi square is greater than 5.991 than reject
the null hypothesis.
• Calculated chi square value is 2.33
• Calculated value is lesser than table value
hence fail to reject null hypotheis.
• Hence there is no difference among the three
group.