Non parametric tests with examples

1. Non – Parametric Test
By
Dr. Dinesh kumar Meena, Pharm.D
Ph.D Research Scholar
Department of Pharmacology,
Jawaharlal Institute of Postgraduate Medical Education & Research (JIPMER), Puducherry, India

2. Parameter
(Something that decides or limits the way in which something can be done)
Introduction
Fixed parameter No fixed parameter
Parametric statistics Non-Parametric statistics
Parametric test Non – Parametric test

3. Parametric test Non-Parametric test
Parameter Information about population and
parameter is known
Information about population and
parameter is known
Assumptions Assumptions are made No assumptions are made
Value for central
tendency
Mean Median
Probability distribution Normal distribution Arbitrary distribution
Power More powerful Less powerful
Applicable for Variables Variables & Attributes
Null hypothesis Made on parameters of population
distribution
Free from parameters
Difference between parametric and non-parametric test

4. Parametric test Non-Parametric test
Used for Ratio or Interval data Nominal or Ordinal data
Independent measures
(2 groups)
Independent measure t – test Mann Whitney U test
Independent measures
(>2 groups)
One way independent measure ANOVA Kruskal Wallis test
Repeated measures
(2 conditions)
Matched pair t-test Wilcoxon signed rank test

5. Non-parametric tests
• Non-parametric test also known as distribution free test.
• They do not assume that outcome is approximately normally distributed.
• Situations in which outcome does not follow normal distribution.
1. When the outcome is an ordinal variable or Rank
2. When there are definite outlier
3. When the outcome has clear limit of detection

6. 1. Mann Whitney U test
2. Kruskal Wallis test
3. Wilcoxon signed rank test
Commonly used non-parametric test

7. Mann Whitney U test

8. Example 1:
With the help of Mann Whitney U test, find if significant difference exist
between the scores of treatment A and Treatment B.
Treatment A 3, 4, 2, 6, 2, 5
Treatment B 9, 7, 5, 10, 6, 8

9. Steps to be followed in Mann Whitney U Test
1. Generate the hypothesis (Null & Alternative)
2. Arrange data
3. Assign ranks
4. Determine the sum of rank for each sample
5. Calculate U (test) for each sample
6. Take the smaller U(test) value
7. Compare smaller U (test) value with U (Critical) to accept/reject null
hypothesis.

10. Step 1. Generate the hypothesis (Null & Alternative)
Null hypothesis = There is no significant difference between Treatment A and Treatment B.
Alternative hypothesis = There is significant difference between Treatment A and
Treatment B
Step 2. Arrange data

11. Step 3. Assign ranks : combine rank for data 1 and data 2 to be assigned in Mann Whitney U
test.
Data Rank
2 1
2 2
3 3
4 4
5 5
5 6
6 7
6 8
7 9
8 10
9 11
10 12
Data Rank
2 1.5
2 1.5
3 3
4 4
5 5.5
5 5.5
6 7.5
6 7.5
7 9
8 10
9 11
10 12
Data 1 R 1
3 3
4 4
2 1.5
6 7.5
2 1.5
5 5.5
Data 2 R 2
9 11
7 9
5 5.5
10 12
6 7.5
8 10
1st
Arrange all
data in
ascending
order and
give ranks
2nd
Wherever any
number is
repeating more
than one time,
calculate the
average of their
ranking
3rd
Arrange
rankings as
per their data
in different
data groups

12. Step 4. Calculate the sum of ranks of each sample
Σ R1= 23 Σ R1= 55
Step 5. Calculate U (test) for each sample
For treatment A = 23 – 6 (6+1) /2 = 2.0
For treatment B = 55 – 6 (6+1) / 2 = 34.0
Data 1 R 1
3 3
4 4
2 1.5
6 7.5
2 1.5
5 5.5
Data 2 R 2
9 11
7 9
5 5.5
10 12
6 7.5
8 10
U (test) = Rank sum – n (n+1)
2

13. Step 4. Take the smaller U(test) value
Smaller U (Test ) value is 2.0 for treatment A
Step 5. Compare smaller U (test) value with U (Critical) to accept/reject null hypothesis.
U (test ) = 2
U (Critical ) = 6
U (test) < U (critical)
Step 6. Interpretation if
Null hypothesis rejected and alternative hypothesis is accepted
“There is significant difference between Treatment A and Treatment B”
H (test) < H (Critical)
H (test) > H (Critical)
Reject null hypothesis
Accept null hypothesis

15. Example 2.
Researchers recruited 19 participants to study the effect of two different
analgesics i.e. A & B. he divided patients in two groups . One group consists of
10 patients and was given drug A while another group consists of 9 patients and
was given drug B. With the help of Mann Whitney U test, find if significant
difference exist between the scores of group 1 and group 2.
Group 1: 25, 44, 23, 50, 48, 29, 60, 75, 49, 66
Group 2: 17, 23, 13, 24, 23, 21, 18, 16, 32

16. Kruskal Wallis test

17. Example 1 :
15 patients having complaint of tooth decay were randomly assigned to 3
groups of 5 each. One group ( group A) was given no treatment. Second groups
(group B) was given drug A and third group (group C) was given drug B. at the
end of six weeks, the extent of tooth decay was evaluated as % of tooth decay.
We wish to know if there is a difference among three groups.
Control group 87 76 65 81 75
Drug A 63 70 87 92 70
Drug B 45 60 43 56 60

18. Steps to be followed in Kruskal Wallis test
1. Generate the hypothesis (Null & Alternative)
2. Arrange data
3. Assign ranks
4. Determine the sum of rank for each sample
5. Calculate H (test)
6. Calculate degree of freedom
7. Compare H (test) value with H (Critical) to accept/reject null hypothesis.

19. Steps 1: Generate the hypothesis
Null hypothesis : There is no difference in extent of tooth decay among three
groups.
Alternative hypothesis: There is a difference in extent of tooth decay among
three groups.
Step 2: Arrange data
Data 1 Data 2 Data 3
87 63 45
76 70 60
65 87 43
81 92 56
75 70 60

20. Step 3. Assign ranks
Step 4.Calculate the sum ΣR1= 53.5 Σ R2 = 51.5 Σ R3 = 15
of rank for each sample
Data 1 R1 Data 2 R2 Data 3 R3
87 13.5 63 6 45 2
76 11 70 8.5 60 4.5
65 7 87 13.5 43 1
81 12 92 15 56 3
75 10 70 8.5 60 4.5

21. Step.5: Calculate H (test)
H (test) = 9.39
Step 6. Calculate degree of freedom = K (No. of groups) -1 = 3-1 = 2
H = [ 12 ] [ Σ(ΣR)2 ] – 3(N + 1)
-------- -------------
N(N+1) n
N = Total No. of participants = 15
n = no. of participant in one group = 5
H = [ 12 ] (53.5)2 (51.5)2 (15)2 - 3(15 + 1)
-------- ----------- + ---------- + --------
15(15+1) 5 5 5

22. Step 7. Compare H (test) value with H (Critical) by using chi-square table
to accept/reject null hypothesis.

23. H (Critical ) H (Test)
α = 0.05 5.99 9.39 H (test) > H (Critical )
There is a difference in extent of tooth decay among three groups.
Step 7. Interpretation
H (test) > H (Critical )
H (test) < H (Critical )
Reject the null hypothesis
Accept the null hypothesis

24. Wilcoxon signed rank test

25. Wilcoxon signed rank test
• Also called the Wilcoxon matched pairs test or the Wilcoxon signed rank
test.
• Appropriate for a repeated measure design where the same subjects are
evaluated under two different conditions
• For example, measurements taken every 15 min after dosing for 2 h, each
animal will be sampled once for baseline, and 8 times for treatment resulting in
8 comparisons. However, these 9 groups are not independent as they are all
obtained from the same set of animals after the same treatment

26. Example 1
The table shows score of pain relief provided by physiotherapy in 13 patients at
4 weeks and 8 weeks suffering from arthritis. Find if significant difference exist
in median of these scores at 4 weeks and 8 weeks ?
Participants 1 2 3 4 5 6 7 8 9 10 11 12 13
4 weeks 18.3 13.3 16.5 12.6 9.5 13.6 8.1 8.9 10 8.3 7.9 8.1 13.4
8 weeks 12.7 11.1 15.3 12.7 10.5 15.6 11.2 14.2 16.2 15.5 19.9 20.4 36.8

27. Step 1. Arrange the data
Step 2. Calculate each paired difference
Step 3. Calculate the absolute difference
Participants 1 2 3 4 5 6 7 8 9 10 11 12 13
Difference -5.6 -2.2 -1.2 0.1 1 2 3.1 5.3 6.2 7.2 12 12.3 23.4
Participants 1 2 3 4 5 6 7 8 9 10 11 12 13
Absolute
Difference
5.6 2.2 1.2 0.1 1 2 3.1 5.3 6.2 7.2 12 12.3 23.4

28. Step 4. Rank the absolute difference
Step 5. Calculate Rank rum of positive value (T+) and negative value (T-)
T(-) = 8+5+3 = 16
T (+) = 1+2+4+6+7+9+10+11+12+13 = 75
Step 6. Smaller T value will be W (test)
W (test) = 16
Step 7. Find out the W (critical ) value by using wilcoxan signed rank test
table
W (critical) = 17
Participants 1 2 3 4 5 6 7 8 9 10 11 12 13
Absolute
Difference
5.6 2.2 1.2 0.1 1 2 3.1 5.3 6.2 7.2 12 12.3 23.4
Ranking 8 5 3 1 2 4 6 7 9 10 11 12 13

30. Step 8. Interpretation
W (test) < W (critical)
“There is a significant difference between the median score of pain relief at 4
hour and 8 hour duration”
W (test) < W (critical)
W (test) > W (critical)
Reject the null hypothesis
Accept the null hypothesis
If

31. Z test statistics
Z test = T – T
SE
Z test = 75 – 45.5 / 14.30 = 2.06
T = T (max) T = T(max) + T(min)/2
SE = n (n + 1) ( 2n+1)
24
T = 75 T = 75 + 16 / 2 = 45.5
SE = 13 (13 + 1) ( 2x 13+1) = 14.30
24

32. If Z (test ) value
is less than 1.96 than accept the null hypothesis at 5 % level of significance
is greater than 1.96 at 5 % level of significance than reject the null hypothesis
In our study Z (test) is 2.06 which is greater than 1.96 thus null hypothesis will be rejected

33. • When parametric tests are not satisfied.
• When testing the hypothesis, it does not have any distribution.
• For quick data analysis.
• When unscaled data is available.
Applications of Non-Parametric Test

34. • Easily understandable
• Short calculations
• Assumption of distribution is not required
• Applicable to all types of data
Advantages of the non-parametric test

35. • Less efficient as compared to parametric test.
• The results may or may not provide an accurate answer because they are
distribution free.
Disadvantages of the non-parametric test

36. Thank You