• Non parametric tests are distribution free methods, which do not rely on assumptions that the data are drawn from a given probability distribution. As such it is the opposite of parametric statistics • In non- parametric tests we do not assume that a particular distribution is applicable or that a certain value is attached to a parameter of the population.  When to use non parametric test??? 1) Sample distribution is unknown. 2) When the population distribution is abnormal  Non-parametric tests focus on order or ranking 1) Data is changed from scores to ranks or signs 2) A parametric test focuses on the mean difference, and equivalent non-parametric test focuses on the difference between medians. 1) Chi – square test • First formulated by Helmert and then it was developed by Karl Pearson • It is both parametric and non-parametric test but more of non - parametric test. • The test involves calculation of a quantity called Chi square. • Follows specific distribution known as Chi square distribution • It is used to test the significance of difference between 2 proportions and can be used when there are more than 2 groups to be compared. Applications 1) Test of proportion 2) Test of association 3) Test of goodness of fit Criteria for applying Chi- square test • Groups: More than 2 independent • Data: Qualitative • Sample size: Small or Large, random sample • Distribution: Non-Normal (Distribution free) • Lowest expected frequency in any cell should be greater than 5 • No group should contain less than 10 items Example: If there are two groups, one of which has received oral hygiene instructions and the other has not received any instructions and if it is desired to test if the occurrence of new cavities is associated with the instructions. 2) Fischer Exact Test • Used when one or more of the expected counts in a 2×2 table is small. • Used to calculate the exact probability of finding the observed numbers by using the fischer exact probability test. 3) Mc Nemar Test • Used to compare before and after findings in the same individual or to compare findings in a matched analysis (for dichotomous variables). Example: comparing the attitudes of medical students toward confidence in statistics analysis before and after the intensive statistics course. 4) Sign Test • Sign test is used to find out the statistical significance of differences in matched pair comparisons. • Its based on + or – signs of observations in a sample and not on their numerical magnitudes. • For each subject, subtract the 2nd score from the 1st, and write down the sign of the difference.  It can be used a. in place of a one-sample t-test b. in place of a paired t-test or c. for ordered categorial data where a numerical scale is inappropriate but where it is possible to rank the observations. 5) Wilcoxon signed rank test • Analogous to paired ‘t’ test 6) Mann Whitney Test • similar to the student’s t test 7) Spearman’s rank correlation - similar to pearson's correlation.

1. TESTSOFSIGNIFICANCE
NON-PARAMETRIC TESTS
By
Dr. Lasya

2. • Non – Parametric tests
• When to use Non-Parametric tests?
• Difference between Parametric and Non-Parametric tests
• Non-Parametric tests:
1) Chi Square test
2) Fischer exact test
3) Mc Nemar test
4) Sign test
5) Wilcoxon signed rank test
CONTENTS

3. 6) Mann Whitney ‘U’ Test
7) Kruskal wallis test
8) Friedman ANOVA
9) Spearman correlation coefficient
10) Kendall’s coefficient of concordance
• Conclusion
• References

4. NON – PARAMETRIC TESTS
• Non-parametric tests are distribution free methods,
which do not rely on assumptions that the data are
drawn from a given probability distribution. As such it is
the opposite of parametric statistics
• In non-parametric tests we do not assume that a
particular distribution is applicable or that a certain
value is attached to a parameter of the population.

5.  When to use non parametric test???
• Sample distribution is unknown.
• When the population distribution is abnormal
i.e. too many variables involved
 Non-parametric tests focus on order or ranking
• Data is changed from scores to ranks or signs
• A parametric test focuses on the mean difference, and
equivalent non-parametric test focuses on the difference
between medians.

7. Chi-square test
• First formulated by Helmert and then it was developed by
Karl Pearson
• It is both parametric and non parametric test but more of non
parametric test.
• The test involves calculation of a quantity called chi square .
• Follows specific distribution known as Chi-square
distribution

8. • It is used to test the significance of difference between 2
proportions and can be used when there are more than 2
groups to be compared.
Applications:
• Test of proportion
• Test of association
• Test of goodness of fit

9. Groups: More than 2 independent
Data: Qualitative
Sample size: Small or Large
Distribution: Non-Normal (Distribution free)

10. Prerequisites:
• Random sampling
• Qualitative data
• Data need not follow normal distribution
• Lowest expected frequency in any cell should be
greater than 5
• No group should contain less than 10 items

11. • For example, if there are two groups, one of which
has received oral hygiene instructions and the other
has not received any instructions and if it is desired to
test if the occurrence of new cavities is associated
with the instructions.

12. Group
Present Absent Total
Number who
received
instructions
10 40 50
Number who
did not
received
instructions
32 8 40
Total 42 48 90
Occurrence of new cavities

13. Steps:
1) Test the null hypothesis
• States that there is no association between oral
hygiene instructions received in dental hygiene and
the occurrence of new cavities
2) Then the χ2 – statistic is calculated as,
χ2 =
Σ(O−E)
𝐸
• Proportion of people with caries = 42/90 = 0.47
• Proportion of people without caries = 48/90 = 0.53

14.  Among those who received instructions
• Expected number attacked = 50×0.47= 23.5
• Expected number not attacked = 50×0.53 = 26.5
 Among those who did not receive instructions
• Expected number attacked = 40×0.47= 18.8
• Expected number not attacked = 40×0.53 = 21.2
Group Attacked Not Attacked
Number who
received
instructions
O= 10
E= 23.5
O-E= 13.5
O= 40
E= 26.5
O-E= 13.5
Number who
received
instructions
O= 32
E= 18.8
O-E= 13.2
O= 8
E= 21.2
O-E= 13.2

15. 3) Applying χ2 test,
χ2 =
Σ(O−E)
𝐸
=
(13.5)2
23.5
+
(13.5)2
26.5
+
(13.2)2
18.8
+
(13.2)2
21.2
=7.76+ 6.88+ 9.27+ 8.22
= 32.13
4) Finding the degree of freedom(d.f)
• Depends upon the number of columns and rows in original
table
• d.f = (column – 1) (row – 1)
= (2-1) (2-1)
= 1

16. • For example, if there are two groups, one of which
has received oral hygiene instructions and the other
has not received any instructions and if it is desired to
test if the occurrence of new cavities is associated
with the instructions.

17. Group
Present Absent Total
Number who
received
instructions
10 40 50
Number who
did not
received
instructions
32 8 40
Total 42 48 90
Occurrence of new cavities

18. Steps:
1) Test the null hypothesis
• States that there is no association between oral
hygiene instructions received in dental hygiene and
the occurrence of new cavities
2) Then the χ2 – statistic is calculated as,
χ2 =
Σ(O−E)
𝐸
• Proportion of people with caries = 42/90 = 0.47
• Proportion of people without caries = 48/90 = 0.53

19.  Among those who received instructions
• Expected number attacked = 50×0.47= 23.5
• Expected number not attacked = 50×0.53 = 26.5
 Among those who did not receive instructions
• Expected number attacked = 40×0.47= 18.8
• Expected number not attacked = 40×0.53 = 21.2
Group Attacked Not Attacked
Number who
received
instructions
O= 10
E= 23.5
O-E= 13.5
O= 40
E= 26.5
O-E= 13.5
Number who
received
instructions
O= 32
E= 18.8
O-E= 13.2
O= 8
E= 21.2
O-E= 13.2

20. 3) Applying χ2 test,
χ2 =
Σ(O−E)
𝐸
=
(13.5)2
23.5
+
(13.5)2
26.5
+
(13.2)2
18.8
+
(13.2)2
21.2
=7.76+ 6.88+ 9.27+ 8.22
= 32.13
4) Finding the degree of freedom(d.f)
• Depends upon the number of columns and rows in original
table
• d.f = (column – 1) (row – 1)
= (2-1) (2-1)
= 1

21. 5) Probability tables
• In the Probability table, with degree of freedom of 1,
the χ2 value for probability of 0.05 is 3.84.
• Since the observed value 32 is much higher
• Null hypothesis is false and there is difference in
occurrence of caries in 2 groups with caries being
lower in those who received instructions

22. Assessment of Musculoskeletal Disorders and Associated
Risk Factors among Dentists in Rajahmundry City: A
Cross-Sectional Study

23. Fischer Exact Test
• Used when one or more of the expected counts in a
2×2 table is small.
• Used to calculate the exact probability of finding the
observed numbers by using the fischer exact
probability test

24. Mc Nemar Test
• Used to compare before and after findings in the
same individual or to compare findings in a matched
analysis (for dichotomous variables)
• Example: Comparing the attitudes of medical
students toward confidence in statistics analysis
before and after the intensive statistics course.

25. Sign Test
• Sign test is used to find out the statistical significance of
differences in matched pair comparisons.
• The sign test is one of the simplest nonparametric tests.
• Its based on + or – signs of observations in a sample and
not on their numerical magnitudes.
• For each subject, subtract the 2nd score from the 1st, and
write down the sign of the difference.

26.  It can be used
• a) in place of a one-sample t-test
• b) in place of a paired t-test or
• c) for ordered categorial data where a numerical scale
is inappropriate but where it is possible to rank the
observations. (Note that the Wilcoxon Signed Rank
Sum Test is also appropriate in these situations and is
a more powerful test than the sign test.)

27. • Eg: Suppose playing 4 rounds of golf at city club. 11
professionals totaled 280, 282, 292, 273, 283, 283,
275, 284, 282, 279 & 281. At 5% level of
significance, test that the professional golfers average
284 for four rounds against they average less

28. No. of ‘+’ signs = 1
No. of ‘-’ signs = 9
Number of 0 = 1
Reduced sample size = 10
S = max. = 9
P value=
= Prob ( observing a value of 9
or higher using B(10,1/2)
= 1- Prob ( observing a value of
8 or less using B(10,1/2)
= 1- 0.9892
= 0.0108

29. Wilcoxon signed rank test
• Analogous to paired ‘t’ test
• Stronger than the sign test for paired observation
• Used for ordered categorical data where a numerical scale
is inappropriate but it is possible to rank the observations.
• We reject null hypothesis at alpha 100% level of
significance if,
o W- ≤ W for a right tailed test
o W+ ≤ W for a left tailed test
o W - ≤ W /2 or W+ ≤ W /2 , for a two tailed test

30. • Eg: Suppose playing 4 rounds of golf at city club. 11
professionals totaled 280, 282, 292, 273, 283, 283,
275, 284, 282, 279 & 281.
• Median - 284

31. • W+ = 8; W- = 47
• W+ + W- = n(n+1)/2 = 55
• The table value of T at 5% level of significance when
n=11 is 11
• The value of T in our example is 8, being less than
11, we reject the null hypothesis
• Concluded that the golfer’s average is less than 284 in
four rounds.

32. Mann Whitney Test
• Mann-Whitney U - similar to the student’s t test
• Data – Ordinal type of data
• In this test, all the observations of two samples are ranked
numerically from smallest to the largest, without regard
to whether the observations are from first sample or from
the second sample
• Now, the ranks of observations from the two samples are
summed separately

33. • Average rank and variance of ranks are determined
• U = n1 . n2 +
n1 (n1 +1)
2
- R1

34. • Eg: The values in sample A are 53, 38, 69, 57, 46, 39,
73, 48, 73, 74, 60 and 78. In sample B are 44, 40, 61,
52, 32, 44, 70, 41, 67, 72, 53 and 72. Test at 10%
level the hypothesis that they come from populations
with the same mean. Apply U-test.

35. n1 = 12
n2 = 12
R1= 167.5
R2= 132.5
U = n1 . n2 +
n1 (n1 +1)
2
- R1
= 12.12 +
12 (12 +1)
2
- 167.5
=144 +78 - 167.5= 54.5

36. • Mean = U =
n1 ×n2
2
= 72
• Standard deviation = ΣU =
n1 . n2(n1+n2+1)
12
= 17.32
• As alternative hypothesis is that means of 2
populations are not equal, a two tailed test is
appropriate,
• Test statistic Zc =
U −U
ΣU
=
54.5 −72
17.32
= -1.0104
• p- value is 2p(Z˃ 1.0104) = 0.3123

37. Kruskal Wallis Test
• It’s more powerful than Chi-square test.
• It is computed exactly like the Mann-Whitney test,
except that there are more groups (>2 groups).
• Applied on independent samples with the same shape
(but not necessarily normal).

38. • Eg: Comparative photoelastic study of dental and
skeletal anchorages in the canine retraction
In the above study to compare both types of anchorage,
the Mann-Whitney test was used in each area evaluated,
whereas to compare the stress between the peri-
radicular regions of the canine, the Kruskal-Wallis test
was used in each type of anchorage.

39. Friedman ANOVA
• When either a matched-subjects or repeated-measure
design is used and the hypothesis of a difference
among three or more (k) treatments is to be tested, the
Friedman ANOVA by ranks test can be used.
• Assumptions - one group that is measured on 3 or
more different occasions

40. • 1 dependent variable either ordinal, interval or ratio
Sample : Random
Distribution: Non-Normal (Distribution free)

41. Spearman’s Rank Correlation
• Developed by Charles Spearman
• Rank correlation is a measure of correlation that exists
between two sets of ranks.
• The procedure consists of ranking the two sets of values
X and Y, and computing the difference of each pair. The
differences are then squared and added.
d= difference in ranks
rs = 1 -
6Σd2
n(n
2
−1)
n= number of pairs

42. • Use to assess the relationship between two ordinal
variables or two skewed continuous variables.
• Non parametric equivalent of the Pearson correlation
• It is a relative measure which varies from -1 (perfect
negative relationship) to +1 (perfect positive
relationship).

43. Serial no.
of patient
Fasting blood
glucose
level(mg/dl)
Rank
(R1)
Systolic BP
level
(mmHg)
Rank
(R2)
d=R1-R2
d2
1
2
3
4
5
6
7
8
9
10
90
92
98
112
120
121
126
132
143
145
1
2
3
4
5
6
7
8
9
10
136
140
142
130
148
135
150
170
145
165
3
4
5
1
7
2
8
10
6
9
-2
-2
-2
+3
-2
+4
-1
-2
+3
+1
4
4
4
9
4
16
1
4
9
1
Eg: Calculating rank correlation between fasting blood
glucose level and systolic blood pressure in 10 diabetics
patients.

44. rs = 1 -
6×56
10×99
= 1 - 0.339
= 0.661
• Referring the table, for n=10, at 5% level of significance we
find the value greater than the table value.
• Hence, the null hypothesis is rejected and it is concluded that
fasting blood sugar and systolic blood pressure are correlated.

47. Kendall’s coefficient of concordance
• It is represented by symbol W.
• It determines the degree of association among several (k)
sets of ranking of N objects or individuals.
• If there are only two sets of ranking N objects, we
generally work out Spearman’s coefficient of correlation,
but Kendall’s coefficient of concordance (W) is
considered an appropriate measure of studying the degree
of association among three or more sets of rankings.

48. • Eg: The correlation between the stages of tooth
calcification and the cervical vertebral maturation in
Iranian females.

49. Chi square test Sign test Mann Whitney
U test
Wilcoxon signed
rank test
Type of data Nominal data Ordinal data Ordinal data Ordinal data
Application To find the
strength of
association
between 2
variables
Analogous to t-
test
Analogous to
unpaired t- test
Analogous to paired
t- test
Example To test the
occurrence of
new cavities is
associated with
the oral hygiene
instructions
Effect of an
electronic
classroom
communication
device on dental
student
examination
scores
Comparing
lithotripsy to
ureteroscopy in
the treatment of
renal calculi
Compare knowledge,
attitude, and practice
measures between
groups in an
educational program
for type 1 diabetes

50. Kruskal wallis test Friedman test Spearman correlation
coefficient
Type of data Ordinal data Ordinal data Ordinal data
Application Analogous to ANOVA Analogous to repeated
measures of ANOVA
Analogous to Pearson
correlation coefficient
Example Analysis of the effects
of electronic medical
record systems on the
quality of
documentation in
primary care
Intragroup comparison of
retention of sealant and
development of caries at
3, 6 and 12 months
Correlation between fasting
blood glucose level and
systolic blood pressure.

51. CONCLUSION
• Tests of significance play an important role in conveying
the results of any research and thus the choice of an
appropriate statistical test is very important as it decides
the fate of outcome of the study.
• Hence the emphasis placed on tests of significance in
clinical research must be tempered with an
understanding that they are tools for analyzing data and
should never be used as a substitute for knowledgeable
interpretation of outcomes.

52. REFERENCES
• Katz DL, Elmore JG, Wild DMG, Lucan SC. Jekel’s
Epidemiology, Biostatistics and Preventive Medicine. 4rd
edition. Philadelphia: Elsevier Publishers; 2014.
• Kothari CR. Research Methodology-Methods and
Techniques: 4th Edition: New Age International
publishers; 2019.
• Mahajan BK. Methods in Biostatistics. 8th ed. New Delhi:
Jaypee Publishers; 2009.

53. • Peter S. Essentials of preventive and community
dentistry. 6th edition Arya publishers; 2017.
• Kim JS and Dailey RJ. Biostatistics for oral healthcare.
1st edition.
• Jaakkola S, Rautava P, Alanen P, Aromaa M,
Pienihäkkinen K, Räihä H, Vahlberg T, Mattila ML,
Sillanpää M. Dental fear: one single clinical question for
measurement. The open dentistry journal. 2009;3:161.

54. • Valizadeh S, Eil N, Ehsani S, Bakhshandeh H.
Correlation between dental and cervical vertebral
maturation in Iranian females. Iranian Journal of
Radiology. 2013 Jan;10(1):1.