Non-parametric tests

1. 1
Introduction
NON-PARAMETRICTESTS
Most of the hypothesis testing (Z, t, F) we have done prior involve inferences concerning population parameters
such as population mean or proportion ( or  ).These tests are called the parametric tests and their test and
statistics are called parametric statistics.
The sampling distributions for the statistics used in these tests depend upon assumptions about the population
from where the samples are obtained from.
All the above tests assume that the population distribution is normal and the variances are equal and the samples
are normally distributed with a mean of zero and variance of 2
 ie 2
(0, )X N :
But there is no way we can tell that the population is normally distributed and has same variance. Even if the
population distribution is known, the normal distribution is just a theoretical assumption to make computations
easier. There is nothing like normally distribution population.
Under Non parametric tests, also known as distribution free tests, there are no assumptions about the distribution
of the present population. They are free of the assumptions regarding the distribution of the populations.
The NPT are concerned with testing hypothesis only and not bothered with estimation of parameters
TYPES OF NON PARAMETRIC TESTS AND THEIR APPLICATION
The Mann- Whitney U-Test
This test is applied when you have two independent randomly selected populations. It tests whether two
independently drawn samples have been drawn from the populations with the same characteristics ie (with equal
means, medians, modes, σ2, σ)
Ho: The two populations are the same HA: the two populations are different
Procedure
i. Combine the two samples and then rank the observations. The lowest value should be given rank 1 and soon.
In case of tie of the observations, get the average rank. Eg assume that the smallest observations are 2 of
them. Average rank will be (1 2) 1.5  and then the next rank will be 3 since position 1 and 2 have already being

2. 2
2
exhausted.
ii. The smaller sample size should be taken as sample one with size ( 1n ) and the larger sample as sample two
with size 2n
iii. Find the sum of the ranks for each sample separately which should be denoted by
∑ R1and ∑ R2 respectively.
Test statistic formula:-
1 1
1 1 2 1
( 1)
2
n n
U n n R

    2 1 2 1U n n U  = 2 2
1 2 2
( 1)
2
n n
n n R

  
Decision rule:
Compare U1 and U2 and choose the smaller value which should be taken as the value for CU (U calculated). , Use
the Mann Whitney table which gives the critical (theoretical) values of U (U ) depending on the level of
significance.
If calU ≤U reject Ho
If cU > Uα accept Ho
Example
A doctor treats 8 patients with Fancider and 10 patients with Malariaquin. He obtains the following results for the
number of days required to cure malaria
AFancider 15, 9, 12, 22, 14, 9, 10, 15
BMalariaquin 7, 8, 10, 6, 7, 7, 4, 13, 11, 5
At 5% level of significance, can we conclude that effectiveness of the two drugs is the same?

3. 3
Solutions
Obvs Sample A Ranks of A Obvs Sample B Ranks of B
1 15 16.5 1 7 5
2 9 8.5 2 8 7
3 12 13 3 10 10.5
4 22 18 4 6 3
5 14 15 5 7 5
6 9 8.5 6 7 5
7 10 10.5 7 4 1
8 15 16.5 8 13 14
9 11 12
10 5 2
Sum 106.5 64.5
∑ R1 = 106.5 and ∑ R2 = 64.5 where n1 = 8 and n2 = 10
Note that ∑ R1 and ∑ R2 are sum ranks for smaller and larger sample respectively.
1
8(8 1)
8*10 106.5 9.5
2
U

    2 10*8 9.5 70.5U   
CalU = 9.5,
Ho: A=B,
HA: A≠B
Uα (tabulated value) = 17 at α =5%
CalU <Uα
Conclusion:- reject Ho at 5% and conclude that the two medicines differ in their effectiveness of treating malaria.
If n1 and n2 are both at least 10, the sampling distributions for the U values will be normally distribution with the
mean and standard deviation of the sampling distribution for the U-statistics calculated as follows.
Mean of the sampling distribution -
1 2 1 2
2 2
u
n n U U


 
For Mann-Whitney U test
Standard deviation of the Sampling - 1 2 1 2( 1)
12
u
n n n n

 

distribution for Mann- Whitney U test

4. 4
Z- Value to normalize the Mann Whitney U test
i u
u
U
Z




Where iU is the appropriate U – value, either 1 2U orU , depending on the nature of the test ie
whether one tailed or two tailed test.
Wilcoxon Rank Sum Test for Independent Samples
This test is used to test for the differences between two independent populations using two samples drawn from
two populations. They are independent in that there are no intrinsic characteristics that make them to be related.
Procedure:-
 Combine data from the two samples.
 Assign ranks to the pooled data. Note:- the smaller value is assigned rank1
The samples are then separated and the sums of the ranks are calculated for each sample.
NB: In case of tie of the observations, get the average rank. Assume that the smallest observations are 2 of
them. Average rank will be (1 2) 1.5  and then the next rank will be 3 since position 1 and 2 have already
being exhausted.
 The rank sums obtained are used as test statistics
Example
We wish to know whether the following samples A and B were picked from population with the same
characteristics.
A 3.1, 4.8, 2.3, 5.6, 0, 2.9
B 4.4, 5.8, 3.9, 8.7, 6.3, 10.5, 10.8
Use Wilcoxon rank sum test and 5% level of significance
A
RA
B
RB
3.1 4 4.4 8
4.8 7 5.8 9
2.3 2 3.9 5
5.6 6 8.7 11
0 1 6.3 10
2.9 3 10.5 12
10.8 13
Sum 23 68

5. 5
Æ
Mostly we use a two tailed test stated as:- two populations have identical distribution
H0::A = B
HA: A≠B
Decision rule
Reject H0if RÆ≤ TLor RB≥TU
To carry out the test we use the sum rank of the smaller sample size. If the sample sizes are the same then we use
either sum. This sum must be represented by RA
The tables to be used depend on the sample size and the level of significance Case I: If 3 ≤ ni≤10.
Use the critical values provided by Wilcoxon rank sum test table for independent samples. In the table n1
represent columns and the rows are represented by n2. The TL and the TU are the boundaries of the lower and
upper regions for the sum rank associated with the sample that has fewer and more observations respectively.
For this example:-
AR = 23 and BR = 68
We use the sum of the ranks for smaller sample size i.e. AR = 23
Where . n1 = 6 and n2 = 7
From Theoretical tables TL = 29, TU = 57
Decision rule: Reject H0 if LA UBR or TRT   ie reject H0 if AR is smaller than or equal to TL or bigger than
or equal to TU.
Conclusion:- we reject H0 since AR = 23 is smaller than TL = 29 and conclude that the two populations are not
same i.e. samples were not picked from same population. The two populations are different
CaseII if ni>10
The sampling distribution of ΣRA can be approximated by a normal distribution with mean and standard deviation
expressed as

6. 6
Mean =   1( 1)
2
n N
E R

 
where N = 1 2n n
Standard deviation of the distribution =
1 2 ( 1)
12
n n N



we use a (Z) test. The calculated Z score is given by the formula:-
A
cal
R
Z





The decision rule is that reject null hypothesis if absolute value of Z calculated is greater than
absolute value of Z tabulated at a certain level of significance
Wilcoxon Signed – Rank Test for a Paired Sample
This test is used two compare two populations for the same set of observations before and after
an event. Eg testing whether the performance of the students in a class remain the same before
and after remedial teaching or whether productivity of land remain the same before and after use
of a chemical fertilizer. It therefore compares the distributions of two related groups.
Procedure
i Obtain data for the matched pairs
ii Calculate their difference of the values of the observation
iii Ignore the signs but take note of original difference and rank the absolute values of the
difference using the same rule as in Mann Whitney or Wilcoxon Sum test. Ignore any zero
difference
iv Determine No. of differences and label that no by N but remember to ignore any zero
difference.
v Find the sum of ranks for the +ve differences denoted by ΣR+ and for the –ve differences
denoted byΣR-

7. 7
Compare ΣR+ and ΣR- and take the smaller one which you denote by V
Decision Rule:-
Refer to Wilcoxon sign table and find the theoretical value or critical value for the Wilcoxon test.
If V or calculated value is less or equal to critical value gotten from the table you reject H0
otherwise do not reject H0.
Example
A Lecturer wishes to find out whether the performance of the students is the same before and
after remedial teaching.
Student (A)
Before
(B)
After
Di
(difference)
Rank
1 20 45 -25 8
2 30 36 -6 3
3 35 23 +12 5
4 38 35 +3 1
5 45 45 0 -
6 45 50 -5 2
7 50 70 -20 7
8 55 65 -10 4
9 25 55 -30 9
10 45 30 +15 6
Hypothesis
Ho : A = B
HA : A ≠ B
From the table above
ΣR+ = 5 + 1 +6 = 12
ΣR- = 8 + 3 + 2 + 7 + 4 + 9 = 33
Therefore: V =12 N= 9 (ignore ties) Tabulated V at α= 5% =5 (Two tailed)
Since Calculated V = 12 > Vα = 5
Decision: accept Ho and conclude that at 5% level there are no differences in performance before
and after remedial teaching. In case N (number of pairs) are greater than 30, the sampling
distribution for V approximate normality with
( 1)
4
N N
mean 

  and standard deviation =
( 1)(2 1)
24
N N N

 


8. 8
N = number of pairs with differences
cal
V
Z




The decision rule is that reject null hypothesis if absolute value of Z calculated is greater than absolute
value of Z tabulated at a certain level of significance. Note that now we have assumed normality, we are
now in the parametric tests and the rule of rejecting or accepting remains the same
Kruskal – Wallis Test
This is one has approach similar to the Mann-Whitney (U) test only that it is applicable when
there are 3 or more populations we wish to compare. It also analogous to ANOVA in the
parametric tests which tested equivalent of three populations with assumption of normality.
Requirements
• Ordinal level
• three or more samples
• independent samples
• simple random sample
Ho: All K populations have the same distribution
HA: Not all K populations have the same distribution
Procedure
• get ranks of the pooled data
• separate the samples and obtain the sum of the respective ranks
• apply the following test statistic
•
Where:
in is the number item in each sample
N is the total number of observations in all samples
2
12
3( 1)
( 1)
i
i
R
K N
N N n
 
   
   


9. 9
R is the sum of the ranks in ith sample
The distribution of K is a approximated by a chi-square distribution with K – 1 degrees of
freedom. If K exceeds the critical value for chi – square, the null is rejected and vice versa
Note:-
Always use 2
 (chi square table) if
i. K > 3 i.e. you have more than 3samples
ii. K = 3 but ni > 5 i.e. there are just three samples but the respective sample sizes are
greater than five for each
Use Kruskal Wallis table if K = 3 but in ≤5 i.e. you have just three samples (3 populations) and
their respective sample sizes are less or equal to five
Decision rule:
i) if using table A:10, reject Ho if Kcal≤Kα (tabulated)
ii) if using x2then reject Ho ifKcal ≥x2α(tabulated)
Example
Samp 1 Rank 1 Samp 2 Rank 2 Samp 3 Rank 3 Samp 4 4Rank 4 Samp 5 Rank 5
85 12 95 19 67 3 90 15 100 21
73 7 54 1 74 8 65 2 101 22
96 20 72 6 84 11 92 17 103 23
91 16 81 10 68 4 94 18 105 24
88 14 69 5 87 13 110 25
77 9
69 41 48 52 115
= 14.178
Kcal = 14.178
X2α (tabulated) =
9.488
Kcal > X2α we reject Ho and accept that at 5% there are differences between the five populations
2 2 2 2
12 69 41 48 52 115
3(25 1)
25(25 1) 5 5 4 4 5
K
 
         

10. 10
Example 2
Samp 1 Rank 1 Samp 2 Rank 2 Samp 3 Rank 3
3 3.5 1 1 9 9
5 5 2 2 8 8
7 7 3 3.5 9 10
6 6
21.5 6.5 27
Note that in the Kuskal Wallis table
the critical value at alpha =5% and n1
n2 n3 being 4 3 3 = 5.791. Since calculate K is greater than tabulated value we reject Ho and
conclude that there are differences between the three populations.
Spearman Rank Correlation Test
This is used to test for correlation between two populations i.e. you test whether two population
are associated. The argument here is that despite sample spearman rank correlation showing that
two variables X and Y are related, there might not be such kind of relationship in the population
where the samples were taken. It may be by mere chance that chosen sample indicate correlation.
We shall hypothesis that, the population correlation coefficient is zero (  =0)
Ho:  =0, no correlation between Y and X H1:  ≠0, there is correlation
Procedure
i) Obtain data for the two samples
ii) Assign ranks separately to the two samples
iii) Calculate the spearman rank correlation coefficient
2
2
6
1
( 1)
i
s
D
r
n n
 


Where Di is differences in the ranks for the corresponding observations. sr is the sample
correlation coefficient
Decision rule: If cal ≥  reject Ho and if opposite accept H0
2 2
12 21.5 6.5 27
3(10 1) 7.7
10(10 1) 4 3 3
K
 
        

11. 11
To get the critical value (  ) we use Spearman rank correlation table
Example
A manager sought to find out whether absenteeism (X) and Performance ranking (Y) of workers
are correlated:
X Y Rank for
X(rx)
Rank for Y(rY Di2
50 12 7 2 25
48 10 6 1 25
30 40 3 5 4
47 13 5 3 4
20 50 1 7 36
25 45 2 6 16
40 20 4 4 0
110
6*110
1 0.9643
7(49 1)
sr    

Negative association between X and Y  = 0.6786 at 10%
Since cal =|−0.96|≥  =0.6786 reject Ho and conclude that there is negative correlation between
absenteeism and performance ranking.
Runs Test
This test is used to test for randomness of the data i.e. to find out whether the sample was
randomly picked from the population. If the data is not random, then one cannot use the various
tests e.g. X, f, t to test for the significance of the data.
Def: - A run ( r ) consists of an unbroken sequence of one or more like symbols The null and
alternate hypothesis is stated as:
Ho: The randomness exist in the sample
H1: The randomness does not prevail
Procedure
• Obtain the sample
• Obtain the number of attributes for each or sample size for each i.e.ni
• Obtain the number of runs = r. If there is unusually large or unusually small number of runs,

12. 12
a pattern is suggested. make a decision by use of table M1 or M2
NB: Table M1 shows the minimum critical number of runs for α = 5% if r ≤ rα you reject Ho at
5%
Table M2 provides maximum critical values of r at 5% level if r ≥ rα we reject Ho at 5%
To complete a run test two symbol say A and B . Assign all observation with one of two
symbols. If categories are grouped into categories of A and B, then the following sequence may
be observed
AA BBB A BB AAA B
1 2 3 4 5 6
Example 1
Suppose a panel of students for a specialized program is selected randomly from school of
economics. The selection panel decides to use gender as a test of randomness such that if
selection does not depend on gender then it is deemed as a random, if the converse is true, the
selection is assumed to be gender based. The following table shows the number of runs
Testing for randomness with figures
mmmm fffff mmm
1 2 3
Number of runs = 3 nm = 7 be the number of males and nf = 5 be the number of females.
• Table M1 shows the minimum critical number of runs for an  -value of 5percent.
• If the number of runs is equal to or less than the value shown in Table MI, it suggests that, at
the 5 percent level, there are too few runs to support the null hypothesis of randomness.
• Given that n1 = 7 and n2 = 5, the critically low value is found to be 3.The decision rule is that
we reject Ho if r ≥ r .Since number of runs (3) in this example are equal to critical value in the
table we reject Ho and conclude that randomness does not exist.
If you have figures which you wish to determine whether there is randomness in how
observations are laid down e.g. marks of students and you want to find out that they are random
we follow the following procedure:

13. 13
i. Find the median of the data.
ii. The median should be excluded from further analysis
iii. Any number greater than the median can be assigned a different letter (notation) or
symbol and those below a different symbol e.g.A and B
Example 2
The following data gives the marks of students. Find out whether the marks were randomly
picked or not. In such a case, we can use the median as a benchmark. The median is 37.
Therefore:
Let observations >37 be categorized in group A and observations <37 be B
Ho: marks are random HA: marks are not random
31, 57, 52, 22, 24, 59, 25, 29, 27, 44, 43, 32, 40, 37, 60
B AA BB A BBB AA B AA
1 2 3 4 5 6 7 8
No of runs = 8, n1=7, n2 =7
.rα = 3 forM1
.rα = 13 forM2
For both we accept Ho, that the marks were randomly picked at 5% level
Large samples
If both n1 and n2 are greater than 20, the sampling distribution for r approximates normality.
The distribution has a mean of:
1 2
1 2
2
1i
n n
n n
  

2*27*25
1 26.96
27 25
i   

The standard deviation of:
1 2 1 2 1 2
2
1 2 1 2
2 (2 )
( ) ( 1)
r
n n n n n n
n n n n

 

  
2
2*27*25(2*27*25 27 25)
3.565
(27 25) (27 25 1)
r
 
 
  
Standardizing the distribution of runs can be accomplished by using the normal deviate:

14. 14
r
cal
r
r
Z




Example
A sales presentation made to a group of 52 buyers resulted in 27 sales, 25 no sales and 18 runs.
At 1% level, is the sample random?
Ho: The sample is random
HA: The sample is not random
At 1% level two tailed test, the Z value is 2.58. The decision rule is that we do not reject null
hypothesis if
2.58 2.58calZ  
2*27*25
1 26.96
27 25
i   

2
2*27*25(2*27*25 27 25)
3.565
(27 25) (27 25 1)
r
 
 
  
18 26.96
2.513
3.565
calZ

  
calZ fall with the acceptance region, we accept Ho and conclude that the sample was random.
ADVANTAGES AND DISADVANTAGES OF NON PARAMETRIC TESTS
Advantages of NPT
• They are easy to comprehend and easy to apply
• No assumptions made about the distribution of the present population
• Applied to data expressed in ranks
• Can be used even with small samples
Disadvantages
i. Can be used only if the data is in the form of ranks or can be expressed in the form of
ranks
ii. They only test the hypothesis only and cannot be used to estimate parameters