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Biostatistics,
statistical software VII.
Non-parametric tests: Wilcoxon’s signed
rank test, Mann-Whitney U-test, KruskalWallis test, Spearman’ rank correlation.
Krisztina Boda PhD
Department of Medical Informatics,
University of Szeged
Parametric tests
Parameter: a parameter is a number
characterizing an aspect of a population
(such as the mean of some variable for the
population), or that characterizes a
theoretical distribution shape.
 Usually, population parameters cannot be
known exactly; in many cases we make
assumptions about them.

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Parameters of the normal distribution: , 
 Parameter of the binomial distribution: n, p
 Parameter of the Poisson distribution: 

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Normal distributions N(, )
N(0,1)
N(1,1)
Probability Dens ity Function
y=norm al(x;0;1)
Probability Dis tribution Function
p=inorm al(x;0;1)
0.6
Probability Dens ity Function
y=norm al(x;1;1)
1.0
Probability Dis tribut
p=inorm al(x
0.6
1.0
0.5
0.8
0.5
0.8
0.4
0.6
0.4
0.6
0.3
0.3
0.4
0.2
0.4
0.2
0.2
0.1
0.2
0.1
0.0
0.0
-3
-2
-1
0
1
2
3
-3
-2
-1
0
1
2
3
0.0
-2
Probability Dis tribution -3
Function
p=inorm al(x;0;2)
Probability Dens ity Function
y=norm al(x;0;2)
0.6
0.0
-1
0
1
2
3
-3
-2
-1
1.0
0.5
0.8
0.4
0.6
0.3
0.4
0.2
0.2
0.1
0.0
0.0
-3
-2
-1
0
1
2
3
-3
,  : parameters
(a parameter is a number that
describes the distribution)
-2
-1
0
1
2
3
N(0,2)
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0
Binomial distributions







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1. Each trial results in one of two possible, mutually exclusive outcome. (success,
failure)
2. The probability of a success, p, remains constant from trial to trial
3. The trials are independent.
We are interested in being able to compute the probability of k successes in n
trials.
The binomial distribution is useful for describing distributions of binomial events,
such as the number of males and females in a random sample of companies, or
the number of defective components in samples of 20 units taken from a
production process. The binomial distribution is defined as:
 n
n!
 n
, n!  1  2...n
 
Pk  P( X  k )    p k q n  k , k  0,1,..., n
k


k
!(
n

k
)!
k
 
p is the probability that the respective event will occur
q is equal to 1-p
n is the maximum number of independent trials.
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Example

Suppose that it is known that 30% of a
certain population are immune to some
disease. If a random sample of size n=10 is
selected from this population, what is the
probability that it will contain exactly k=4
immune persons?
10  4 6 10! 4 6
P( X  4)   0.3 0.7 
0.3 0.7
4!6!
4 
 210  0.0081 0.117649  0.200121
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Number of Success
0
1
2
3
4
5
6
7
8
9
10
Összesen
Probabilty
distribution
0.028247525
0.121060821
0.233474441
0.266827932
0.200120949
0.102919345
0.036756909
0.009001692
0.001446701
0.000137781
5.9049E-06
Distribution function
0.028247525
0.149308346
0.382782786
0.649610718
0.849731667
0.952651013
0.989407922
0.998409614
0.999856314
0.999994095
1
Probabilty of
"success"
0.3
1
Probabilty distribution
Distribution function
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
1
2
3
4
5
6
7
8
9
10
11
0
0
1
2
3
4
5
6
7
8
9
10
11
Binomial distribution n=10, p can be changfed, k=0,1,…,10
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Poisson distribution


The Poisson distribution is also sometimes referred to as
the distribution of rare events. Examples of Poisson
distributed variables are number of accidents per
person, number of sweepstakes won per person, or the
number of catastrophic defects found in a production
process.
If n tends to infinity, but at the same time np= is kept
constant the binomial distribution approaches a fixed
distribution
 n k n   k
k  
lim Pk  lim   p q
 f (k ) 
e
n 
k!
n   k 
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
Example. In a certain disease the number of new occurrences in a month is 3
in average. Assuming that the number of new occurrences follows a Poisson
distribution, what is the probability that
 Nobody becomes ill (0.0498)
 There are exactly 2 new occurrences (0.224)
Number of events
0
1
2
3
4
5
6
7
8
9
10
Total
Probability
Distribution function
0.049787068
0.049787068
0.149361205
0.199148273
0.224041808
0.423190081
0.224041808
0.647231889
0.168031356
0.815263245
0.100818813
0.916082058
0.050409407
0.966491465
0.021604031
0.988095496
0.008101512
0.996197008
0.002700504
0.998897512
0.000810151
0.999707663
0.999707663
Average number of
events
3
Probability
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
0
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Distribution function
1
2
3
4
5
6
7
8
9
10
11
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0
1
2
3
4
5
6
7
8
9
10
11
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Parametric tests
The null hypothesis contains a parameter
of a distribution. The assumptions of the
tests are that the samples are drawn from
a normally distributed population.
 One sample t-test: H0: =c,
 Two sample t-test: H0: 1=2,assumptions:
1=2

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Nonparametric tests
We do not need to make specific
assumptions about the distribution of data.
 They can be used when

 The distribution is not normal
 The shape of the distribution is not evident
 Data are measured on an ordinal scale (lownormal-high, passed – acceptable – good –
very good)
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Ranking data

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Nonparametric tests can't use the estimations of
population parameters. They use ranks instead. Instead of
the original sample data we have to use its rank.
To show the ranking procedure suppose we have the
following sample of measurements: 199, 126, 81, 68, 112,
112.
Sort the data in ascending order: 68, 81,112,112,126,199
Give ranks from 1 to n:
1, 2, 3, 4, 5, 6
Cases 5 and 6 are equal, they are assigned a rank of 3.5,
the average rank of 3 and 4. We say that case 5 and 6 are
tied.
Ranks corrected for ties:
1, 2, 3.5, 3.5, 5, 6
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Result of ranking data
Case Data Rank Ranks corrected for ties
4 68
1
1
3
81
2
2
5 112
3
3.5
6 112
4
3.5
2 126
5
5
1 199
6
6
n

The sum of all ranks must be
 ri 
i 1


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n(n  1)
2
Using this formula we can check our computations.
Now the sum of ranks is 21, and 6(7)/2=21.
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Nonparametric tests for paired data
(nonparametric alternatives of paired t-test)
Sign test
 Wilcoxon’s matched pairs test
 Null hypothesis: the paired samples are
drawn from the same population

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The sign test




Example: 13 students
were measured in
reading speed and
comprehension at a
course ending and after
1 month. Suppose we
have reason to believe
that the two
distributions of reading
scores are not normal.
Number of positive
signs: 6
Number of negative
signs: 5
Cases with no change
are omitted
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Student Score
at course
ending
1
50
2
48
3
46
4
50
5
62
6
80
7
23
8
30
9
45
10
53
11
49
12
51
13
46
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Score
after
1 month
52
51
46
49
50
70
21
33
46
53
48
48
48
Difference
-2
-3
0
1
2
10
2
-3
-1
0
1
3
-2
Sign
+
+
+
+
+
+
-
15
Table of the sign test

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The table contains the
acceptance region for
given sample size
and 
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Decision based on table

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Krisztina Boda
If the distributions of the two variables are the same (If
the null hypothesis is true), the numbers of positive and
negative differences should be similar.
The null hypothesis is accepted if both numbers lie in the
interval given it table for the sign test
Number of positive signs: 6
Number of negative signs: 5
For n=11 and =0.05, this interval is 1-10.
As both 5 and 6 lies in the interval 1-10, we accept the
null hypothesis at 5% level.
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The Wilcoxon signed rank test

Example: 13 students
were measured in
reading speed and
comprehension at a
course ending and after
1 month. Suppose we
have reason to believe
that the two
distributions of reading
scores are not normal.

Sum of ranks belonging
to positive signs:
R+=40.5

Sum of ranks belonging
to negative signs:
R-=25.5

Cases with no change
are omitted
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Student Score
at course
ending
1
50
2
48
3
46
4
50
5
62
6
80
7
23
8
30
9
45
10
53
11
49
12
51
13
46
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Score
after
1 month
52
51
46
49
50
70
21
33
46
53
48
48
48
Difference Rank
ignoring
signs
-2
5.5
-3
9
0
1
2
2
5.5
10
11
2
-3
9
-1
2
0
1
2
3
9
-2
5.5
18
Table of the Wilcoxon
signed rank test

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The table contains the
acceptance region for
given sample size
and 
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Decision based on table



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

Krisztina Boda
If the distributions of the two variables are the same (If the null
hypothesis is true), the sum of positive and negative ranks should be
similar.
The null hypothesis is accepted if both numbers lie in the interval
given it table for the test
Sum of ranks belonging to positive signs: R+=40.5
Sum of ranks belonging to negative signs: R-=25.5
For n=11 and =0.05, this interval is 10-56.
As both rank sums are in this interval, we do not reject the null
hypothesis and claim that the difference is not significant at 5%
level.
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The case of large samples

When the sample size is large, we can
count the mean and standard deviation of
the ranks and use the normal distribution
to get the p-value. Computer packages
use this normal approximation also in
case of small sample size
n
R   Ri

z
i 1
n
R
i 1
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2
R   n(n  1) / 4

~ N (0,1)
(n(n  1)( 2n  1) / 24)
i
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Nonparametric test for data in independent groups
(nonparametric alternatives of two sample t-test)

Mann-Whitney U test

Null hypothesis: the samples are drawn
from the same population
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Hypothetical example


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The change of body weight are compared in two
groups: patients having a special diet and
control patients.
Null hypothesis: the diet is not effective, data are
drawn from the same population.
The original data are ranked and the sum of
ranks in each group is computed. If the null
hypothesis is true, the sum of ranks in the two
groups are similar.
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Krisztina Boda
Patient
Change in body
weight (kg)
Group
Rank
Rank corrected for
ties
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Sum of ranks, R1
11.
12.
13.
14.
15.
16.
17.
18.
19
20.
21.
Sum of ranks R2
-1
5
3
10
6
4
0
1
6
6
Diet
Diet
Diet
Diet
Diet
Diet
Diet
Diet
Diet
Diet
3
16
12
21
18
15
4
8
19
20
2
0
1
0
3
1
5
0
-2
-2
3
Control
Control
Control
Control
Control
Control
Control
Control
Control
Control
Control
11
5
9
6
13
10
17
7
1
2
14
3
16.5
13
21
19
15
5.5
9
19
19
140
11
5.5
9
5.5
13
9
16.5
5.5
1.5
1.5
13
91
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Table of the MannWhitney U test
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Decision based on table







Krisztina Boda
If the distributions of the two variables are the same (If the null
hypothesis is true), the sum of ranks in the two groups should be
similar.
The test statistic T is the sum of the ranks in the smaller group. The
null hypothesis is accepted T lies in the interval given it table for the
test
Sum of ranks in the first group (n=10): R1=140
Sum of ranks in the second group (n=11): R2=91
The test statistic T is the sum of the ranks in the smaller group.
T=140. For n1=10 and n2=11 and =0.05, this interval is 81-139.
As T lies outside of this interval, we reject the null hypothesis and
claim that the difference is significant at 5% level.
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An alternative test statistic

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The statistic U (due to Mann Whitney) is the
number of all possible pairs of observations
comprising one from each sample, say xi and yi
, for which xi<yi. This if the sample sizes are n1
and n2, the U/n1n2 is the proportion of all such
pairs, and so is also the estimated probability
that a new observation from the first population
will be less than a new observation sampled
from the second population.
1
U  n1 n2  n1 (n1  1)  T
2
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The case of large samples

When the sample size is large, T test statistic T has an
approximately Normal distribution And we can calculate
the test statistic z according to the following formula: (ns
and nL are the sample sizes in the smaller and larger
group respectively).
z

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Rs  ns (ns  n L  1) / 2
ns n L (ns  n L  1)
12
~ N (0,1)
Computer packages use this normal approximation also
in case of small sample size
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Comparing several independent groups:
the Kruskal-Wallis test




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It is also called nonparametric one-way ANOVA
It tests whether k independent samples that are defined
by a grouping variable are from the same population.
This test assumes that there is no a priori ordering of the
k populations from which the samples are drawn.
As a result, it gives one p-value.
If the null hypothesis is rejected, further tests are
required to make pairwise comparisons. These pairwise
comparisons are generally not available in standard
statistical packages. Pairwise comparisons can be
performed by Mann Whitney U tests and p-values can be
corrected by Bonferroni correction.
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Comparison of several related samples:
the Friedman test



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The Friedman test is the nonparametric equivalent of a one-sample
repeated measures design or a two-way analysis of variance with
one observation per cell.
Friedman tests the null hypothesis that k related variables come
from the same population. For each case, the k variables are ranked
from 1 to k. The test statistic is based on these ranks.
As a result, it gives one p-value.
If the null hypothesis is rejected, further tests are required to make
pairwise comparisons. These pairwise comparisons are generally
not available in standard statistical packages. Pairwise comparisons
can be performed by Wilxocon signed rank tests and p-values can
be corrected by Bonferroni correction.
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Review questions and exercises

Problems to be solved by handcalculations
 ..\Handouts\Problems hand VII.doc

Solutions
 ..\Handouts\Problems hand VII solutions.doc

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Problems to be solved using computer none
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Useful WEB pages




Krisztina Boda
http://www-stat.stanford.edu/~naras/jsm
http://www.ruf.rice.edu/~lane/rvls.html
http://my.execpc.com/~helberg/statistics.html
http://www.math.csusb.edu/faculty/stanton/m26
2/index.html
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