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6.3. Nonparametric Methods
(I) One sample problem: Wilcoxon sign rank test:
(not necessarily normal
x1 ,, x n are the data from some population
population). Suppose m is the population median. We want to test if H 0 :
m  m0 , where m0 is some number. We now introduce the Wilcoxon sign rank
test. Let zi  zi  m0 and ri is the rank of z i . Let s  
all the ranks corresponding to xi  m0 >0 while s  
r
r
i
( xi  m0 ) 0
i
( xi  m0 )0
be the sum of
is the sum of all the
ranks corresponding to xi  m0 <0.
Example 1:
H 0 : m  m0  8
xi
7.1
13.5
5.2
4.2
6.4
10.4
5.7
5.4
6.3
7.5
zi
0.9
5.5
2.8
3.8
1.6
2.4
2.3
2.6
1.7
0.5
ri
2
10
8
9
3
6
5
7
4
1
Sign
of
xi  8
_
+
_
_
_
+
_
_
_
_
Then, s   10  6  16, s   2  8  9  3  5  7  4  1  39 .
The following are the Wilcoxon sign rank tests under three alternatives:
(i) H 0 : m  m0 vs H a : m  m0
 s   wn, , then reject H 0 , where wn , can be found by table.
(ii)
H 0 : m  m0 vs H a : m  m0
 s   wn, , then reject H 0 , where wn , can be found by table.
(iii) H 0 : m  m0 vs H a : m  m0
 s  min( s  , s  )  wn, , then reject H 0 , where wn , can be found by
table.
Note: s   s  
n(n  1)
.
2
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95
Example 1 (continue):
Test H 0 : m  m0  8 vs H 1 : m  8 ,   0.05.
[sol]: s   16, w10,0.05  11  s   16  11  not reject H 0 .
As the sample size is large, p-value can be obtained vis a normal approximation. As n
n(n  1) n(n  1)( 2n  1)
is large, T   N (
,
)
4
24
n(n  1)
4
 N (0,1). Thus, as
n(n  1)( 2n  1)
24
T 
H 0 : M  M 0 vs H 1 : M  M 0 and T   c, then
p-value= P( Z 
n(n  1)
n(n  1)
c
4
4
)  (
) . The p-values under the
n(n  1)( 2n  1)
n(n  1)( 2n  1)
24
24
c
other alternatives can be obtained similarly.
(II) Two samples problem: Wilcoxon rank sum test:
x1 ,, x n and y1 ,, y m are the data from two populations. Let m1 and m2 be
the population medians the two populations, respectively. We want to test H 0 :
m1 = m2 .
Wilcoxon rank sum test:
1. Combine x1 ,, x n , y1 ,, y m and find all the ranks of these data.
2. Find the sum of the ranks w corresponding to x1 ,, x n .
The following are the Wilcoxon rank sum tests under three alternatives:
(i) H 0 : m1  m2 vs H a : m1  m2
 reject H 0 as w  wu , where wu can be found by table.
(ii) H 0 : m1  m2 vs H a : m1  m2
 reject H 0 as w  wL , where wL can be found by table.
(iii) H 0 : m1  m2 vs H a : m1  m2
 reject H 0 as w  wu or w  wL .
Example 2:
Area A
60
100
150
290
Area B
110
130
140
170
200
310
The above data are the incomes in two areas. Test H 0 : m1  m2 vs H a :
m1  m2 with   0.05.
[sol]:
95
96
1.
Income
60
100
150
290
110
130
140
Rank
1
2
6
9
3
4
5
2. n1  4, n2  6, w  1  2  6  9  18, wL  12, wu  32.
Since 12  w  18  32  not reject H 0 .
170
200
310
7
8
10
Conclusion: there is no difference in income between the two areas.
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