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Transcript of Nonparametric Statistics. In previous testing, we assumed that our samples were drawn from normally...
Nonparametric Statistics
In previous testing, we assumed that our samples were drawn from normally distributed populations.
This chapter introduces some techniques that do not make that assumption.
These methods are called distribution-free or nonparametric tests.
In situations where the normal assumption is appropriate, nonparametric tests are less efficient than traditional parametric methods.
Nonparametric tests frequently make use only of the order of the observations and not the actual values.
In this section, we will discuss four nonparametric tests:
the Wilcoxon Rank Sum Test (or Mann-Whitney U test),
the Wilcoxon Signed Ranks Test,
the Kruskal-Wallis Test, and
the one sample test of runs.
The Wilcoxon Rank Sum Testor Mann-Whitney U Test
This test is used to test whether 2 independent samples have been drawn from populations with the same median.
It is a nonparametric substitute for the t-test on the difference between two means.
Based on the following samples from two universities, test at the 10% level whether graduates from the two schools have the same average grade on an aptitude test.
Wilcoxon Rank Sum Test Example:
university
A B
50 70
52 73
56 77
60 80
64 83
68 85
71 87
74 88
89 96
95 99
university
A B
50 70
52 73
56 77
60 80
64 83
68 85
71 87
74 88
89 96
95 99
First merge and rank the grades.Sum the ranks for each sample.
rank grade university
1 50 A
2 52 A
3 56 A
4 60 A
5 64 A
6 68 A
7 70 B
8 71 A
9 73 B
10 74 A
11 77 B
12 80 B
13 83 B
14 85 B
15 87 B
16 88 B
17 89 A
18 95 A
19 96 B
20 99 B
rank sum for university A: 74rank sum for university B: 136
Note: If there are ties, each value gets the average rank. For example, if 2 values tie for 3th and 4th place, both are ranked 3.5. If three differences would be ranked 7, 8, and 9, rank them all 8.
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st
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1 1 21 T
n (n 1)The mean of T is ,
2
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1
1 2 1 2T
n n (n 1)and the standard deviation is .
12
n
1 2 1If n and n are each at least 10, T is approximately normal.
1
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1 T
T
T - So, Z has a standard normal distribution.
st1Define T sum of the ranks for 1 sample .
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1For our example, T 74.
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n n (n 1) (10)(10)(20 1) 13.229
12 12
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T
T - 74 - 105Z -2.343.
13.229
-1.645 0 1.645 Z
.45.45.05 .05
critical region
critical region
Since the critical values for a 2-tailed Z test at the 10% level are 1.645 and -1.645, we reject H0 that the medians are the same and accept H1 that the medians are different.
For small sample sizes, you can use Table E.6 in your textbook, which provides the lower and upper critical values for the Wilcoxon Rank Sum Test.
That table shows that for our 10% 2-tailed test, the lower critical value is 82 and the upper critical value is 128.
Since our smaller sample’s rank sum is 74, which is outside the interval (82, 128) indicated in the table, we reject the null hypothesis that the medians are the same and conclude that they are different.
Equivalently, since the larger sample’s rank sum is 136, which is also outside the interval (82, 128), we again reject the null hypothesis that the medians are the same and conclude that they are different.
The Wilcoxon Signed Rank Test
This test is used to test whether 2 dependent samples have been drawn from populations with the same median.
It is a nonparametric substitute for the paired t-test on the difference between two means.
Wilcoxon Signed Rank Test Procedure1. Calculate the differences in the paired values (Di=X1i – X2i)2. Take absolute values of the differences and rank them (Discard
all differences that equal 0.)
3. Assign ranks Ri with the smallest rank equal to 1. As in the rank sum test, if two or more of the differences are
equal, each difference gets the average rank. (That is, if two differences would be ranked 3 and 4, rank them both 3.5. If three differences would be ranked 7, 8, and 9, rank them all 8.)
4. Assign the symbol + to positive differences and – to negative differences.
5. Calculate the Wilcoxon statistic W as the sum of the positive ranks. So,
iW R
Wilcoxon Signed Rank Test Procedure (cont’d)
is W statistic Wilcoxon theofmean The4
)1(
nnW
is W statistic Wilcoxon theofdeviation standard The
24
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W
WW
Z
well.)as used sometimes ision approximat Z thesizes, sample small(For
s.difference zero-non ofnumber the torefersn following, In the
exam1 exam2diff
(ex2-ex1)rank (+)
rank (-)
exam1 exam2diff
(ex2-ex1)rank (+)
rank (-)
95 97 72 68
76 76 78 94
82 75 58 55
48 54 73 75
27 31 71 70
34 39 69 66
58 61 57 62
98 97 84 92
45 45 91 81
77 94 83 90
27 36 67 73
Example
Suppose we have a class with 22 students, each of whom has two exam grades.
We want to test at the 5% level whether there is a difference in the median grade for the two exams.
exam1 exam2diff
(ex2-ex1)rank (+)
rank (-)
exam1 exam2diff
(ex2-ex1)rank (+)
rank (-)
95 97 2 72 68 -4
76 76 0 78 94 16
82 75 -7 58 55 -3
48 54 6 73 75 2
27 31 4 71 70 -1
34 39 5 69 66 -3
58 61 3 57 62 5
98 97 -1 84 92 8
45 45 0 91 81 -10
77 94 17 83 90 7
27 36 9 67 73 6
We calculate the difference between the
exam grades: diff = exam2 – exam 1.
exam1 exam2diff
(ex2-ex1)rank (+)
rank (-)
exam1 exam2diff
(ex2-ex1)rank (+)
rank (-)
95 97 2 72 68 -4
76 76 0 78 94 16
82 75 -7 58 55 -3
48 54 6 73 75 2
27 31 4 71 70 -1 1.5
34 39 5 69 66 -3
58 61 3 57 62 5
98 97 -1 1.5 84 92 8
45 45 0 91 81 -10
77 94 17 83 90 7
27 36 9 67 73 6
Then we rank the absolute values of the differences from smallest to largest, omitting the two zero differences.
The smallest non-zero |differences| are the two |-1|’s. Since they are tied for ranks 1 and 2, we rank them both 1.5.
Since the differences were negative, we put the ranks in the negative column.
exam1 exam2diff
(ex2-ex1)rank (+)
rank (-)
exam1 exam2diff
(ex2-ex1)rank (+)
rank (-)
95 97 2 3.5 72 68 -4
76 76 0 78 94 16
82 75 -7 58 55 -3
48 54 6 73 75 2 3.5
27 31 4 71 70 -1 1.5
34 39 5 69 66 -3
58 61 3 57 62 5
98 97 -1 1.5 84 92 8
45 45 0 91 81 -10
77 94 17 83 90 7
27 36 9 67 73 6
The next smallest non-zero |differences| are the two |2|’s. Since they are tied for ranks 3 and 4, we rank them both 3.5.
Since the differences were positive, we put the ranks in the positive column.
exam1 exam2diff
(ex2-ex1)rank (+)
rank (-)
exam1 exam2diff
(ex2-ex1)rank (+)
rank (-)
95 97 2 3.5 72 68 -4
76 76 0 78 94 16
82 75 -7 58 55 -3 6
48 54 6 73 75 2 3.5
27 31 4 71 70 -1 1.5
34 39 5 69 66 -3 6
58 61 3 6 57 62 5
98 97 -1 1.5 84 92 8
45 45 0 91 81 -10
77 94 17 83 90 7
27 36 9 67 73 6
The next smallest non-zero |differences| are the two |-3|’s and the |3|. Since they are tied for ranks 5, 6, and 7, we rank them all 6.
Then we put the ranks in the appropriately signed columns.
exam1 exam2diff
(ex2-ex1)rank (+)
rank (-)
exam1 exam2diff
(ex2-ex1)rank (+)
rank (-)
95 97 2 3.5 72 68 -4 8.5
76 76 0 78 94 16 19
82 75 -7 14.5 58 55 -3 6
48 54 6 12.5 73 75 2 3.5
27 31 4 8.5 71 70 -1 1.5
34 39 5 10.5 69 66 -3 6
58 61 3 6 57 62 5 10.5
98 97 -1 1.5 84 92 8 16
45 45 0 91 81 -10 18
77 94 17 20 83 90 7 14.5
27 36 9 17 67 73 6 12.5
We continue until we have ranked all the non-zero |differences| .
exam1 exam2diff
(ex2-ex1)rank (+)
rank (-)
exam1 exam2diff
(ex2-ex1)rank (+)
rank (-)
95 97 2 3.5 72 68 -4 8.5
76 76 0 78 94 16 19
82 75 -7 14.5 58 55 -3 6
48 54 6 12.5 73 75 2 3.5
27 31 4 8.5 71 70 -1 1.5
34 39 5 10.5 69 66 -3 6
58 61 3 6 57 62 5 10.5
98 97 -1 1.5 84 92 8 16
45 45 0 91 81 -10 18
77 94 17 20 83 90 7 14.5
27 36 9 17 67 73 6 12.5
154 56
Then we total the signed ranks. We get 154 for the sum of the positive ranks and 56 for the sum of the negative ranks.
The Wilcoxon test statistic is the sum of the positive ranks. So W = 154.
Since we had 22 students and 2 zero differences, the number of non-zero differences n = 20.
is W ofmean that theRecall 1054
)21)(20(
4
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is W ofdeviation standard The
786.2624
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24
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nnnW
:have weSo 829.1786.26
105154
W
WW
Z
-1.96 0 1.96 Z
.475.475.025 .025
critical region
critical region
Since the critical values for a 2-tailed Z test at the 5% level are 1.96 and -1.96, we can not reject the null hypothesis H0 and so we conclude that the medians are the same.
For small sample sizes, you can use Table 12.19 in the online material associated with section 12.8 of your textbook, which provides the lower and upper critical values for the Wilcoxon Signed Rank Test.
This table is shown on the next slide.
Lower & Upper Critical Values, W, of Wilcoxon Signed Ranks Test
ONE-TAIL α = 0.05 α = 0.025 α = 0.01 α = 0.005TWO-TAIL α = 0.10 α = 0.05 α = 0.02 α = 0.01
n (Lower, Upper)5 0,15 —,— —,— —,—6 2,19 0,21 —,— —,—7 3,25 2,26 0,28 —,—8 5,31 3,33 1,35 0,369 8,37 5,40 3,42 1,44
10 10,45 8,47 5,50 3,5211 13,53 10,56 7,59 5,6112 17,61 13,65 10,68 7,7113 21,70 17,74 12,79 10,8114 25,80 21,84 16,89 13,9215 30,90 25,95 19,101 16,10416 35,101 29,107 23,113 19,11717 41,112 34,119 27,126 23,13018 47,124 40,131 32,139 27,14419 53,137 46,144 37,153 32,15820 60,150 52,158 43,167 37,173
Recall that we have 20 non-zero differences and are performing a 5% 2-tailed test.Here we see that the lower critical value is 52 and the upper critical value is 158.Our statistic W, the sum of the positive ranks, is 154, which is inside the interval (52, 158) indicated in the table. So we can not reject the null hypothesis and we conclude that the medians are the same.
The Kruskal-Wallis Test
This test is used to test whether several populations have the same median.
It is a nonparametric substitute for a one-factor ANOVA F-test.
, 1)3(n - n
R
1)n(n
12 K is statistic test The
j
2j
where nj is the number of observations in the jth sample,n is the total number of observations, and Rj is the sum of ranks for the jth sample.
groups. sample ofnumber theis c where 1,-c dof with isK ofon distributi then the
true,is hypothesis null theand 5neach If2
j
In the case of ties, a corrected statistic should be computed:
nn
)t(t-1
KK
3j
3j
c where tj is the number of ties in the jth sample.
Kruskal-Wallis Test Example: Test at the 5% level whether average employee performance is the same at 3 firms, using the following standardized test scores for 20 employees.
Firm 1 Firm 2 Firm 3
score rank score rank score rank
78 68 82
95 77 65
85 84 50
87 61 93
75 62 70
90 72 60
80 73
n1 = 7 n2 = 6 n3 =7
We rank all the scores. Then we sum the ranks for each firm.
Then we calculate the K statistic.
Firm 1 Firm 2 Firm 3
score rank score rank score rank
78 12 68 6 82 14
95 20 77 11 65 5
85 16 84 15 50 1
87 17 61 3 93 19
75 10 62 4 70 7
90 18 72 8 60 2
80 13 73 9
n1 = 7 R1 = 106 n2 = 6 R2 = 47 n3 =7 R3 = 57
1)3(n - n
R
1)n(n
12 K
j
2j
6.6413(21) -
7
57
6
47
7
106
20(21)
12
222
f(2)
acceptance region
crit. reg.
.05
5.99122
From the 2 table, we see that the 5% critical value for a 2 with 2 dof is 5.991.
Since our value for K was 6.641, we reject H0 that the medians are the same and accept H1 that the medians are different.
One sample test of runs
a test for randomness of order of occurrence
A run is a sequence of identical occurrences that are followed and preceded by different occurrences.
Example: The list of X’s & O’s below consists of 7 runs.
x x x o o o o x x o o o o x x x x o o x
Suppose r is the number of runs, n1 is the number of type 1 occurrences and n2 is the number of type 2 occurrences.
. 1nn
n2nμ
is runs ofnumber mean The
21
21r
. 1)n(n)n(n
)n-n-n(2nn2n
is runs ofnumber theofdeviation standard The
212
21
212121r
If n1 and n2 are each at least 10, then r is approximately normal.
variable.normal standard a is
-r
Z So,r
r
Example: A stock exhibits the following price increase (+) and decrease () behavior over 25 business days. Test at the 1% whether the pattern is random.
+ + + + + + + + + + + + + r =16, n1 (+) = 13, n2 () = 12
-r
Zr
r
1nn
n2nμ
21
21r
1)n(n)n(n
)n-n-n(2nn2n
212
21
212121r
11213
2(13)(12)
48.13
1)21(13)21(13
12]-13-)[(2(13)(12 2(13)(12)2
44.2
03.12.44
13.48 - 16
Since the critical values for a 2-tailed 1% test are 2.575 and -2.575, we accept H0 that the pattern is random.
-2.575 0 2.575
Z
critical region.005
critical region.005acceptance
region
.495.495