Transcript Nonparametric Statistical Procedures

Nonparametric
Statistical Methods
Definition
When the data is generated from process
(model) that is known except for finite
number of unknown parameters the model is
called a parametric model.
Otherwise, the model is called a nonparametric model
Statistical techniques that assume a nonparametric model are called non-parametric.
For example
If you assume that your data has come from a
normal distribution with mean m and standard
deviation s (both unknown) then the data is
generated from process (model) that is known
except for two of parameters.(m and s )
The model is called a parametric model.
Models that do not assume normality (or
some other distribution with a finite no. of
paramters) are non-parametric
We will consider two nonparametric tests
1. The sign test
2. Wilcoxon’s signed rank test
These are tests for the central location of a
population.
They are alternatives to the z-test and the t-test
for the mean of a normal population
z
x  m0
s
n
t
x  m0
s
n
Nonparametric
Statistical Methods
Single sample nonparametric tests for central
location
1. The sign test
2. Wilcoxon’s signed rank test
These are tests for the central location of a
population.
They are alternatives to the z-test and the t-test
for the mean of a normal population
z
x  m0
s
n
t
x  m0
s
n
Both the z-test and the t-test assumes the data
is coming from a normal population
If the data is not coming from a normal
population, properties of the z-test and the ttest that require this assumption will no longer
be true.
The probability of a type I error may be
different than the desired value (0.05 or 0.01)
Single sample non parametric tests
If the data is not coming from a normal
population we should then use one of the two
nonparametric tests
1. The sign test
2. Wilcoxon’s signed tank test
These tests do not assume the data is coming
from a normal population
The sign test
A nonparametric test for the central
location of a distribution
We want to test:
H0: median = m0
against
HA: median ≠ m0
(or against a one-sided alternative)
The Sign test:
1. The test statistic:
S = the number of observations
that exceed m0
Comment: If H0: median = m0 is true
we would expect 50% of the
observations to be above m0, and 50%
of the observations to be below m0,
50%
0
0
50%
median = m0
If H 0 is true then S will have a binomial distribution
with p = 0.50, n = sample size.
m0 > median
p < 0.50
p
median
m0
If H 0 is not true then S will still have a binomial
distribution. However p will not be equal to 0.50.
m0 < median
p > 0.50
p
m0
median
p = the probability that an observation is greater
than m0.
Summarizing: If H 0 is true then S will have a
binomial distribution with p = 0.50, n = sample size.
n = 10
x
0
1
2
3
4
5
6
7
8
9
10
0.3
p(x)
0.0010
0.0098
0.0439
0.1172
0.2051
0.2461
0.2051
0.1172
0.0439
0.0098
0.0010
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
8
9
10
The critical and acceptance region:
n = 10
x
0
1
2
3
4
5
6
7
8
9
10
p(x)
0.0010
0.0098
0.0439
0.1172
0.2051
0.2461
0.2051
0.1172
0.0439
0.0098
0.0010
0.3000
0.2500
0.2000
0.1500
0.1000
0.0500
0.0000
0
1
2
3
4
5
6
7
8
9
Choose the critical region so that a is close to 0.05 or 0.01.
e. g. If critical region is {0,1,9,10} then a = .0010 + .0098 +
.0098 +.0010 = .0216
10
e. g. If critical region is {0,1,2,8,9,10} then a = .0010 + .0098
+.0439+.0439+ .0098 +.0010 = .1094
n = 10
x
0
1
2
3
4
5
6
7
8
9
10
p(x)
0.0010
0.0098
0.0439
0.1172
0.2051
0.2461
0.2051
0.1172
0.0439
0.0098
0.0010
0.3000
0.2500
0.2000
0.1500
0.1000
0.0500
0.0000
0
1
2
3
4
5
6
7
8
9
10
Example
Suppose that we are interested in determining
if a new drug is effective in reducing
cholesterol.
Hence we administer the drug to n = 10
patients with high cholesterol and measure the
reduction.
The data
Case
1
2
3
4
5
6
7
8
9
10
Initial
240
237
264
233
236
234
264
241
261
256
Cholesterol
Final
Reduction
228
12
222
15
262
2
224
9
240
-4
237
-3
264
0
219
22
252
9
254
2
Suppose we want to test
H0: the drug is not effective
median reduction ≤ 0
against
HA: the drug is effective
median reduction > 0
The Sign test
S = the no. of positive obs
The Sign test
The test statistic
S = the no. of positive obs = 8
We will use the p-value approach
p-value = P[S ≥ 8]
= 0.0439 + 0.0098 + 0.0010
= 0.0547
Since p-value > 0.05 we cannot reject H0
Summarizing: To carry out Sign Test
We
1. Compute S = The # of observations greater than m0
2. Let sobserved = the observed value of S.
3. Compute the p-value = P[S ≤ sobserved]
(2 P[S ≤ sobserved] for a two-tailed test).
Use the table for the binomial dist’n (p = ½ , n = sample size)
4.
Conclude HA (Reject H0) if p-value is less than 0.05
(or 0.01).
Sign Test for Large Samples
If n is large we can use the Normal approximation
to the Binomial.
Namely S has a Binomial distribution with p = ½
and n = sample size.
Hence for large n, S has approximately a Normal
distribution with
mean
and
n
m S  np 
2
standard deviation
n
 1  1 
s S  npq  n   
2
 2  2 
Hence for large n,use as the test statistic (in
place of S)
n
z
S  mS
sS

S
n
2
2
Choose the critical region for z from the
Standard Normal distribution.
i.e. Reject H0 if z < -za/2 or z > za/2
two tailed ( a one tailed test can also be set up.
Nonparametric
Confidence Intervals
Assume that the data, x1, x2, x3, … xn is a sample
from an unknown distribution.
Now arrange the data x1, x2, x3, … xn in increasing
order
x(1) < x(2) < x(3) < … < x(n)
Hence
x(1) = the smallest observation
x(2) = the 2nd smallest observation
x(n) = the largest observation
Consider the kth smallest observation and the kth
largest observation in the data x1, x2, x3, … xn
x(k) and x(n – k + 1)
Hence
P[x(k) < median < x(n – k + 1) ]
= P[at least k observations lie below the
median and at least k observations lie
above the median ]
If at least k observations lie below the median than
x(k) < median
If at least k observations lie above the median than
median < x(n – k + 1)
Thus
P[x(k) < median < x(n – k + 1) ]
= P[at least k observations lie below the
median and at least k observations lie
above the median ]
= P[The number of observations below the
median is at least k and at most n-k]
= P[k ≤ S ≤ n-k]
where S = the number of observations below the
median
S has a binomial distribution with
n = the sample size and p =1/2.
Hence
P[x(k) < median < x(n – k + 1) ]
= P[k ≤ S ≤ n-k]
= p(k) + p(k + 1) + … + p(n-k) = P
where p(i)’s are binomial probabilities with
n = the sample size and p =1/2.
This means that x(k) to x(n – k + 1) is a P100%
confidence interval for the median
Summarizing
x(k) to x(n – k + 1)
is a P100% confidence interval for the
median
where P = p(k) + p(k + 1) + … + p(n-k)
and p(i)’s are binomial probabilities with
n = the sample size and p =1/2.
Example: n = 10 and k =2
Binomial probabilities
x
0
1
2
3
4
5
6
7
8
9
10
p(x)
0.0010
0.0098
0.0439
0.1172
0.2051
0.2461
0.2051
0.1172
0.0439
0.0098
0.0010
0.3000
0.2500
0.2000
0.1500
0.1000
0.0500
0.0000
0
1
2
3
4
5
6
7
P = p(2) + p(3) + p(4) + p(5) + p(6) + p(7) + p(8)= .9784
Hence x(2) to x(9) is a 97.84% confidence interval for the
median
8
9
10
Example
Suppose that we are interested in determining
if a new drug is effective in reducing
cholesterol.
Hence we administer the drug to n = 10
patients with high cholesterol and measure the
reduction.
The data
Case
1
2
3
4
5
6
7
8
9
10
Initial
240
237
264
233
236
234
264
241
261
256
Cholesterol
Final
Reduction
228
12
222
15
262
2
224
9
240
-4
237
-3
264
0
219
22
252
9
254
2
The data arranged in order
k
x (k )
1
2
3
4
5
6
7
8
9
10
-4
-3
0
2
2
9
9
12
15
22
x(2) = -3 to x(9) =15 is a
97.84% confidence interval
for the median
Example
In the previous example to repeat the study
with n = 20 patients with high cholesterol.
The
data
Case
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Initial
231
241
270
264
256
240
269
242
269
230
234
266
253
267
248
238
266
266
244
249
Cholesterol
Final
Reduction
212
19
235
6
268
2
255
9
252
4
234
6
256
13
243
-1
264
5
221
9
220
14
267
-1
252
1
242
25
248
0
239
-1
268
-2
240
26
247
-3
250
-1
The binomial distribution with n = 20, p = 0.5
x
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
p (x )
0.0000
0.0000
0.0002
0.0011
0.0046
0.0148
0.0370
0.0739
0.1201
0.1602
0.1762
0.1602
0.1201
0.0739
0.0370
0.0148
0.0046
0.0011
0.0002
0.0000
0.0000
Note:
p(6) + p(7) + p(8) + p(9) + p(10) +
p(11) + p(12) + p(13) + p(14)
= 0.037 + 0.0739 + 0.1201 + 0.1602
+ 0.1762 + 0.1602 + 0.1201 + 0.0739
+ 0.037 = 0.9586
Hence
x(6) to x(15) is a 95.86% confidence
interval for the median reduction in
cholesterol
The data arranged in order
k
x (k )
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
-3
-2
-1
-1
-1
-1
0
1
2
4
5
6
6
9
9
13
14
19
25
26
x(6) = -1 to x(15) = 9 is a
95.86% confidence interval
for the median
For large values of n one can use the normal
approximation to the Binomial to find the value of k so
that x(k) to x(n – k + 1) is a 95% confidence interval for the
median.
i.e. we want to find k so that

n
n
k 
S 
2k  n 

2
2
PS  k   P 

  P Z 
  0.025
n 
n 

 n
 2
2 
but PZ  1.960  0.025
2k  n
n  1.96 n
hence
 1.960 or k 
2
n
n  1.96 n
Using k 
2
n
20
40
100
200
k
5.6
13.8
40.2
86.1
Next we will consider:
1. The Wilcoxon signed rank test
The Wilcoxon signed rank test is an
alternative to the Sign test, a test for the
central location of a single population
The sign test
A nonparametric test for the central
location of a distribution
We want to test:
H0: median = m0
against
HA: median ≠ m0
(or against a one-sided alternative)
The Sign test:
1. The test statistic:
S = the number of observations
that exceed m0
Comment: If H0: median = m0 is true
then
• The distribution of S is binomial
-
n = sample size,
p = 0.50
To carry out the The Sign test:
1. Compute the test statistic:
S = the number of observations
that exceed m0 = sobserved
2. Compute the p-value of test
statistic, sobserved :
p-value = P [S ≥ sobserved ]
( = 2 P [S ≥ sobserved ] for 2-tailed test)
where S is binomial, n = sample size, p = 0.50
3. Reject H0 if p-value low (< 0.05)
Non-parametric confidence intervals for the median
of a population
x(k) to x(n – k + 1)
is a (1 – a)100% = P100% confidence interval
for the median
where
x(k) = kth smallest xi and
x(n – k + 1) = kth largest xi
P = p(k) + p(k + 1) + … + p(n-k)
and p(i)’s are binomial probabilities with
n = the sample size and p =1/2.
The Wilcoxon Signed Rank Test
An Alternative to the sign test
Situation
• A sample of size n , (x1 , x2 , … , xn) from an
unknown distribution and we want to test
H0 : the centre of the distribution, m = m0 , against
HA : m ≠ m0 ,
• For the sign test we would count S, the number
of positive values of (x1 – m0 , x2 – m0 , … , xn –
m0).
• We would reject H0 if S was not close to n/2
• For Wicoxon’s signed-Rank test we would
assign ranks to the absolute values of (x1 – m0 , x2
– m0 , … , xn – m0).
• A rank of 1 to the value of xi – m0 which is
smallest in absolute value.
• A rank of n to the value of xi – m0 which is
largest in absolute value.
W+ = the sum of the ranks associated with
positive values of xi – m0.
W- = the sum of the ranks associated with negative
values of xi – m0.
Note: W+ + W- = 1 + 2+ 3+ …n = n(n + 1)/2
If H0 is true then W+ 
 W- 
 n(n + 1)/4
If H0 is not true then either
1. W+ will be small (W- large) or
2. W+ will be large (W- small)
m0 > True median
True
median
W- large
m0
W+ small
m0 < True median
m0
W- small
True
median
W+ large
Note: It is possible to work out the sampling
distribution of W+ ( and W-) when H0 is true.
Note: We use the fact that if H0 is true that there
is an equal probability (1/2) that the sign attached
to any rank is plus (+) or minus (-).
Example: n = 4.
ranks
1
2
3
4
W+
W-
Prob
-
-
-
-
0
10
1/16
+
-
-
-
1
9
1/16
-
+
-
-
2
8
1/16
-
-
+
-
3
7
1/16
-
-
-
+
4
6
1/16
+
+
-
-
3
7
1/16
+
-
+
-
4
6
1/16
+
-
-
+
5
5
1/16
-
+
+
-
5
5
1/16
-
+
-
+
6
4
1/16
-
-
+
+-
7
3
1/16
+
+
+
-
6
4
1/16
+
+
-
+
7
3
1/16
+
-
+
+
8
2
1/16
-
+
+
+
9
1
1/16
+
+
+
+
10
0
1/16
The distribution of W+ and W- : n = 4.
0
Prob
1/16
W0
Prob
1/16
1
2
3
4
1/16
1/16
2/16
2/16
1
2
3
4
1/16
1/16
2/16
2/16
5
6
7
2/16
2/16
216
5
6
7
2/16
2/16
216
8
9
10
1/16
1/16
1/16
8
9
10
1/16
1/16
1/16
W+
If T = W+ or W- : n = 4.
T
0
P[T = t]
1/16
P[T ≤ t]
0.0625
1
2
3
4
1/16
1/16
2/16
2/16
0.1250
0.1875
0.3125
0.4375
5
6
7
2/16
2/16
2/16
0.5625
0.6875
0.8125
8
9
10
1/16
1/16
1/16
0.8750
0.9375
1.000
These are the values
found in the table A.6
in the textbook
table A.6 Page A- 15
Distribution of the test statistic for Wilcoxon signed-rank test
Sample Size
T
2
3
4
5
6
1
0.5000
0.2500
0.1250
0.0625
0.0313
2
0.3750
0.1875
0.0938
0.0469
3
0.6250
0.3125
0.1563
0.0782
4
0.4375
0.2188
0.1094
5
0.5625
0.3125
0.1563
6
0.4063
0.2188
7
0.5000
0.2813
8
0.3438
9
0.4219
10
0.5000
table A.6 Page A- 15
Only goes up to n = sample size = 12
For sample sizes, n > 12 we can use the fact that T (W+ or W-)
has approximately a normal distribution with
nn  1
mean mT 
4
nn  12n  1
standarddeviation s T 
24
 T  mT t  mT 

t  mT 
and PT  t   P 

  P Z 

sT 
sT 
 sT

PT  t 
1
2
3
4
5
12
0.0005
0.0008
0.0013
0.0018
0.0025
0.0014
0.0019
0.0024
0.0030
0.0038
6
7
8
9
10
0.0035
0.0047
0.0062
0.0081
0.0105
0.0048
0.0060
0.0075
0.0093
0.0115
11
12
13
14
15
0.0135
0.0171
0.0213
0.0262
0.0320
0.0140
0.0171
0.0207
0.0249
0.0299
Exact Values

t  mT 
 P Z 

sT 

Normal
Approximation
Example
• In this example we are interested in the
quantity FVC (Forced Vital Capacity) in
patients with cystic fibrosis
• FVC (Forced Vital Capacity) = the volume of
air that a person can expel from the lungs in a
6 sec period.
• This will be reduced with time for cystic
fibrosis patients
The research question: Will this reduction be
less when a new experimental drug is
administered?
The Experimental Design
• The design will be a matched pair design
• Pairs of patients are matched (Using initial FVC
readings)
• One member of the pair is given the new drug the
other member is given a placebo
• We measure the reduction in FVC for each member
and compute the difference
– xi = Reduction in FVC (placebo) – Reduction in FVC
(drug)
These values will be generally positive if the drug is
effective in minimizing the deterioration in Forced
Vital Capacity (FVC).
W+ will be large (W- will be small)
Table: Reduction in forced vital capacity (FVC) for a
matched pair sample of patients with cystic fibrosis
Reduction in FVC
Subject
Placebo
Drug
xi
Difference
1
224
213
11
1
2
8
95
-15
2
3
75
33
42
3
3
4
541
440
101
4
4
5
74
-32
106
5
5
6
85
-28
113
6
6
7
293
445
-152
7
8
-23
-178
155
8
8
9
525
367
158
9
9
10
-38
140
-179
10
11
508
323
185
11
11
12
255
10
245
12
12
13
525
65
460
13
13
14
1023
343
680
14
14
Rank
Signed Rank
1
-2
-7
-10
W+ = 86
W- = 19
We have to judge if
W+ = 86 is large (or W- = 19 is small)
mW
sW


nn  1 1415 


 52.5
4
4
nn  12n  1
141529


 15.92953
24
24


W
 52.5 19  52.5 

P W  19  P 


 15.92953 15.92953
 PZ  2.10  0.0179 p - value


Since the p-value is small (< 0.05) we conclude the
drug is effective in reducing the deterioration of FVC
Summarizing: To carry out Wilcoxon’s signed rank test
We
1. Compute T = W+ or W- (usually it would be the smaller
of the two)
2. Let tobserved = the observed value of T.
3. Compute the p-value = P[T ≤ tobserved] (2 P[T ≤ tobserved]
for a two-tailed test).
i.
ii.
4.
For n ≤ 12 use the table.
For n > 12 use the Normal approximation.
Conclude HA (Reject H0) if p-value is less than 0.05 (or
0.01).
Alternative tests for this example
1.
The t – test
x  m0
x
test statistic t 

s
s
n
n
2.
The sign test
test statistic S  # of positivexi
Comments
The t – test
1.
i.
ii.
This test requires the assumption of normality.
If the data is not normally distributed the test is invalid
•
iii.
2.
The probability of a type I error may not be equal to its desired value
(0.05 or 0.01)
If the data is normally distributed, the t-test commits type II errors
with a smaller probability than any other test (In particular
Wilcoxon’s signed rank test or the sign test)
The sign test
i.
ii.
This test does not require the assumption of normality (true also for
Wilcoxon’s signed rank test).
This test ignores the magnitude of the observations completely.
Wilcoxon’s test takes the magnitude into account by ranking them
Nonparametric
Statistical Methods
Single sample nonparametric tests for central
location
1. The sign test
2. Wilcoxon’s signed rank test
These are tests for the central location of a
population.
They are alternatives to the z-test and the t-test
for the mean of a normal population
z
x  m0
s
n
t
x  m0
s
n
The Sign test
Summarizing: To carry out Sign Test
We
1. Compute S = The # of observations greater than m0
2. Let sobserved = the observed value of S.
3. Compute the p-value = P[S ≤ sobserved]
(2 P[S ≤ sobserved] for a two-tailed test).
Use the table for the binomial dist’n (p = ½ , n = sample size)
4.
Conclude HA (Reject H0) if p-value is less than 0.05
(or 0.01).
m0 = True median
True
median
S ≈ n/2
 m0
m0 > True median
True
median
m0
S small (close to 0)
m0 < True median
m0
True
median
S large (close to n)
Wilcoxon’s signed-Rank test
• For Wilcoxon’s signed-Rank test we would
assign ranks to the absolute values of (x1 – m0 , x2
– m0 , … , xn – m0).
• A rank of 1 to the value of xi – m0 which is
smallest in absolute value.
• A rank of n to the value of xi – m0 which is
largest in absolute value.
W+ = the sum of the ranks associated with
positive values of xi – m0.
W- = the sum of the ranks associated with negative
values of xi – m0.
Note: W+ + W- = 1 + 2+ 3+ …n = n(n + 1)/2
If H0 is true then W+ 
 W- 
 n(n + 1)/4
If H0 is not true then either
1. W+ will be small (W- large) or
2. W+ will be large (W- small)
m0 = True median
True
median
 m0
W+ ≈ W- ≈ n(n + 1)/4
m0 > True median
True
median
W- large
m0
W+ small
m0 > True median
True
median
W- large
m0
W+ small
m0 < True median
m0
W- small
True
median
W+ large
Summarizing: To carry out Wilcoxon’s signed rank test
We
1. Compute T = W+ or W- (usually it would be the smaller
of the two)
2. Let tobserved = the observed value of T.
3. Compute the p-value = P[T ≤ tobserved] (2 P[T ≤ tobserved]
for a two-tailed test).
i.
ii.
4.
For n ≤ 12 use the table.
For n > 12 use the Normal approximation.
Conclude HA (Reject H0) if p-value is less than 0.05 (or
0.01).
Two-sample – Non-parametic
tests
Mann-Whitney Test
A non-parametric two sample test for
comparison of central location
The Mann-Whitney Test
• This is a non parametric alternative to the two
sample t test (or z test) for independent samples.
• These tests (t and z) assume the data is normal
• The Mann- Whitney test does not make this
assumption.
• Sample of n from population 1
x1, x2, x3, … , xn
• Sample of m from population 2
y1, y2, y3, … , ym
The Mann-Whitney test statistic U1 counts the
number of times an observation in sample 1
precedes an observation in sample 2.
An Equivalent statistic U2 that counts the number
of times an observation in sample 2 precedes an
observation in sample 1 can also be computed
Example
n = m = 4 measurements of bacteria counts per
unit volume were made for two type of cultures.
The n = 4 measurements for culture 1 were
27, 31, 26, 25
The m = 4 measurements for culture 2 were
32, 29, 35, 28
To compute the Mann-Whitney test statistics U1 and
U2, arrange the observations from the two samples
combined in increasing order (retaining sample
membership).
25(1), 26(1), 27(1) , 28(2) , 29(2), 31(1), 32(2) , 35(2)
For each observation in sample 2 let ui demote the
number of observations in sample 1 that precede that
value.
u1 = 3, u2 = 3, u3 = 4, u4 = 4,
Then U1 = u1 + u2 + u3 + u4 = 3 + 3 + 4 + 4 =14
To compute U2, repeat the process for the second
sample
25(1), 26(1), 27(1) , 28(2) , 29(2), 31(1), 32(2) , 35(2)
For each observation in sample 1 let vi demote the
number of observations in sample 2 that precede that
value.
v1 = 0, v2 = 0, v3 = 0, v4 = 2,
Then U2 = v1 + v2 + v3 + v4 = 0 + 0 + 0 + 2 =2
Note: U1 + U2 = mn = 16.
This is true in general
For each pair (xi,yj) either xi < yj or xi > yj (Assume no
ties)
In one case U1 will be increased by 1 while in the other
case U2 will be increased by 1.
There are mn such pairs.
An Alternative way of o computing the Mann-Whitney
test statistic U
Arrange the observations from the two samples
combined in increasing order (retaining sample
membership) and assign ranks to the observations.
25(1) 26(1) 27(1) 28(2) 29(2) 31(1) 32(2) 35(2)
1
2
3
4
5
6
7
8
Let W1 = the sum of the ranks for sample 1.
= 1 + 2 + 3 + 6 = 12
Let W2 = the sum of the ranks for sample 2.
= 4 + 5 + 7 + 8 = 24
It can be shown that
n  n  1
U1  nm 
U 2  nm 
and
Note: W  W  1  2  3 
1
2
U1  U 2  2nm 

n  n  1
2
m  m  1
2
  n  m
m  m  1

 W2
n  m  n  m  1


2
 n  m  n  m  1
2
2
2
4nm  n  n  1  m  m  1   n  m  n  m  1
2
4nm  n  n  m  m  n2  2nm  m2  n  m
2


 W1
2

2
  nm
• The distribution function of U (U1 or U2) has
been tabled for various values of n and m (<n)
when the two observations are coming from
the same distribution.
• These tables can be used to set up critical
regions for the Mann-Whitney U test.
Example
A researcher was interested in comparing “brightness” of
paper prepared by two different processes
A measure of brightness in paper was made for n = m = 9
samples drawn randomly from each of the two processes. The
data is presented below:
Process A
Process B
6.1
9.1
9.2
8.2
8.7
8.6
8.9
6.9
7.6
7.5
7.1
7.9
9.5
8.3
8.3
7.8
9.0
8.9
Process A
xi
rank
6.1
1
9.2
17
8.7
12
8.9
13.5
7.6
5
7.1
3
9.5
18
8.3
9.5
9.0
15
Process B
yj
rank
9.1
16
8.2
8
8.6
11
6.9
2
7.5
4
7.9
7
8.3
9.5
7.8
6
8.9
13.5
Ranks averaged
because
observations are
tied
WA = 94
WB = 77
It can be shown that
U A  nm 
n  n  1
2
 WA   9  9  
9 10 
2
 94  32
and
U B  nm 
m  m  1
2
 WA   9  9  
9 10 
2
From table for n = m = 9
P Ui  18  .025
Hence we will reject H0 if either UA or UB ≤ 18.
Thus H0 is accepted
 77  49
The Mann-Whitney test for large
samples
For large samples (n > 10 and m >10) the
statistics U1 and U2 have approximately a
Normal distribution with mean and standard
nm
deviation mU 
i
2
sU 
i
nm  n  m  1
12
Thus we can convert Ui to a standard normal
statistic
nm
Ui 
U i  mUi
2
z

s Ui
nm  n  m  1
12
And reject H0 if z < -za/2 or z > za/2 (for a two
tailed test)
The Kruskal Wallis Test
• Comparing the central location for k populations
• An nonparametric alternative to the one-way
ANOVA F-test
Situation:
Data is collected from k populations.
The sample size from population i is ni.
The data from population i is:
xi1, xi 2 ,
, xini
i  1, 2, .k
The computation of The Kruskal-Wallis statistic
We group the N = n1 + n2 + … + nk observation from
k populations together and rank these observations
from 1 to N.
Let rij be the rank associated with with the observation
xij.
Handling of “tied” observations
If a group of observations are equal the ranks that
would have been assigned to those observations are
averaged
The computation of The Kruskal-Wallis statistic
Let
ri 
ri1  ri 2 
 rini
 the average rank for the ith sample
ni
r  the average rank for all the observations.
1 2  3 

N
N
N 1

2
Note: If the k populations do not differ in
central location the
r1 , r2 ,
, rk should be approximately equal
and close to r .
The Kruskal-Wallis statistic
k
12
2
K
ni  ri  r 

N  N  1 i 1
3  N  1
U
12



N  N  1 i 1 ni
 N  1
k
ni
where Ui   rij  ri1 
j 1
2
i
2
 rini
= the sum of the ranks for the ith
sample
The Kruskal-Wallis test
Reject
H0: the k populations have same
central location
if K  a with d. f .  k 1
2
Example
In this example we are measuring an enzyme
level in three groups of patients who have
received “open heart” surgery. The three groups
of patients differ in age:
1. Age 30 – 45
2. Age 46 – 60
3. Age 61+
The data
Age Group
30 - 45
46 - 60
218.3
166.1
140.8
124.1
268.4
84.3
201
124.2
70.2
70.7
61+
197.9
120.6
81.1
181.6
118.9
Computation of the Kruskal-Wallis
statistic
The raw data
Age Group
30 - 45
46 - 60
218.3
166.1
140.8
124.1
268.4
84.3
201
124.2
70.2
70.7
The data ranked
61+
197.9
120.6
81.1
181.6
118.9
ri
Ui
Age Group
30 - 45
46 - 60
14
10
9
7
15
4
13
8
1
2
12.75
5.33
51
32
Ui2 512 322 372



 1094.717

4
6
5
i 1 ni
3
61+
12
6
3
11
5
7.40
37
The Kruskal-Wallis statistic
3  N  1
U
12
K


N  N  1 i 1 ni
 N  1
k
2
i
2
3 16 
12

 7.698
1094.717  
15 14 
14
2
since K  
2
0.05
 5.99 for df  k 1  2
H0 is rejected.
There are significant differences in the central
enzyme levels between the three age groups