Download Lecture 6 (May 14)

Ch10 Nonparametric Tests
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Outline
Introduction
The sign test
Rank-sum tests
Tests of randomness
The Kolmogorov-Smirnov and AndersonDarling Tests
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Introduction
Previous methods of inference that we have
studied are based on the assumption that the
observation come from normal population.
However, since there are many situation where it
is doubtful whether the assumption of normality
can be met.
Alternative techniques based on less stringent
assumptions – nonparametric tests.
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10.2 The sign test
The sign test: when we sample a
continuous symmetrical population, the
probability of getting a sample value less
than the mean and the probability of
getting a sample value greater than the
mean are both ½.
We can formulate the hypotheses in terms
of the population median.
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Conducting a sign test
The following data constitute a random sample
of 15 measurements of the octane rating a
certain kind of gasoline:
99.0 102.3 99.8 100.5 99.7 96.2 99.1 102.5 103.3
97.4 100.4 98.9 98.3 98 101.6
Test the null hypothesis   98.0 against the alternative
hypothesis   98.0 at the 0.01 level of significance.
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Solution
1. Null hypothesis:
  98.0 ( p  0.5)
Alternative hypothesis
  98.0 ( p  0.5)
2. Level of significance: 0.01
3. Criterion: based on the number of plus signs or
the number of minus signs. Using the number of
plus signs, denoted by x, reject the null
hypothesis if the probability of getting x or more
plus is less than or equal to 0.01.
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Solution
4. Calculation: replacing each value greater than
98.0 with a plus sign and each value less than
98.0 with a minus sign, the 14 sample values
yield
+++++-+++-++++
Thus x=12, and from the binomial distribution of n=14,
p=0.5, we get
P( X  12)  1  B(11;14, 0.5)  1  0.9935  0.0065
5. Since 0.0065 is less than 0.01, the null
hypothesis must be rejected. We conclude tha the
median exceeds 98.0.
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10.3 Rank-sum Tests
Rank sums: the U test and the H test.
The U test will be presented as a
nonparametric alternative to the twosample t test.
The H test will be presented as a
nonparametric alternative to the one-way
analysis of variance.
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The U test
The U test (also called Wilcoxon test or the
MannWhitney test)
Suppose in a study of sedimentary rocks,
the following diameters (in millimeters)
were obtained fro two kinds of sand:
sand I: 0.63 0.17 0.35 0.49 0.18 0.43 0.12 0.20
0.47 1.36 0.51 0.45 0.84 0.32 0.40
sand II: 1.13 0.54 0.96 0.26 0.39 0.88 0.92 0.53
1.01 0.48 0.89 1.07 1.11 0.58
The problem is to decide whether the two populations are the same of if
one is more likely to produce larger observations than the other. 9/23
Let X1 be a random variable having the first
distribution and X2 be a random variable having the
second distribution.
If P(a  X1 )  P(a  X 2 ) for all a, with strict inequality
for some a, we say that the second population is
stochastically larger than the first population.
The U Test: ranking the data jointly, as if they
comprise one sample, in an increasing order
of magnitude, and for our data we get
0.12 0.17 0.18 0.20 0.26 0.32 0.35 0.39 0.40 0.43
I
I
I
I
II I
I
II I
I
0.45 0.47 0.48 0.49 0.51 0.53 0.54 0.58 0.63 0.84
I
I
II
I
I
II II
II
I
I
0.88 0.89 0.92 0.96 1.01 1.07 1.11 1.13 1.36
II
II
II
II
II
II
II
II
I
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The values of the first sample (Sand I):
1, 2, 3, 4, 6, 7, 9, 10,11,12,14,15,19,20, and 29.
If there were tie among values, we would assign to
each of the tied observations the mean of the ranks
which they jointly occupy. For instance, the third and
the fourth are identical, we would assign each the rank
(3+4)/2 = 3.5.
The sums of the ranks are
Statistics:
W1  162, and W2  273
n1 (n1  1)
U1  W1 
2
n2 (n2  1)
U 2  W2 
2
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Under the null hypothesis that the two samples come
from identical populations, it can be shown that the
mean and the variance of the sampling distribution of
U1 are
U 
1
n1n2 2 n1n2 (n1  n2  1)
,  U1 
2
12
If there are ties in rank, these formulas provide only
approximations, but if the number of ties is small, these
approximations will generally be good.
Z
U1  U1
U
1
Is a random variable having approximately the standard
normal distribution.
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Solution of the example
1. Null hypothesis: Populations are identical.
Alternative hypothesis: The populations are not
identical.
2. Level of significance: 0.01
3. Criterion: Reject the null hypothesis if Z<-2.575 or
Z>2.575.
4. Calculations: since n1=15 and n2=14, we have
15 16
15 14
 42, U1 
 105,  U21  525
2
2
42  105
z
 2.75
525
U1  162 
5. The null hypothesis must be rejected. There is a
difference in the populations of grain size.
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The H test
(Kruskal-Wallis test)
The H test is a generalization of the U test in that it
enables us to test the null hypothesis that k independent
random samples come from identical populations.
If Ri is the sum of the ranks occupied by the ni
observations of the i-th sample, and n1  n2   nk  n
the test is based on the statistic
n
Ri2
12
H
 3(n  1)

n(n  1) i 1 ni
When ni  5, for all i
and the null hypothesis is
true, the sampling distribution of the H statistic is well
approximated by the chi-square distribution with k-1
degrees of freedom.
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EX.
An experiment designed to compare three preventive
methods against corrosion yielded the following
maximum depths of pits in pieces of wire subjected to
the respective treatments:
Method A: 77 54 67 74 71 66
Method B: 60 41 59 65 62 64 52
Method C: 49 52 69 47 56
Use the 0.05 level of significance to test the null hypothesis
that the three samples come from the identical
populations.
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Solution of the example
1. Null hypothesis: Populations are identical.
Alternative hypothesis: The populations are not
identical.
2. Level of significance: 0.05
3. Criterion: Reject the null hypothesis if H>5.991
4. Calculations:
R1  84, R2  55.5, R3  31.5
12 842 55.52 31.52
H
(


)  3 19  6.7
18 19 6
7
5
5. The null hypothesis must be rejected.
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6. The P-value is 1-0.9649=0.0351 < 0.05
10.4 Tests of Randomness
Remind that in Ch6, we have discussed
some assurance that a sample taken will
be random.
Provide a technique for testing whether a
sample may be looked upon as random
after it has actually been obtained.
It is based on the number of runs exhibited
in the sample results. EX. 8 runs
TT HH TT HHH T HHH TTTT HHH
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If a sequence contains n1 symbols of one kind and n2 of
another kind (and neither n1 nor n2 is less than 10), the
sampling distribution of the total number of runs, u, can
be approximated closely by a normal distribution with
2n1n2
2n1n2 (2n1n2  n1  n2 )
u 
 1, and  u 
n1  n2
(n1  n2 )2 (n1  n2  1)
Thus the test of the null hypothesis is that the
arrangement of the symbols is random can be based
on the statistic
Z
u  u
u
which has approximately the standard normal
distribution.
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EX
The following is the arrangement of
defective, d, and nondefective, n, pieces
produced in the given order by a certain
machine:
nnnnn dddd nnnnnnnnnn dd nn dddd
Test for randomness at the 0.01 level of
significance.
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Solution of the example
1. Null hypothesis: Arrangement is random.
Alternative hypothesis: Arrangement is not random.
2. Level of significance: 0.01
3. Criterion: Reject the null hypothesis if Z<-2.575 or
Z>2.575.
4. Calculations: since n1=10, n2=17, and u=6, we have
u 
z
2 10 17
2 10 17(2 10 17  10  17)
 1  13.59,  u 
 2.37
10  17
(10  17) 2 (10  17  1)
6  13.59
 3.20
2.37
5. The null hypothesis must be rejected. The arrange is
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not random.
10.5 The Kolmogorov-Smirnov and
Anderson-Darling Tests
The Kolmogorov-Smirnov tests are nonparametric
tests for differences between cumulative distributions.
The Kolmogorov-Smirnov one-sample test is generally
more efficient than the chi-square tests for goodness
of fit for small samples, and it can be used for very
small samples where the chi-square test does not
apply.
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Solution of the example
1. Null hypothesis:
0
x

F ( x)  
 30
1
for x  0
for 0  x  30
for x  30
Alternative hypothesis: are not uniformly distributed.
2. Level of significance: 0.05
3. Criterion: Reject the null hypothesis if D>0.410,
where D is the maximum difference between the
empirical cumulative distribution and the cumulative
distribution assumed under the null hypothesis.
4. Calculations: The difference is greatest at x=6.2
D  0.4 
6.2
 0.193
30
5. The null hypothesis cannot be rejected.
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Anderson-Darling test
Difference in the tails can be easier to
detect if the difference between the
empirical cumulative distribution Fn and F
is divided by. In particular it is based on
the statistic

A   [ Fn ( x)  F ( x)]2
2

1
f ( x)dx
F ( x)(1  F ( x))
or
n
A2 
[ (2i  1)(ln(ui )  ln(1  un 1i ))]
i 1
n
n
where
ui  F ( xi )
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