Transcript Document
Ch10 Nonparametric Tests
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Outline
Introduction The sign test Rank-sum tests Tests of randomness The Kolmogorov-Smirnov and Anderson Darling Tests 2/23
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
98.0
0.01
level of significance. 5/23
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
12)
B
(11;14, 0.5) 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
two sample 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 X 1 be a random variable having the first distribution and X 2 be a random variable having the second distribution. If
X
1 )
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 10/23
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
n n
1 ( 1
W
1 162,
and W
2 1) Statistics :
U
1
W
1 2
U
2
W
2
n n
2 ( 2 2 1) 273 11/23
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 U 1 are
U
1
n n
1 2 , 2
U
1 2
n n n
1 2 ( 1
n
2 1) 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
U
1
U
1
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 n 1 =15 and n 2 =14, we have
U
1 162
z
525 42, 2 2.75
U
1 2 105, 2
U
1 525 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
R i
is the sum of the ranks occupied by the
n i
observations of the i-th sample, and
n
1
n
2 the test is based on the statistic
H
12 1)
i n
1
R i
2
n i
3(
n
1)
n k
n
true, the sampling distribution of the H statistic is well approximated by the
chi-square
distribution with
k-1
degrees of freedom. 14/23
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:
R
1 84,
R
2 55.5,
R
3 31.5
H
12 ( 84 2 6 55.5
2 7 31.5
2 5 5. The null hypothesis must be rejected. 6. The P-value is 1-0.9649=0.0351 < 0.05
6.7
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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 17/23
If a sequence contains
n 1
symbols of one kind and
n 2
another kind (and neither
n 1
nor
n 2
of is less than 10), the sampling distribution of the total number of runs,
u
, can be approximated closely by a normal distribution with
u
2
n
1
n
2 1,
and
u
2 (
n
1 (2
n
2
n
1 2
n
2 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 n 1 =10, n 2 =17, and u=6, we have
u
u
2 2.37
z
3.20
2.37
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. 21/23
Solution of the example
1. Null hypothesis: 0
x
1 30
for x
0
for
0
for x
30 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
30 0.193
5. The null hypothesis cannot be rejected. 22/23
Anderson-Darling test
Difference in the tails can be easier to detect if the difference between the empirical cumulative distribution F n and F is divided by. In particular it is based on the statistic
A
2 [
n
F x
2 1 or
A
2 [
i n
1 (2
i
u i
n
u n i
))]
n
where
u i
F x i
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