Ch10 Nonparametric Tests

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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

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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 against
  the alternative hypothesis at the
  0.01 level of significance.

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Solution

• 1. Null hypothesis

Alternative hypothesis
2. Level of significance 0.01

• 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

• 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 x12, and from the binomial distribution of
  n14, p0.5, we get

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.

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Let X1 be a random variable having the first
distribution and X2 be a random variable having
the second distribution.
If 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 (34)/2 3.5. The sums of
the ranks are
Statistics
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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
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.
Is a random variable having approximately the
standard normal distribution.
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Solution of the example

• 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
Zlt-2.575 or Zgt2.575.
4. Calculations since n115 and n214, we have
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
• the test is based on the statistic

When 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

• 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
Hgt5.991
4. Calculations
5. The null hypothesis must be rejected.
6. The P-value is 1-0.96490.0351 lt 0.05
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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

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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
Thus the test of the null hypothesis is that the
arrangement of the symbols is random can be based
on the statistic
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

• Null hypothesis Arrangement is random.
• Alternative hypothesis Arrangement is not
  random.

2. Level of significance 0.01
3. Criterion Reject the null hypothesis if
Zlt-2.575 or Zgt2.575.
4. Calculations since n110, n217, and u6, we
have
5. The null hypothesis must be rejected. The
arrange is not random.
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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

• Null hypothesis
• Alternative hypothesis are not uniformly
  distributed.

2. Level of significance 0.05
3. Criterion Reject the null hypothesis if
Dgt0.410, 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
x6.2
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

or
where