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Index to Module 4 Notes
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Recall in Module Notes 4.3 we
introduced the technique for comparing means of multiple samples using the
One-Factor ANOVA model. The assumptions required for this model are normality
and homogeneity of variance. When the assumptions of normality and homogeneity
of variance are not met, especially with small samples, and/or especially when
there is not an equal sample size in each group, a nonparametric alternative to
the Single or One-Factor ANOVA is the Kruskal-Wallis Rank Test. This test is
similar to the nonparametric test used as an alternative to the two independent
group t-Test presented in Module Notes 4.3 in that it uses ranks of the
observations rather than the observations themselves.
If nothing is known about the variability of the shape of the distributions
from which the multiple group samples are drawn, then the hypotheses statements
are (Aczel, p. 656):
H0:
All of the multiple populations have the same distribution
Ha: Not all of the multiple populations have the same distribution
One of the reference texts (Levine) notes that if the populations are assumed to have the same variability and shape, then location hypotheses can be stated as:
H0:
All of the population medians are equal
Ha: At least two of the medians are not equal
In
either case, we also assume that the samples are randomly and independently
drawn from the respective populations.
Let's demonstrate the Kruskal-Wallis Rank Test with the miles per gallon
example used in Module Notes 4.3. Worksheet 4.6.1 provides the sorted mpg data
for the seven automobiles randomly and independently assigned to Brand of gas
Groups A, B and C.
Worksheet 4.6.1
|
Brand A |
Rank A |
Brand B |
Rank B |
Brand C |
Rank C |
|
14 |
1 |
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15 |
2 |
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|
16 |
3 |
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|
17 |
4 |
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|
18 |
5 |
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18.5 |
6 |
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|
19 |
8 |
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|
19 |
8 |
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|
19 |
8 |
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|
20 |
12 |
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|
20 |
12 |
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|
20 |
12 |
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|
20 |
12 |
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20 |
12 |
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20.5 |
15 |
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23 |
17 |
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|
23 |
17 |
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|
23 |
17 |
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|
24 |
19 |
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|
25 |
20 |
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|
26 |
21 |
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|
Rank Sum |
42 |
66 |
123 |
||
|
TA |
TB |
TC |
Worksheet 4.6.1 also contains the rankings for the observations within each
group, using the same ranking technique as in Module Notes 4.3.
A check on the rankings is provided by Equation 4.6.1. :
Eq. 4.6.1: TA + TB + TC = n ( n + 1 ) / 2
The
sum of the ranks is 231 (42 + 66 +123). The total sample size, n, is 21 so the
right hand side of Eq. 4.6.1 is 21 (21 + 1) / 2 which equals 231. Therefore,
the rankings check.
The Kruskal-Wallis Test Statistics is shown in Eq. 4.6.2:
Eq. 4.6.2: H = {12/[n(n+1)]} *
[TA2/n + TB2/n + TC2/n] - 3(n+1)
=
{12/[21(21+1)]} * [422/7 + 662/7 + 1232/7] - 3 (21 + 1) = 12.846
This
Test Statistic is compared to a critical or threshold value for the chi-square
Statistic. If the Test Statistic is greater than the chi-square critical value,
we reject the null hypothesis and conclude that the medians are not equal. I
will present and discuss the chi-square Distribution in Module 5. For now, we
will let Excel simply return the Chi Square Critical Value.
The Excel function for this is =CHIINV(alpha, degrees of freedom). We will use
alpha of 0.05, and number of groups being compared minus one for the degrees of
freedom. So, in an active cell in an Excel Worksheet, enter =CHIINV(0.05,3-1)
and Excel will return 5.99. Since 12.846 is greater than 5.99, we reject the
null hypothesis and conclude that the groups are different (the population distributions
are different or the medians of the three groups are not equal).
Now, what do we do after we reject the null hypothesis? That is, we know that
the three groups are not equal, but where do the differences lie? Recall in
Module Notes 4.4, we introduced the Bonferroni Post Hoc procedure to determine
which pairs of means were significantly different in single factor ANOVA.
That's what we can do with the following post hoc procedure to the Kruskal
Wallis Test.
Suppose we want to determine if the Brand A population was different from the
Brand C population. First, find the average ranks of the A and C groups:
Eq.
4.6.3: Mean RankA = TA/7 = 42 / 7 = 6
Eq. 4.6.4: Mean RankC = TC/7 = 123/7 = 17.6
Now find the absolute difference between the two mean ranks:
Eq. 4.6.5: Difference = | 17.6 - 6 | = 11.6
Now find the critical point for this paired difference. If the paired difference is greater than the critical point, reject the null hypothesis of equality and conclude that the Brand A population distribution is different than the Brand C population distribution. The equation for the critical point is:
Eq. 4.6.6: CKW = Sq Rt{Chi
Square*[n(n+1)/12]*[(1/nA)+(1/nB)]}
= Sq Rt{5.99*[21(21+1)/12]*[(1/7) + (1/7)]} = 8.11
Since 11.6 is greater than
8.11, we reject the null hypothesis and conclude that Brand A population
distribution is different than the Brand C population distribution. If we have
prior knowledge that the variability of the populations is the same, we may
extend this conclusion to state that the population medians are not the same.
This ends Module 4: a set of tools to use for the comparison of multiple
samples.
References:
Ken
Black. Business Statistics for Contemporary Decision Making. Fourth Edition,
Wiley. Chapter 10 & 11
D. Groebner, P. Shannon, P.
Fry & K. Smith. Business Statistics:
A Decision Making Approach, Fifth Edition, Prentice Hall,
Chapter 15
Aczel, A. (1993). Complete
Business Statistics (2nd. ed.). Homewood, IL: Irwin.
Levine, D., Berenson,
M. & Stephan, D. (1999). Statistics for Managers Using Microsoft Excel (2nd.
ed.). Upper Saddle River, NJ: Prentice-Hall, Chapter 10.
Mason, R., Lind, D. &
Marchal, W. (1999). Statistical Techniques in Business and Economics (10th.
ed.). Boston: Irwin McGraw Hill, Chapter 15.