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Chapter
Chi-Square Tests
15
Chi-Square Test for Independence
Contingency Tables
Chi-Square Test for
Independence
Chi-Square Tests for Goodnessof-Fit
Uniform Goodness-of-Fit Test
Poisson Goodness-of-Fit Test
Normal Chi-Square Goodnessof-Fit Test
ECDF Tests (Optional)
McGraw-Hill/Irwin
Copyright © 2009 by The McGraw-Hill Companies, Inc.
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15-2
Chi-Square Test for Independence
Chi-Square Test for Independence
Contingency Tables
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Chi-Square Test
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For example:
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Table 15.1
15-3
A contingency table is a cross-tabulation of n
paired observations into categories.
Each cell shows the count of observations that
fall into the
B
category
A
defined by its
row (r) and
column (c)
heading.
15-4
In a test of independence for an r x c
contingency table, the hypotheses are
H0: Variable A is independent of variable B
H1: Variable A is not independent of variable B
Use the chi-square test for independence to
test these hypotheses.
This non-parametric test is based on
frequencies.
The n data pairs are classified into c columns
and r rows and then the observed frequency fjk
is compared with the expected frequency ejk.
Chi-Square Test for Independence
Chi-Square Test for Independence
Chi-Square Distribution
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Chi-Square Distribution
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The critical value comes from the chi-square
probability distribution with ν degrees of
freedom.
ν = degrees of freedom = (r – 1)(c – 1)
where r = number of rows in the table
c = number of columns in the table
Appendix E contains critical values for righttail areas of the chi-square distribution.
The mean of a chi-square distribution is ν with
variance 2ν.
15-5
Figure 15.1
15-6
Chi-Square Test for Independence
Chi-Square Test for Independence
Expected Frequencies
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15-7
Consider the shape of the chi-square
distribution:
Steps in Testing the Hypotheses
Assuming that H0 is true, the expected
frequency of row j and column k is:
ejk = RjCk/n
where Rj = total for row j (j = 1, 2, …, r)
Ck = total for column k (k = 1, 2, …, c)
n = sample size
15-8
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Step 1: State the Hypotheses
H0: Variable A is independent of variable B
H1: Variable A is not independent of variable B
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Step 2: Specify the Decision Rule
Calculate ν = (r – 1)(c – 1)
For a given α, look up the right-tail critical
value (χ2R) from Appendix E or by using Excel.
Reject H0 if χ2R > test statistic.
Chi-Square Test for Independence
Steps in Testing the Hypotheses
Chi-Square Test for Independence
Steps in Testing the Hypotheses
For example, for ν = 6 and α = .05, χ2.05 = 12.59.
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Here is the rejection region.
Figure 15.2
15-9
Figure 15.3
15-10
Chi-Square Test for Independence
Steps in Testing the Hypotheses
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Chi-Square Test for Independence
Steps in Testing the Hypotheses
Step 3: Calculate the Expected Frequencies
ejk = RjCk/n
For example,
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Step 4: Calculate the Test Statistic
The chi-square test statistic is
calc
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15-11
15-12
Step 5: Make the Decision
Reject H0 if χ2R > test statistic or if the
p-value < α.
Chi-Square Test for Independence
Small Expected Frequencies
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Chi-Square Test for Independence
Cross-Tabulating Raw Data
The chi-square test is unreliable if the
expected frequencies are too small.
Rules of thumb:
• Cochran’s Rule requires that ejk > 5 for all
cells.
• Up to 20% of the cells may have ejk < 5
Most agree that a chi-square test is infeasible
if ejk < 1 in any cell.
If this happens, try combining adjacent rows or
columns to enlarge the expected frequencies.
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Chi-square tests for independence can also be
used to analyze quantitative variables by
coding them into categories.
For example, the
variables Infant
Deaths per 1,000
and Doctors per
100,000 can
each be coded
into various
categories:
Figure 15.6
15-13
15-14
Chi-Square Test for Goodness-ofFit
Why Do a Chi-Square Test on Numerical
Data?
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15-15
Chi-Square Test for Independence
3-Way Tables and Higher
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The researcher may believe there’s a
relationship between X and Y, but doesn’t
want to use regression.
There are outliers or anomalies that prevent
us from assuming that the data came from
a normal population.
The researcher has numerical data for one
variable but not the other.
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15-16
More than two variables can be compared
using contingency tables.
However, it is difficult to visualize a higher
order table.
For example, you could visualize a cube as
a stack of tiled 2-way contingency tables.
Major computer packages permit 3-way
tables.
Chi-Square Test for Goodness-ofFit
Purpose of the Test
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Chi-Square Test for Goodness-ofFit
Multinomial GOF Test
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The goodness-of-fit (GOF) test helps you
decide whether your sample resembles a
particular kind of population.
The chi-square test will be used because
it is versatile and easy to understand.
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A multinomial distribution is defined by any k
probabilities π1, π2, …, πk that sum to unity.
For example, consider the following “official”
proportions of M&M colors.
calc
15-17
15-18
Chi-Square Test for Goodness-ofFit
Chi-Square Test for Goodness-ofFit
Multinomial GOF Test
Hypotheses for GOF
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The hypotheses are
H0: π1 = .30, π2 = .20, π3 = .10, π4 = .10, π5 = .10, π6 = .20
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15-19
The hypotheses are:
H0: The population follows a _____ distribution
H1: The population does not follow a ______
distribution
H1: At least one of the πj differs from the
hypothesized value
No parameters are estimated (m = 0) and there
are c = 6 classes, so the degrees of freedom are
ν=c–m–1=6–0-1
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15-20
The blank may contain the name of any
theoretical distribution (e.g., uniform, Poisson,
normal).
Chi-Square Test for Goodness-ofFit
Chi-Square Test for Goodness-ofFit
Test Statistic and Degrees of Freedom for
GOF
Test Statistic and Degrees of Freedom for
GOF
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Assuming n observations, the observations
are grouped into c classes and then the chisquare test statistic is found using:
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calc
where
fj = the observed frequency of
observations in class j
ej = the expected frequency in class j if
H0 were true
15-21
15-22
Chi-Square Test for Goodness-ofFit
Chi-Square Test for Goodness-ofFit
Test Statistic and Degrees of Freedom for
GOF
Data-Generating Situations
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v = c − m = c − 0 −1 = c −1
v = c − m = c −1−1 = c − 2
Instead of “fishing” for a good-fitting
model, visualize a priori the characteristics
of the underlying data-generating process.
Mixtures: A Problem
v = c − m = c − 2 −1 = c − 3
15-23
If the proposed distribution gives a good fit
to the sample, the test statistic will be near
zero.
The test statistic follows the chi-square
distribution with degrees of freedom
ν=c–m–1
where c is the no. of classes used in the
test m is the no. of parameters estimated
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15-24
Mixtures occur when more than one datagenerating process is superimposed on top
of one another.
Chi-Square Test for Goodness-ofFit
Eyeball Tests
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Uniform Goodness-of-Fit Test
Uniform Distribution
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A simple “eyeball” inspection of the
histogram or dot plot may suffice to rule
out a hypothesized population.
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Small Expected Frequencies
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Goodness-of-fit tests may lack power in
small samples. As a guideline, a chisquare goodness-of-fit test should be
avoided if n < 25.
15-25
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15-26
Uniform Goodness-of-Fit Test
Uniform GOF Test: Grouped Data
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15-27
The uniform goodness-of-fit test is a special
case of the multinomial in which every value
has the same chance of occurrence.
The chi-square test for a uniform distribution
compares all c groups simultaneously.
The hypotheses are:
H0: π1 = π2 = …, πc = 1/c
H1: Not all πj are equal
Uniform Goodness-of-Fit Test
Uniform GOF Test: Raw Data
The test can be performed on data that are
already tabulated into groups.
Calculate the expected frequency ej for each
cell.
The degrees of freedom are ν = c – 1 since there
are no parameters for the uniform distribution.
Obtain the critical value χ2α from Appendix E for
the desired level of significance α.
The p-value can be obtained from Excel.
Reject H0 if p-value < α.
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15-28
First form c bins of equal width and create a
frequency distribution.
Calculate the observed frequency fj for each bin.
Define ej = n/c.
Perform the chi-square calculations.
The degrees of freedom are ν = c – 1 since there
are no parameters for the uniform distribution.
Obtain the critical value from Appendix E for a
given significance level α and make the
decision.
Uniform Goodness-of-Fit Test
Uniform GOF Test: Raw Data
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Uniform Goodness-of-Fit Test
Uniform GOF Test: Raw Data
Maximize the test’s power by defining bin
width as
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As a result, the expected frequencies will
be as large as possible.
15-29
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15-30
Poisson Goodness-of-Fit Test
Poisson Data-Generating Situations
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15-31
Calculate the mean and standard deviation of
the uniform distribution as:
µ = (a + b)/2
σ=
[(b – a + 1)2 – 1)/12
If the data are not skewed and the sample size
is large (n > 30), then the mean is
approximately normally distributed.
So, test the hypothesized uniform mean using
Poisson Goodness-of-Fit Test
Poisson Goodness-of-Fit Test
In a Poisson distribution model, X
represents the number of events per unit of
time or space.
X is a discrete nonnegative integer (X = 0, 1,
2, …)
Event arrivals must be independent of each
other.
Sometimes called a model of rare events
because X typically has a small mean.
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15-32
The mean λ is the only parameter.
Assuming that λ is unknown and must be
estimated from the sample, the steps are:
Step 1: Tally the observed frequency fj of
each X-value.
Step 2: Estimate the mean λ from the
sample.
Step 3: Use the estimated λ to find the
Poisson probability P(X) for each value of X.
Poisson Goodness-of-Fit Test
Poisson Goodness-of-Fit Test
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Poisson Goodness-of-Fit Test
Poisson GOF Test: Tabulated Data
Step 4: Multiply P(X) by the sample size n
to get expected Poisson frequencies ej.
Step 5: Perform the chi-square
calculations.
Step 6: Make the decision.
You may need to combine classes until
expected frequencies become large enough
for the test (at least until ej > 2).
15-33
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c
^
λ=
•
Σx f
j =1 j j
n
Using this estimate mean, calculate the
Poisson probabilities either by using the
Poisson formula
P(x) = (λxe-λ)/x! or Excel.
15-34
Normal Chi-Square
Goodness-of-Fit Test
Poisson Goodness-of-Fit Test
Poisson GOF Test: Tabulated Data
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15-35
Calculate the sample mean as:
Normal Data Generating Situations
For c classes with m = 1 parameter
estimated, the degrees of freedom are
ν=c–m–1
Obtain the critical value for a given α from
Appendix E.
Make the decision.
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15-36
Two parameters, µ and σ, fully describe the
normal distribution.
Unless µ and σ are know a priori, they must
be estimated from a sample by using x and s.
Using these statistics, the chi-square
goodness-of-fit test can be used.
Normal Chi-Square
Goodness-of-Fit Test
Normal Chi-Square
Goodness-of-Fit Test
Method 1: Standardizing the Data
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Transform the sample observations x1, x2,
…, xn into standardized values.
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Count the sample observations fj within
intervals of the form x + ks and compare
them with the known frequencies ej based
on the normal distribution.
Method 1: Standardizing the Data
Advantage is a
standardized
scale.
Disadvantage is
that data are no
longer in the
original units.
Figure 15.14
15-37
15-38
Normal Chi-Square
Goodness-of-Fit Test
Method 2: Equal Bin Widths
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Normal Chi-Square
Goodness-of-Fit Test
Method 2: Equal Bin Widths
To obtain equal-width bins, divide the exact
data range into c groups of equal width.
Step 1: Count the sample observations in
each bin to get observed frequencies fj.
Step 2: Convert the bin limits into
standardized z-values by using the formula.
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15-40
Step 3: Find the normal area within each
bin assuming a normal distribution.
Step 4: Find expected frequencies ej by
multiplying each normal area by the
sample size n.
Classes may need to be collapsed from the
ends inward to enlarge expected
frequencies.
Normal Chi-Square
Goodness-of-Fit Test
Normal Chi-Square
Goodness-of-Fit Test
Method 3: Equal Expected Frequencies
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Method 3: Equal Expected Frequencies
Define histogram bins in such a way that an
equal number of observations would be
expected within each bin under the null
hypothesis.
Define bin limits so that ej = n/c
A normal area of 1/c in each of the c bins is
desired.
The first and last classes must be open-ended
for a normal distribution, so to define c bins,
we need c – 1 cutpoints.
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The upper limit of bin j can be found
directly by using Excel.
Alternatively, find zj for bin j using Excel
and then calculate the upper limit for bin j
as x + zjs
Once the bins are defined, count the
observations fj within each bin and
compare them with the expected
frequencies ej = n/c.
15-42
Normal Chi-Square
Goodness-of-Fit Test
Normal Chi-Square
Goodness-of-Fit Test
Method 3: Equal Expected Frequencies
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Histograms
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Standard normal cutpoints for equal area bins.
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The fitted normal histogram gives visual clues
as to the likely outcome of the GOF test.
Histograms reveal any outliers or other nonnormality issues.
Further tests are needed since histograms
vary.
Table 15.16
Figure 15.15
15-43
15-44
Normal Chi-Square
Goodness-of-Fit Test
Critical Values for Normal GOF Test
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ECDF Tests
Kolmogorov-Smirnov and Lilliefors Tests
Since two parameters, m and s, are
estimated from the sample, the degrees of
freedom are ν = c – m – 1
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Table 15.19
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At least 4 bins are needed to ensure 1 df.
15-45
There are many alternatives to the chi-square
test based on the Empirical Cumulative
Distribution Function (ECDF).
The Kolmogorov-Smirnov (K-S) test statistic D
is the largest absolute difference between the
actual and expected cumulative relative
frequency of the n data values:
D = Max |Fa – Fe|
The K-S test is not recommended for grouped
data.
15-46
ECDF Tests
Kolmogorov-Smirnov and Lilliefors Tests
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ECDF Tests
Kolmogorov-Smirnov and Lilliefors Tests
Fa is the actual cumulative frequency at
observation i.
Fe is the expected cumulative frequency at
observation i under the assumption that the
data came from the hypothesized distribution.
The K-S test assumes that no parameters are
estimated.
If parameters are estimated, use a Lilliefors
test.
Both of these tests are done by computer.
K-S test for
uniformity.
Figure 15.20
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15-48
ECDF Tests
Kolmogorov-Smirnov and Lilliefors Tests
ECDF Tests
Anderson-Darling Tests
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K-S test for
normality.
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The Anderson-Darling (A-D) test is widely used
for non-normality because of its power.
The A-D test is based on a probability plot.
When the data fit the hypothesized distribution
closely, the probability plot will be close to a
straight line.
The A-D test statistic measures the overall
distance between the actual and the
hypothesized distributions, using a weighted
squared distance.
Figure 15.21
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Applied Statistics in
Business & Economics
ECDF Tests
Anderson-Darling Tests with MINITAB
End of Chapter 15
Figure 15.22
15-51
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McGraw-Hill/Irwin
Copyright © 2009 by The McGraw-Hill Companies, Inc.