Chapter 16
Chi-Squared Tests
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.1
A Common Theme…
What to do?
Data Type?
Number of
Categories?
Statistical
Technique:
Describe a
population
Nominal
Two or more
X2 goodness of fit
test
Compare two
populations
Nominal
Two or more
X2 test of a
contingency table
Compare two or
more populations
Nominal
--
X2 test of a
contingency table
Analyze relationship
between two
variables
Nominal
--
X2 test of a
contingency table
One data type…
…Two techniques
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16.2
Two Techniques…
The first is a goodness-of-fit test applied to data produced by
a multinomial experiment, a generalization of a binomial
experiment and is used to describe one population of data.
The second uses data arranged in a contingency table to
determine whether two classifications of a population of
nominal data are statistically independent; this test can also
be interpreted as a comparison of two or more populations.
In both cases, we use the chi-squared (
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
) distribution.
16.3
The Multinomial Experiment…
Unlike a binomial experiment which only has two possible
outcomes (e.g. heads or tails), a multinomial experiment:
• Consists of a fixed number, n, of trials.
• Each trial can have one of k outcomes, called cells.
• Each probability pi remains constant.
• Our usual notion of probabilities holds, namely:
p1 + p2 + … + pk = 1, and
• Each trial is independent of the other trials.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.4
Chi-squared Goodness-of-Fit Test…
We test whether there is sufficient evidence to reject a
specified set of values for pi.
To illustrate, our null hypothesis is:
H0: p1 = a1, p2 = a2, …, pk = ak
(where a1, a2, …, ak are the values of interest)
Our research hypothesis is:
H1: At least one pi ≠ ai
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.5
Chi-squared Goodness-of-Fit Test…
The test builds on comparing actual frequency and the
expected frequency of occurrences in all the cells.
Example 16.1…
We compare market share before and after an advertising
campaign to see if there is a difference (i.e. if the advertising
was effective in improving market share).
H0: p1 = a1, p2 = a2, …, pk = ak
Where ai is the market share before the campaign. If there
was no change, we’d expect H0 to not be rejected. If there is
evidence to reject H0 in favor of: H1: At least one pi ≠ ai,
what’s a logical conclusion?
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.6
Example 16.1…
IDENTIFY
Market shares before the advertising campaign…
Company A – 45%
Company B – 40%
All Others – 15 %
200 customers surveyed after the campaign. The results:
Company A – 102 customers preferred their product.
Company B – 82 customers…
All Others – 16 customers.
Before the campaign, we’d expect 45% of 200 customers
(i.e. 90 customers) to prefer company A’s product. After the
campaign, we observe its 102 customers. Does this mean the
campaign was effective? (i.e. at a 5% significance level).
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.7
Example 16.1…
Expected Frequency
Observed Frequency
A
B
A
B
Are these changes
statistically
significant?
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.8
Example 16.1…
IDENTIFY
Our null hypothesis is:
H0: pCompanyA = .45, pCompanyB = .40, pOthers = .15
(i.e. the market shares pre-campaign), and our alternative
hypothesis is:
H1: At least one pi ≠ ai
In order to complete our hypothesis testing we need a test
statistic and a rejection region…
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.9
Chi-squared Goodness-of-Fit Test…
Our Chi-squared goodness of fit test statistic is given by:
observed
frequency
expected
frequency
Note: this statistic is approximately Chi-squared with k–1
degrees of freedom provided the sample size is large. The
rejection region is:
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.10
Example 16.1…
COMPUTE
In order to calculate our test statistic, we lay-out the data in a
tabular fashion for easier calculation by hand:
Observed
Frequency
Expected
Frequency
Delta
Summation
Component
fi
ei
(fi – ei)
(fi – ei)2/ei
A
102
90
12
1.60
B
82
80
2
0.05
Others
16
30
-14
6.53
Total
200
200
Company
8.18
Check that these are equal
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.11
Example 16.1…
INTERPRET
Our rejection region is:
Since our test statistic is 8.18 which is greater than our
critical value for Chi-squared, we reject H0 in favor of H1,
that is,
“There is sufficient evidence to infer that the proportions
have changed since the advertising campaigns were
implemented”
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.12
Example 16.1…
COMPUTE
Note: Table 5 in Appendix B does not allow for the direct
calculation of
, so we have to use Excel:
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16.13
Example 16.1…
p-value
Note: There are a couple of different ways to calculate the
p-value of the test:
Computed manually
from our table
Computed directly
from the data
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.14
Required Conditions…
In order to use this technique, the sample size must be large
enough so that the expected value for each cell is 5 or more.
(i.e. n x pi ≥ 5)
If the expected frequency is less than five, combine it with
other cells to satisfy the condition.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.15
Identifying Factors…
Factors that Identify the Chi-Squared Goodness-of-Fit Test:
ei=(n)(pi)
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16.16
Chi-squared Test of a Contingency Table
The Chi-squared test of a contingency table is used to:
• determine whether there is enough evidence to infer
that two nominal variables are related, and
• to infer that differences exist among two or more
populations of nominal variables.
In order to use use these techniques, we need to classify the
data according to two different criteria.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.17
Example 16.2…
IDENTIFY
The demand for an MBA program’s optional courses and
majors is quite variable year over year.
The research hypothesis is that the academic background of
the students (i.e. their undergrad degrees) affects their choice
of major.
A random sample of data on last year’s MBA students was
collected and summarized in a contingency table…
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.18
Example 16.2
The Data
MBA Major
Undergrad
Degree
Accounting
Finance
Marketing
Total
BA
31
13
16
60
BEng
8
16
7
31
BBA
12
10
17
39
Other
10
5
7
22
Total
61
44
47
152
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16.19
Example 16.2…
Again, we are interesting in determining whether or not the
academic background of the students affects their choice of
MBA major. Thus our research hypothesis is:
H1: The two variables are dependent
Our null hypothesis then, is:
H0: The two variables are independent.
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16.20
Example 16.2…
In this case, our test statistic is:
(where k is the number of cells in the contingency table, i.e.
rows x columns)
Our rejection region is:
where the number of degrees of freedom is (r–1)(c–1)
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16.21
Example 16.2…
COMPUTE
In order to calculate our
test statistic, we need the
calculate the expected frequencies for each cell…
The expected frequency of the cell in row i and column j is:
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.22
Contingency Table Set-up…
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16.23
Example 16.2
COMPUTE
Compute expected frequencies…
MBA Major
Undergrad
Degree
Accounting
Finance
Marketing
Total
BA
31
13
16
60
BEng
8
16
31 x 47
152
31
BBA
12
10
17
39
Other
10
5
7
22
Total
61
44
47
152
e23 = (31)(47)/152 = 9.59 — compare this to f23 = 7
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.24
Example 16.2…
We can now compare observed with expected frequencies…
MBA Major
Undergrad
Degree
Accounting
Finance
Marketing
BA
31
24.08
13
17.37
16
18.55
BEng
8
12.44
16
8.97
7
9.59
BBA
12
15.65
10
11.29
17
12.06
Other
10
8.83
5
6.37
7
6.80
and calculate our test statistic:
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16.25
Example 16.2…
We compare
INTERPRET
= 14.70 with:
Since our test statistic falls into the rejection region, we
reject
H0: The two variables are independent.
in favor of
H1: The two variables are dependent.
That is, there is evidence of a relationship between
undergrad degree and MBA major.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.26
Example 16.2…
COMPUTE
We can also leverage the tools in Excel to process our data:
Tools >
Data Analysis Plus >
Contingency Table
Compare
p-value
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.27
Required Condition – Rule of Five…
In a contingency table where one or more cells have
expected values of less than 5, we need to combine rows or
columns to satisfy the rule of five.
Note: by doing this, the degrees of freedom must be changed
as well.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
16.28
Identifying Factors…
Factors that identify the Chi-squared test of a contingency
table:
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16.29