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Chapter 14
Created by Bethany Stubbe and Stephan Kogitz
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Chapter Fourteen
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Chi-Square Applications for Nominal
Data
GOALS
When you have completed this chapter, you will be able to:
ONE
List the characteristics of the chi-square distribution.
TWO
Conduct a test of hypothesis comparing an observed set of
frequencies to an expected distribution.
THREE
Conduct a test of hypothesis to determine whether two
classification criteria are related.
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Characteristics of the ChiSquare Distribution
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The major characteristics of the chi-square
distribution are:
•It is positively skewed.
•It is never negative.
•It is based on degrees of freedom.
•When the degrees of freedom change a new
distribution is created.
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df = 3
df = 5
df = 10
c2
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Goodness-of-Fit Test: Equal
Expected Frequencies
Let f0 and fe be the observed and expected
frequencies respectively.
H0: There is no difference between the
observed and expected frequencies.
H1: There is a difference between the
observed and the expected frequencies.
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Goodness-of-fit Test: Equal
Expected Frequencies
The test statistic is:
c 
2

  f o  f e 2

fe




The
critical value is a chi-square value with
(k-1) degrees of freedom, where k is the
number of categories
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EXAMPLE 1
The following information shows the
number of employees absent by day of
the week at a large a manufacturing
plant. At the .05 level of significance, is
there a difference in the absence rate by
day of the week?
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Example 1 continued
Day
Monday
Tuesday
Wednesday
Thursday
Friday
Total
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Frequency
120
45
60
90
130
445
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EXAMPLE 1 continued
Assume equal expected frequency:
(120+45+60+90+130)/5=89.
The degrees of freedom is (5-1)=4.
The critical value is 9.488. Use Appendix B.
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Example 1 continued
Day
Frequency
Monday
120
Tuesday
45
Wednesday
60
Thursday
90
Friday
130
Total
445
McGraw-Hill-Ryerson
Expected
89
89
89
89
89
445
(fo – fe)2/fe
10.80
21.75
9.45
0.01
18.89
60.90
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EXAMPLE 1 continued
Because the computed value of chi-square is
greater than the critical value, H0 is rejected.
We conclude that there is a difference in the
number of workers absent by day of the week.
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EXAMPLE 2
The results of the census for a province
indicated that 63.9% of the population is
married, 7.7% widowed, 6.9% divorced (and not
re-married), and 21.5% single (never been
married). A sample of 500 adults from one city
in that province showed that 310 were married,
40 widowed, 30 divorced, and 120 single. At
the .05 significance level can we conclude that
the city sampled is different from the province
as a whole?
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EXAMPLE 2
continued
Status
f0
fe
( f 0  f e )2 / f e
Married
310
319.5
.2825
Widowed
40
38.5
.0584
Divorced
30
34.5
.5870
Single
120
107.5
1.4535
Total
500
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2.3814
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EXAMPLE 2
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continued
Step 1: H0: The distribution has not changed
H1: The distribution has changed.
2: H0 is rejected if χ2 >7.815, df=3, α = .05
Step 3: χ2 = 2.3814
Step 4: The null hypothesis is rejected. The
distribution regarding marital status in the city in
question is different from the rest of the province.
Step
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Contingency Table Analysis
A contingency table is used to investigate whether
two traits or characteristics are related.
Each observation is classified according to two
criteria.
We use the usual hypothesis testing procedure.
The degrees of freedom is equal to: (number of
rows-1)(number of columns-1).
The expected frequency is computed as: Expected
Frequency = (row total)(column total)/grand total
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EXAMPLE 4
Is there a relationship between the
location of an accident and the gender of
the person involved in the accident? A
sample of 150 accidents reported to the
police were classified by type and
gender. At the .05 level of significance,
can we conclude that gender and the
location of the accident are related?
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EXAMPLE 4
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continued
Sex
Work
Home
Other
Total
Male
60
20
10
90
Female
20
30
10
60
Total
80
50
20
150
The expected frequency for the work-male
intersection is computed as (90)(80)/150=48.
Similarly, you can compute the expected
frequencies for the other cells.
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EXAMPLE 4
continued
H0: Gender and location are not related.
H1: Gender and location are related.
H0 is rejected if the computed value of χ2 is
greater than 5.991. There are (3- 1)(2-1) =
2 degrees of freedom.
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Example 4 continued
Step 3:
c2
Find the value of χ2.

60  48 2

48

10  82
 ... 
8
 16 .667
H0 is rejected. We conclude that gender
and location are related.
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