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Used
for comparing frequencies (counts) of nominal or ordinal level data for two samples across
two or more subgroups displayed in a crosstabulation table. More
common and more flexible than z-tests of proportions.
Problem:
You wish to evaluate the association
between a person's sex and their attitudes toward school spending on athletic programs. A
random sample of adults in your school district produced the following table (counts).
|
Female |
Male |
Row
Total |
Spend
more money |
15 |
25 |
40 |
Spend
the same |
5 |
15 |
20 |
Spend
less money |
35 |
10 |
45 |
Column
Total |
55 |
50 |
105 |
Assumptions
Independent
random sampling
Nominal/Ordinal
level data
No
more than 20% of the cells have an expected frequency less than 5
No
empty cells
State
the Hypothesis
Ho:
There is no association
between a person's sex and their attitudes toward spending on athletic programs.
Ha:
There is an association
between a person's sex and their attitudes toward spending on athletic programs.
Set
the Rejection Criteria
Determine
degrees of freedom df=(# of rows - 1)(# of columns - 1)
df=(3
- 1)(2 - 1) or df=2
Establish
the confidence level (.05, .01, etc.); Alpha
= .05
Based
on the chi-square distribution table, the critical value = 5.991
Compute
the Test Statistic
 where
Fo=
observed frequency
Fe=
expected frequency for each cell
Fe=(frequency
for the column)(frequency for the row)/n
Frequency
Observed
|
Female |
Male |
Row
Total |
Spend
more |
15 |
25 |
40 |
Spend
same |
5 |
15 |
20 |
Spend
less |
35 |
10 |
45 |
Column
Total |
55 |
50 |
105 |
Frequency
Expected
|
Female |
Male |
Row
Total |
Spend
more |
55*40/105
= 20.952 |
50*40/105
= 19.048 |
40 |
Spend
same |
55*20/105
= 10.476 |
50*20/105
= 9.524 |
20 |
Spend
less |
55*45/105
= 23.571 |
50*45/105
= 21.429 |
45 |
Column
Total |
55 |
50 |
105 |
Chi-square
Calculations
|
Female |
Male |
Spend
more |
(15-20.952)2/20.952 |
(25-19.048)2/19.048 |
Spend
same |
(5-10.476)2/10.476 |
(15-9.524)2/9.524 |
Spend
less |
(35-23.571)2/23.571 |
(10-21.429)2/21.429 |
Chi-square
|
Female |
Male |
Spend
more |
1.691 |
1.860 |
Spend
same |
2.862 |
3.149 |
Spend
less |
5.542 |
6.096 |
|
21.200 |
|
Decide
Results of Null Hypothesis
Since
the chi-square test statistic 21.2 exceeds the critical value of 5.991, you may conclude
there is a statistically significant association
between a person's sex and their attitudes toward spending on athletic programs. As is
apparent in the contingency table, males are more likely to support spending than females.
Software Output Example
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