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Contents  Introduction Descriptive Hypothesis Tables Appendix

Example: Chi-Square Test of Independence

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.
 

 
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

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=(3 - 1)(2 - 1) or df=2

Alpha = .05

Based on the chi-square distribution table, the critical value = 5.991

 Compute the Test Statistic
Frequency Observed
 
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

Frequency Expected


 
Female
Male
Row Total
Spend more money
55*40/105 = 20.952
50*40/105 = 19.048
40
Spend the same
55*20/105 = 10.476
50*20/105 = 9.524
20
Spend less money
55*45/105 = 23.571
50*45/105 = 21.429
45
Column Total
55
50
105

Chi-square Calculations


 
Female
Male
Spend more money
(15-20.952)2/20.952
(25-19.048)2/19.048
Spend the same
(5-10.476)2/10.476
(15-9.524)2/9.524
Spend less money
(35-23.571)2/23.571
(10-21.429)2/21.429

Chi-square


 
Female
Male
Spend more money
1.691
1.860
Spend the same
2.862
3.149
Spend less money
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.