Chi-Square Test for Independence
This lesson explains how to conduct a
chi-square test for independence.
The test is applied when you have two
from a single population. It is used to determine whether
there is a significant association between the two variables.
For example, in an election survey, voters might be classified
by gender (male or female) and voting preference (Democrat,
Republican, or Independent). We could use a chi-square test for
independence to determine whether gender is related to
voting preference. The
sample problem at the end of the lesson
considers this example.
When to Use Chi-Square Test for Independence
The test procedure described in this lesson is appropriate when the
following conditions are met:
- If sample data are displayed in a
the expected frequency count for each cell of the table is
at least 5.
This approach consists of four steps: (1) state the hypotheses,
(2) formulate an analysis plan, (3) analyze sample data, and
(4) interpret results.
State the Hypotheses
Suppose that Variable A has r levels, and
Variable B has c levels.
null hypothesis states that knowing the level of
Variable A does not help you predict the level of
Variable B. That is, the variables are independent.
H0: Variable A and Variable B are independent.
Ha: Variable A and Variable B are not independent.
alternative hypothesis is that knowing the level of
Variable A can help you predict the level of
Note: Support for the alternative hypothesis suggests that the variables
are related; but the relationship is not necessarily causal, in
the sense that one variable "causes" the other.
Formulate an Analysis Plan
The analysis plan describes
how to use sample data to accept or reject the null
hypothesis. The plan should specify the following elements.
- Significance level. Often, researchers choose
0.01, 0.05, or 0.10; but any value between 0 and
1 can be used.
- Test method. Use the
chi-square test for independence
to determine whether there is a significant relationship
between two categorical variables.
Analyze Sample Data
Using sample data, find the
degrees of freedom, expected frequencies,
test statistic, and the P-value associated with the test statistic.
The approach described in this section is illustrated in the
sample problem at the end of this lesson.
If the sample findings are unlikely, given
the null hypothesis, the researcher rejects the null hypothesis.
Typically, this involves comparing the P-value to the
and rejecting the null hypothesis when the P-value is less than
the significance level.
Test Your Understanding of This Lesson
A public opinion poll surveyed a simple random sample of 1000 voters.
Respondents were classified by gender (male or female) and by
voting preference (Republican, Democrat, or Independent).
Results are shown in the
contingency table below.
Is there a gender gap? Do the men's voting preferences differ
significantly from the women's preferences? Use a 0.05
level of significance.
The solution to this problem takes four steps:
(1) state the hypotheses, (2) formulate an analysis plan,
(3) analyze sample data, and (4) interpret results.
We work through those steps below:
State the hypotheses. The first step is to
null hypothesis and an alternative hypothesis.
H0: Gender and voting preferences are
Ha: Gender and voting preferences are not
Formulate an analysis plan. For this analysis,
the significance level is 0.05. Using sample data, we will
chi-square test for independence.
Analyze sample data. Applying the chi-square
test for independence to sample data, we compute
the degrees of freedom,
the expected frequency counts, and
the chi-square test statistic. Based on the
chi-square statistic and the
degrees of freedom, we determine the
DF = (r - 1) * (c - 1) = (2 - 1) * (3 - 1) = 2
Er,c = (nr * nc) / n
E1,1 = (400 * 450) / 1000 = 180000/1000 = 180
E1,2 = (400 * 450) / 1000 = 180000/1000 = 180
E1,3 = (400 * 100) / 1000 = 40000/1000 = 40
E2,1 = (600 * 450) / 1000 = 270000/1000 = 270
E2,2 = (600 * 450) / 1000 = 270000/1000 = 270
E2,3 = (600 * 100) / 1000 = 60000/1000 = 60
Χ2 = Σ
[ (Or,c - Er,c)2 / Er,c ]
(200 - 180)2/180 + (150 - 180)2/180 + (50 - 40)2/40
(250 - 270)2/270 + (300 - 270)2/270 + (50 - 60)2/60
400/180 + 900/180 +
100/40 + 400/270 +
900/270 + 100/60
2.22 + 5.00 + 2.50 + 1.48 + 3.33 + 1.67 = 16.2
DF is the degrees of freedom,
r is the number of levels of gender,
c is the number of levels of the voting preference,
nr is the number of observations from level r
nc is the number of observations from level c
of voting preference,
n is the number of observations in the sample,
Er,c is the expected frequency count when gender is level
r and voting preference is level c, and
Or,c is the observed frequency count when gender is level
r voting preference is level c.
The P-value is the probability that a chi-square statistic
having 2 degrees of freedom is more extreme than 16.2.
We use the
Chi-Square Distribution Calculator
to find P(Χ2 > 16.2) = 0.0003.
Interpret results. Since the P-value (0.0003) is
less than the significance level (0.05), we cannot accept the
null hypothesis. Thus, we conclude that there is a relationship
between gender and voting preference.
Note: If you use this approach on an exam, you may also want to mention
why this approach is appropriate. Specifically, the approach is
appropriate because the sampling method was simple random sampling,
each population was more than 10 times larger than its
respective sample, the
variables under study were categorical, and the expected frequency count
was at least 5 in each cell of the contingency table.