Chi-Square Test of Independence

This lesson explains how to conduct a chi-square test of independence. The test is applied when you have two categorical variables 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 of independence

The test procedure described in this lesson is appropriate when the following conditions are met:

The sampling method is simple random sampling.
The variables under study are each categorical.
If sample data are displayed in a contingency table, the expected frequency count for each cell of the table is at least 5.

General Procedure for Hypothesis Testing

To test any hypothesis, the same five-step procedure is used: (1) state the hypotheses, (2) choose the significance level, (3) compute the test statistic, (4) find the P-value, and (5) interpret results. Here, we apply the general procedure to the chi-square test of independence.

State the Hypotheses

Suppose that Variable A has r levels, and Variable B has c levels. The null hypothesis (H₀) 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.

H_a: Variable A and Variable B are not independent.

The alternative hypothesis (H_a) is that knowing the level of Variable A can help you predict the level of Variable B.

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.

Choose the Significance Level

The significance level is the probability of rejecting the null hypothesis when it is actually true. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10.

Compute the Test Statistic

The chi-square test of independence requires that you compute the degrees of freedom (df) for the test statistic, an expected frequency count (E_r,c) for each cell of a contingency table (variable A versus variable B), and a chi-square test statistic (χ²). Formulas for all the necessary computations appear below.

Degrees of freedom. The degrees of freedom (df) is equal to:
df = (r - 1) * (c - 1)
where r is the number of levels for one catagorical variable, and c is the number of levels for the other categorical variable.
Expected frequency counts. The expected frequency counts are computed separately for each level of one categorical variable at each level of the other categorical variable. Compute r * c expected frequencies, according to the following formula.
E_r,c = (n_r * n_c) / n
where E_r,c is the expected frequency count for level r of Variable A and level c of Variable B, n_r is the total number of sample observations at level r of Variable A, n_c is the total number of sample observations at level c of Variable B, and n is the total sample size.
Test statistic. The test statistic is a chi-square random variable (χ²) defined by the following equation.
χ² = Σ [ (O_r,c - E_r,c)² / E_r,c ]
where O_r,c is the observed frequency count at level r of Variable A and level c of Variable B, and E_r,c is the expected frequency count at level r of Variable A and level c of Variable B.

Find the P-Value

The P-value is the probability of observing a sample statistic as extreme as the test statistic. To find the probability for a chi-square test statistic with degrees of freedom equal to df, use a chi-square distribution table or a statistical software. (The sample problem at the end of this lesson uses Stat Trek's Chi-Square Distribution Calculator to find the P-value for a chi-square test statistic.)

Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. This involves comparing the P-value to the significance level, and rejecting the null hypothesis when the P-value is less than the significance level.

Test Your Understanding

Problem

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.

	Voting Preferences			Row total
	Rep	Dem	Ind	Row total
Male	200	150	50	400
Female	250	300	50	600
Column total	450	450	100	1000

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.

Solution

To solve this problem, we conduct a chi-square test of independence. Like any hypothesis test, a chi-square test of independence consists of five steps: (1) state the hypotheses, (2) choose the significance level, (3) compute the test statistic, (4) find the P-value, and (5) interpret results.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
H₀: Gender and voting preferences are independent.

H_a: Gender and voting preferences are not independent.
Choose the significance level. For this analysis, the significance level is 0.05.
Compute the test statistic. Applying the chi-square test of independence to sample data, we compute the degrees of freedom (df), the expected frequency counts (E_r,c), and the chi-square test statistic (χ²).
df = (r - 1) * (c - 1) = (2 - 1) * (3 - 1) = 2

E_r,c = (n_r * n_c) / n
E_1,1 = (400 * 450) / 1000 = 180000/1000 = 180
E_1,2 = (400 * 450) / 1000 = 180000/1000 = 180
E_1,3 = (400 * 100) / 1000 = 40000/1000 = 40
E_2,1 = (600 * 450) / 1000 = 270000/1000 = 270
E_2,2 = (600 * 450) / 1000 = 270000/1000 = 270
E_2,3 = (600 * 100) / 1000 = 60000/1000 = 60

χ² = Σ [ (O_r,c - E_r,c)² / E_r,c ]
χ² = (200 - 180)²/180 + (150 - 180)²/180 + (50 - 40)²/40
+ (250 - 270)²/270 + (300 - 270)²/270 + (50 - 60)²/60
χ² = 400/180 + 900/180 + 100/40 + 400/270 + 900/270 + 100/60
χ² = 2.22 + 5.00 + 2.50 + 1.48 + 3.33 + 1.67 = 16.2

where df is the degrees of freedom, r is the number of levels of gender, c is the number of levels of the voting preference, n_r is the number of observations from level r of gender, n_c is the number of observations from level c of voting preference, n is the number of observations in the sample, E_r,c is the expected frequency count when gender is level r and voting preference is level c, O_r,c is the observed frequency count when gender is level r voting preference is level c, and χ² is the chi-square test statistic.
Find the P-value. The P-value is the probability that a chi-square statistic having 2 degrees of freedom would be more extreme (bigger) than the test statistic of 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, the variables under study were categorical, and the expected frequency count was at least 5 in each cell of the contingency table.

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