Chi Square Test for Homogeneity

Everyone's been in the situation before: you and your significant other can't agree on what to watch for date night! While the two of you are debating over which movie to watch, a question arises in the back of your mind; do different types of people (for instance, men vs. women) have different movie preferences? The answer to this question, and others like it, can be found using a specific Chi-square test – the Chi-square test for homogeneity.

Chi-Square Test for Homogeneity Definition

When you want to know if two categorical variables follow the same probability distribution (like in the movie preference question above), you can use a Chi-square test for homogeneity.

In this test, you randomly collect data from a population to determine if there is a significant association between \(2\) or more categorical variables.

Conditions for a Chi-Square Test for Homogeneity

All the Pearson Chi-square tests share the same basic conditions. The main difference is how the conditions apply in practice. A Chi-square test for homogeneity requires a categorical variable from at least two populations, and the data needs to be the raw count of members of each category. This test is used to check if the two variables follow the same distribution.

To be able to use this test, the conditions for a Chi-square test of homogeneity are:

• The variables must be categorical.

  • Because you are testing the sameness of the variables, they have to have the same groups. This Chi-square test uses cross-tabulation, counting observations that fall in each category.

• Groups must be mutually exclusive; i.e., the sample is randomly selected.

  • Each observation is only allowed to be in one group. A person can live in a house or an apartment, but they cannot live in both.

Table 1. Table of contingency, Chi-Square test for homogeneity.

• Expected counts must be at least \(5\).

  • This means the sample size must be large enough, but how large is difficult to determine beforehand. In general, making sure there are more than \(5\) in each category should be fine.

• Observations must be independent.

  • This assumption is all about how you collect the data. If you use simple random sampling, that will almost always be statistically valid.

Chi-Square Test for Homogeneity: Null Hypothesis and Alternative Hypothesis

The question underlying this hypothesis test is: Do these two variables follow the same distribution?

The hypotheses are formed to answer that question.

• The null hypothesis is that the two variables are from the same distribution.\[ \begin{align}H_{0}: p_{1,1} &= p_{2,1} \text{ AND } \\p_{1,2} &= p_{2,2} \text{ AND } \ldots \text{ AND } \\p_{1,n} &= p_{2,n}\end{align} \]

• The null hypothesis requires every single category to have the same probability between the two variables.

• The alternative hypothesis is that the two variables are not from the same distribution, i.e., at least one of the null hypotheses is false.\[ \begin{align}H_{a}: p_{1,1} &\neq p_{2,1} \text{ OR } \\p_{1,2} &\neq p_{2,2} \text{ OR } \ldots \text{ OR } \\p_{1,n} &\neq p_{2,n}\end{align} \]

• If even one category is different from one variable to the other, then the test will return a significant result and provide evidence to reject the null hypothesis.

The null and alternative hypotheses in the heart attack survival study are:

Expected Frequencies for a Chi-Square Test for Homogeneity

You must calculate the expected frequencies for a Chi-square test for homogeneity individually for each population at each level of the categorical variable, as given by the formula:

\[ E_{r,c} = \frac{n_{r} \cdot n_{c}}{n} \]

where,

• \(E_{r,c}\) is the expected frequency for population \(r\) at level \(c\) of the categorical variable,

• \(r\) is the number of populations, which is also the number of rows in a contingency table,

• \(c\) is the number of levels of the categorical variable, which is also the number of columns in a contingency table,

• \(n_{r}\) is the number of observations from population \(r\),

• \(n_{c}\) is the number of observations from level \(c\) of the categorical variable, and

• \(n\) is the total sample size.

Continuing with the heart attack survival study:

Degrees of Freedom for a Chi-Square Test for Homogeneity

There are two variables in a Chi-square test for homogeneity. Therefore, you are comparing two variables and need the contingency table to add up in both dimensions.

Since you need the rows to add up and the columns to add up, the degrees of freedom is calculated by:

\[ k = (r - 1) (c - 1) \]

where,

• \(k\) is the degrees of freedom,

• \(r\) is the number of populations, which is also the number of rows in a contingency table, and

• \(c\) is the number of levels of the categorical variable, which is also the number of columns in a contingency table.

Chi-Square Test for Homogeneity: Formula

The formula (also called a test statistic) of a Chi-square test for homogeneity is:

\[ \chi^{2} = \sum \frac{(O_{r,c} - E_{r,c})^{2}}{E_{r,c}} \]

where,

• \(O_{r,c}\) is the observed frequency for population \(r\) at level \(c\), and

• \(E_{r,c}\) is the expected frequency for population \(r\) at level \(c\).

How to Calculate the Test Statistic for a Chi-Square Test for Homogeneity

Step \(1\): Create a Table

Starting with your contingency table, remove the “Row Totals” column and the “Column Totals” row. Then, separate your observed and expected frequencies into two columns, like so:

Table 3. Table of observed and expected frequencies, Chi-Square test for homogeneity.

Step \(2\): Subtract Expected Frequencies from Observed Frequencies

Add a new column to your table called “O – E”. In this column, put the result of subtracting the expected frequency from the observed frequency:

Table 4. Table of observed and expected frequencies, Chi-Square test for homogeneity.

Step \(3\): Square the Results from Step \(2\)Add another new column to your table called “(O – E)²”. In this column, put the result of squaring the results from the previous column:

Table 5. Table of observed and expected frequencies, Chi-Square test for homogeneity.

Step \(4\): Divide the Results from Step \(3\) by the Expected FrequenciesAdd a final new column to your table called “(O – E)²/E”. In this column, put the result of dividing the results from the previous column by their expected frequencies:

Table 6. Table of observed and expected frequencies, Chi-Square test for homogeneity.

Step \(5\): Sum the Results from Step \(4\) to get the Chi-Square Test StatisticFinally, add up all the values in the last column of your table to calculate your Chi-square test statistic:

\[ \begin{align}\chi^{2} &= \sum \frac{(O_{r,c} - E_{r,c})^{2}}{E_{r,c}} \\&= 0.322 + 0.013 + 3.831 + 0.150 + 5.074 + 0.199 \\&= 9.589.\end{align} \]

The Chi-square test statistic for the Chi-square test for homogeneity in the heart attack survival study is:

\[ \chi^{2} = 9.589. \]

Steps to Perform a Chi-Square Test for Homogeneity

To determine whether the test statistic is large enough to reject the null hypothesis, you compare the test statistic to a critical value from a Chi-square distribution table. This act of comparison is the heart of the Chi-square test of homogeneity.

Follow the \(6\) steps below to perform a Chi-square test of homogeneity.

Step \(1\): State the Hypotheses

• The null hypothesis is that the two variables are from the same distribution.\[ \begin{align}H_{0}: p_{1,1} &= p_{2,1} \text{ AND } \\p_{1,2} &= p_{2,2} \text{ AND } \ldots \text{ AND } \\p_{1,n} &= p_{2,n}\end{align} \]

• The alternative hypothesis is that the two variables are not from the same distribution, i.e., at least one of the null hypotheses is false.\[ \begin{align}H_{a}: p_{1,1} &\neq p_{2,1} \text{ OR } \\p_{1,2} &\neq p_{2,2} \text{ OR } \ldots \text{ OR } \\p_{1,n} &\neq p_{2,n}\end{align} \]

Step \(2\): Calculate the Expected Frequencies

Reference your contingency table to calculate the expected frequencies using the formula:

\[ E_{r,c} = \frac{n_{r} \cdot n_{c}}{n} \]

Step \(3\): Calculate the Chi-Square Test Statistic

Use the formula for a Chi-square test for homogeneity to calculate the Chi-square test statistic:

\[ \chi^{2} = \sum \frac{(O_{r,c} - E_{r,c})^{2}}{E_{r,c}} \]

Step \(4\): Find the Critical Chi-Square Value

To find the critical Chi-square value, you can either:

1. use a Chi-square distribution table, or

2. use a critical value calculator.

No matter which method you choose, you need \(2\) pieces of information:

1. the degrees of freedom, \(k\), given by the formula:

   \[ k = (r - 1) (c - 1) \]

2. and the significance level, \(\alpha\), which is usually \(0.05\).

Find the critical value of the heart attack survival study.

Step \(5\): Compare the Chi-Square Test Statistic to the Critical Chi-Square Value

Is your test statistic large enough to reject the null hypothesis? To find out, compare it to the critical value.

Compare your test statistic to the critical value in the heart attack survival study:

Step \(6\): Decide Whether to Reject the Null Hypothesis

Finally, decide if you can reject the null hypothesis.

• If the Chi-square value is less than the critical value, then you have an insignificant difference between the observed and expected frequencies; i.e., \( p > \alpha \).

  • This means you do not reject the null hypothesis.

• If the Chi-square value is greater than the critical value, then you have a significant difference between the observed and expected frequencies; i.e., \( p < \alpha \).

  • This means you have sufficient evidence to reject the null hypothesis.

Now you can decide whether to reject the null hypothesis for the heart attack survival study:

Continuing from the example in the introduction, you decide to find an answer to the question: do men and women have different movie preferences?