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Chi-Square Test for Association and Independence

Chi-Square Test for Association and Independence

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This lesson will explain the chi-square test for association/independence.

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Tutorial

What's Covered

This tutorial will cover the chi-square test of independence. You'll learn about:

  1. The chi-square test of independence

1. THE CHI-SQUARE TEST OF INDEPENDENCE

The chi-square test of independence is sometimes called a chi-square test of association.

Term to Know

Chi-Square Test of Independence/Association

A hypothesis test that tests whether two qualitative variables have an association or not.

We can test this by looking at a null hypothesis in a problem that says that two variables are independent.

Example 335 students of different backgrounds (rural, suburban, and urban schools) were asked if they had to pick one thing about school that was most important to them, would it be: getting good grades, being popular, or being good at sports. And here's the distribution of responses:

The question is, does there appear to be an association between the geographic location of the school and the answer choice to the question, the goal? This is an ideal time to run a chi-square test of independence. This can tell you if the distribution of answer choices- (grades, popular, and sports) differ significantly for each school location. Are they associated or are they independent?

Think About It

  1. In the null hypothesis, you're going to say that school location and goal are independent. That is, they don't have an association with each other.
  2. The alternative hypothesis is that they do have an association with each other. At least one of these distributions of grades, popularity, and sports is different for suburban, urban, or rural than the others.

For the test of independence, the conditions and the way that chi-square and p-value are calculated are the same as in a test of homogeneity. The expected value for each cell is equal to that particular cell's row total, times its column total, divided by the grand total for all the cells.

This gives you the expected table of results:

The 74.72, for example, was calculated by multiplying the row total for grades, times the column total for rural, divided by the 335 grand total for all students. This process is repeated for all the cells.

What you are interested in is whether or not all the expected counts are at least 5. The smallest one is 7.21. So the conditions are met. Choose an alpha level of 0.05 in this problem and use technology to calculate the chi-square statistic and the p-value.

  • In this case, chi-square is equal to 18.564. That is big.
  • That causes the p-value to be very small: 0.001.

You need to link your p-value to a decision about the null hypothesis. Since the p-value is smaller than 0.05, you reject the null hypothesis in favor of the alternative and conclude that there is an association between the two categorical variables, school location and goal.


Summary

The chi-square test of independence tests whether two qualitative variables have an association or not, so it's sometimes call the chi-square test of association.  The expected value for each cell is equal to that particular cell's row total, times its column total, divided by the grand total for all the cells.


Thank you and good luck!

Source: THIS WORK IS ADAPTED FROM SOPHIA AUTHOR JONATHAN OSTERS

TERMS TO KNOW
  • Chi-Square Test of Independence/Association

    A hypothesis test that tests whether two qualitative variables have an association or not.