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The Chi-Square Distribution: Test of Independence

Module by: Susan Dean, Barbara Illowsky, Ph.D.. E-mail the authors

Summary: This module describes how the chi-square distribution can be used to test for independence.

Tests of independence involve using a contingency table of observed (data) values. You first saw a contingency table when you studied probability in the Probability Topics chapter.

The test statistic for a test of independence is similar to that of a goodness-of-fit test:

Σ ( i j ) ( O - E ) 2 E Σ ( i j ) ( O - E ) 2 E
(1)

where:

  • OO = observed values
  • EE = expected values
  • ii = the number of rows in the table
  • jj = the number of columns in the table

There are i j ij terms of the form ( O - E ) 2 E ( O - E ) 2 E .

A test of independence determines whether two factors are independent or not. You first encountered the term independence in Chapter 3. As a review, consider the following example.

Note:

The expected value for each cell needs to be at least 5 in order to use this test.

Example 1

Suppose A A = a speeding violation in the last year and B B = a cell phone user while driving. If A A and B B are independent then P ( A AND B ) = P ( A ) P ( B ) P(A AND B)=P(A)P(B). A AND B A AND B is the event that a driver received a speeding violation last year and is also a cell phone user while driving. Suppose, in a study of drivers who received speeding violations in the last year and who uses cell phones while driving, that 755 people were surveyed. Out of the 755, 70 had a speeding violation and 685 did not; 305 were cell phone users while driving and 450 were not.

Let yy = expected number of drivers that use a cell phone while driving and received speeding violations.

If AA and BB are independent, then P ( A AND B ) = P ( A ) P ( B ) P(A AND B)=P(A)P(B). By substitution,

y 755 = 70 755 305 755 y 755 = 70 755 305 755

Solve for y : y = 70 305 755 = 28.3 y:y= 70 305 755 =28.3

About 28 people from the sample are expected to be cell phone users while driving and to receive speeding violations.

In a test of independence, we state the null and alternate hypotheses in words. Since the contingency table consists of two factors, the null hypothesis states that the factors are independent and the alternate hypothesis states that they are not independent (dependent). If we do a test of independence using the example above, then the null hypothesis is:

H o H o : Being a cell phone user while driving and receiving a speeding violation are independent events.

If the null hypothesis were true, we would expect about 28 people to be cell phone users while driving and to receive a speeding violation.

The test of independence is always right-tailed because of the calculation of the test statistic. If the expected and observed values are not close together, then the test statistic is very large and way out in the right tail of the chi-square curve, like goodness-of-fit.

The degrees of freedom for the test of independence are:

df = (number of columns - 1)(number of rows - 1) df = (number of columns - 1)(number of rows - 1)

The following formula calculates the expected number (EE):

E = (row total)(column total) total number surveyed E= (row total)(column total) total number surveyed

Example 2

In a volunteer group, adults 21 and older volunteer from one to nine hours each week to spend time with a disabled senior citizen. The program recruits among community college students, four-year college students, and nonstudents. The following table is a sample of the adult volunteers and the number of hours they volunteer per week.

Table 1: Number of Hours Worked Per Week by Volunteer Type (Observed)
The table contains observed (O) values (data).
Type of Volunteer 1-3 Hours 4-6 Hours 7-9 Hours Row Total
Community College Students 111 96 48 255
Four-Year College Students 96 133 61 290
Nonstudents 91 150 53 294
Column Total 298 379 162 839

Problem 1

Are the number of hours volunteered independent of the type of volunteer?

Example 3

De Anza College is interested in the relationship between anxiety level and the need to succeed in school. A random sample of 400 students took a test that measured anxiety level and need to succeed in school. The table shows the results. De Anza College wants to know if anxiety level and need to succeed in school are independent events.

Table 3: Need to Succeed in School vs. Anxiety Level
Need to Succeed in School High
Anxiety
Med-high
Anxiety
Medium
Anxiety
Med-low
Anxiety
Low
Anxiety
Row Total
High Need 35 42 53 15 10 155
Medium Need 18 48 63 33 31 193
Low Need 4 5 11 15 17 52
Column Total 57 95 127 63 58 400

Problem 1

How many high anxiety level students are expected to have a high need to succeed in school?

Problem 2

If the two variables are independent, how many students do you expect to have a low need to succeed in school and a med-low level of anxiety?

Glossary

Contingency Table:
The method of displaying a frequency distribution as a table with rows and columns to show how two variables may be dependent (contingent) upon each other. The table provides an easy way to calculate conditional probabilities.

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