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

Let [latex]y =[/latex] expected number of drivers who used a cell phone while driving and received speeding violations.

If [latex]A[/latex] and [latex]B[/latex] are independent, then [latex]P(A \text{ and } B) = P(A)P(B)[/latex]. By substitution, [latex]\frac{y}{755}=\left(\frac{70}{755}\right)\left(\frac{305}{755}\right)[/latex].

Solve for [latex]y[/latex]: [latex]y = \frac{\left(70\right)\left(305\right)}{755}=28.3[/latex]

About 28 people from the sample are expected to use cell phones while driving and to receive speeding violations.

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

[latex]H_0:[/latex] 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 use cell phones 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, as it is in a goodness-of-fit.

The number of degrees of freedom for the test of independence is:

[latex]df = (\text{number of columns} - 1)(\text{number of rows} - 1)[/latex]

The following formula calculates the expected number ([latex]E[/latex]):

[latex]E=\frac{\text{(row total)(column total)}}{\text{total number surveyed}}[/latex]

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 table below shows a sample of the adult volunteers and the number of hours they volunteer per week.

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

Solution

The observed table and the question at the end of the problem, “Is the number of hours volunteered independent of the type of volunteer?” tell you this is a test of independence. The two factors are number of hours volunteered and type of volunteer. This test is always right-tailed.

[latex]H_0:[/latex] The number of hours volunteered is independent of the type of volunteer.

[latex]H_a:[/latex] The number of hours volunteered is dependent on the type of volunteer.

The expected results are in the following table.

Number of Hours Worked per Week by Volunteer Type (Expected). The table contains expected (E) values (data).

Type of Volunteer	1-3 Hours	4-6 Hours	7-9 Hours
Community College Students	90.57	115.19	49.24
Four-Year College Students	103.00	131.00	56.00
Nonstudents	104.42	132.81	56.77

For example, the calculation for the expected frequency for the top left cell is [latex]E=\frac{\left(\text{row total}\right)\left(\text{column total}\right)}{\text{total number surveyed}}=\frac{\left(255\right)\left(298\right)}{839}=90.57[/latex]

Calculate the test statistic: [latex]\chi^2 = 12.99[/latex] (You can find this using your calculator or computer.)

Distribution for the test: [latex]{\chi }_{4}^{2}[/latex]

[latex]df = (3 \text{columns} – 1)(3 \text{rows} – 1) = (2)(2) = 4[/latex]

Graph:

Probability statement: [latex]\text{p-value} = P(\chi^2 > 12.99) = 0.0113[/latex]

Compare [latex]\alpha[/latex] and the p-value: Since no [latex]\alpha[/latex] is given, assume [latex]\alpha = 0.05[/latex]. [latex]\text{p-value} = 0.0113[/latex], so [latex]\alpha > \text{p-value}[/latex].

Make a decision: Since [latex]\alpha > p-value[/latex], reject [latex]H_0[/latex]. This means that the factors are not independent.

Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the number of hours volunteered and the type of volunteer are dependent on one another.

For the example in the expected results table, if there had been another type of volunteer, teenagers, what would the degrees of freedom be?

Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 3 ENTER 3 ENTER. Enter the table values by row from the expected results table. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 12.9909 and the p-value = 0.0113. Do the procedure a second time, but arrow down to Draw instead of calculate.