11.2 The chi-square distribution: test of independence  (Page 2/4)

Suppose $A$ = a speeding violation in the last year and $B$ = a cell phone user while driving. If $A$ and $B$ are independent then $P(AANDB)=P\left(A\right)P\left(B\right)$ . $AANDB$ 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 speedingviolation and 685 did not; 305 were cell phone users while driving and 450 were not.

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

If $A$ and $B$ are independent, then $P(AANDB)=P\left(A\right)P\left(B\right)$ . By substitution,

$\frac{y}{755}=\frac{70}{755}\cdot \frac{305}{755}$

Solve for $y:y=\frac{70\cdot 305}{755}=28.3$

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:

$H_o$ : 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 teststatistic 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:

$\text{df = (number of columns - 1)(number of rows - 1)}$

The following formula calculates the expected number ( $E$ ):

$E=\frac{\text{(row total)(column total)}}{\text{total number surveyed}}$