Question

One criticism of racial profiling studies is that people’s driving frequency is often unaccounted for. This is a problem because, all else being equal, people who spend more time on the road are more likely to get pulled over eventually. The following table contains PPCS data narrowed down to black male respondents. The variables measure driving frequency and whether these respondents had been stopped by police for traffic offenses within the past 12 months. With an alpha of .01, conduct a five-step hypothesis test to determine if the variables are independent.

Driving Frequency	Yes	No	Raw Marginal
Almost Every Day	214	946	1,160
Often	32	238	270
Rarely	6	360	366
Column Marginal	252	1,544	N=1,796

Answer 1

We have to perform Chi-square test for independence of two random variables (driving frequency and stop by police).

We have to test for null hypothesis

Ho : Driving frequency and stop by police are independent

against the alternative hypothesis

Hi : Driving frequency and stop by police are not independent

Observed frequencies $\tiny \left(f_{ij}\right)$ are as follows.

Row total Observed frequencies Driving Almost everyday frequency Often Rarely Column total Stopped by police Yes No 214 946 32 238 6 360 252 1544 1160 270 366 1796

Under null hypothesis, we obtain expected frequencies $\tiny \left(e_{ij}\right)$ by multiplying corresponding row total and column total and dividing it by grand total as follows.

Stopped by police Expected frequencies Row total Yes No Almost everyday 162.76 997.24 1160.00 Driving Often 37.88 232.12 270.00 frequency Rarely 51.35 314.65 366.00 Column total 252.00 1544.00 1796.00

Our Chi-square test statistic is given by

$\tiny \chi^2=\sum_{i=1}^{m}\sum_{j=1}^{n}\frac{\left(f_{ij}-e_{ij}\right)^2}{e_{ij}}$

Here,

Number of rows $\tiny m=3$

Number of columns $\tiny n=2$

Corresponding calculations (chi square component for each cell, which are to be added later) are as follows.

Row total Chi_square Stopped by police Yes No Almost everyday 16.13 2.63 Driving Often 0.91 0.15 frequency Rarely 40.06 6.54 Column total 57.10 9.32 18.76 1.06 46.59 66.42

$\tiny \therefore \chi^2_{calculated}=\sum_{i=1}^{3}\sum_{j=1}^{2}\frac{\left(f_{ij}-e_{ij}\right)^2}{e_{ij}}=66.42$

Degrees of freedom

v = (m 1) (n − 1) = 2

$\tiny \therefore \text{p-value}=P\left(\chi^2_{2}>66.42\right)=3.774758*10^{-15}$ [Using R-code '1-pchisq(66.42,2)']

Level of significance

a = 0.01

We reject our null hypothesis if $\tiny \text{p-value}<\alpha$

Here, we observe that $\tiny \text{p-value}=3.774758*10^{-15}<0.01=\alpha$

So, we reject our null hypothesis.

Hence, based on the given data we can conclude that there is significant evidence that driving frequency and stop by police are not independent.