Question

A researcher is interested in seeing if traffic stops in her city by local police changes based on the race of the individual. She obtains the racial distribution of drivers in the city from the census and then takes a sample of stops by race. The data is presented below. Using Chi-Square goodness of fit test the model that the data is matches the actual distribution of drivers of that race in the city at the .10 level. What do you conclude?

Model Data E O-E (O-E)2 (O-E)2/E

White 68% 154

Black 22% 64

Other 10% 32

Answer 1

Total Frequency = 154 + 64 + 32 = 250

The expected frequencies now are computed for each race as:
E(White) = 0.68*250 = 170
E(Black) = 0.22*250 = 55
E(Other) = 0.1*250 = 25

The chi square test statistic now is computed here as:

Therefore 4.939 is the required chi square test statistic value here.

For n - 1 = 2 degrees of freedom, we have from the chi square distribution tables:

As the p-value here is 0.0846 < 0.1 which is the level of significance, therefore the test is significant here and we can reject the null hypothesis here. Therefore we have sufficient evidence here that the actual distribution of drivers is different from the expected one.