Question

In a city, the racial make up is 68% White, 24% Black, 5% Asian and the remainder are classified as Other. A report on traffic stops by police officers in this city is being used to determine if the racial makeup of the motorists stopped reflect the racial makeup of the city. The race of drivers stopped by police officers over a 4 month period is recorded in the table. Determine if there is sufficient evidence to warrant the claim that the racial makeup of drivers in traffic stops significantly differs from the city's racial makeup.

Race	White	Black	Asian	Other
Drivers	675	276	40	39

At the 0.05 significance level, test the claim that the racial distribution of drivers stopped in traffic stops conforms to the city's distribution of races.
The test statistic is ?2=
The p-value is  
The conclusion is
A. There is sufficient evidence to claim the racial makeup of drivers pulled over in traffic stops does not reflect the racial makeup of the city.
B. There is not sufficient evidence to claim the racial makeup of drivers pulled over in traffic stops does not reflect the racial makeup of the city.

Answer 1

Null Hypothesis : H_o : The racial distribution of drivers stopped in traffic stops conforms to the city's distribution of races

Alternate Hypothesis : H_a:  The racial distribution of drivers stopped in traffic stops does not conform to the city's distribution of races

Where O : Observed Frequency

E : Expected Frequency

Given :

Race	White	Black	Asian	Other	Total
Drivers:Observe Frequency	675	276	40	39	1030
City Racial Makeup	68%	24%	5%	100-68-24-5=3	%

Expected frequency : City Racial makeup percentage x Total number drivers.

Race	White	Black	Asian	Other	Total
Drivers: Observe Frequency :O	675	276	40	39	1030
City Racial Makeup	68%	24%	5%	3%
Expected Frequnecy: E	68% of 1030	24% of 1030	5% of 1030	3% of 1030
Expected Frequnecy: E	700.4	247.2	51.5	30.9	1030

O	E	O-E	(O-E)2	(O-E)2/E
675	700.4	-25.4	645.16	0.921131
276	247.2	28.8	829.44	3.35534
40	51.5	-11.5	132.25	2.567961
39	30.9	8.1	65.61	2.123301
			Total	8.967733

Degrees of freedom = Number of Categories -1 = 4-1 = 3

p-Value = P( > Test Statistic ) = P( > 8.96773) = 0.0297

Level of significance : = 0.05

As p-value : 0.0297 < : 0.05 ; Reject null hypothesis ;

Therefore,

The conclusion is
A. There is sufficient evidence to claim the racial makeup of drivers pulled over in traffic stops does not reflect the racial makeup of the city.

p-value computation using excel:

CHISQ.DIST.RT function

Returns the right-tailed probability of the chi-squared distribution.

The ?2 distribution is associated with a ?2 test. Use the ?2 test to compare observed and expected values. For example, a genetic experiment might hypothesize that the next generation of plants will exhibit a certain set of colors. By comparing the observed results with the expected ones, you can decide whether your original hypothesis is valid.

Syntax

CHISQ.DIST.RT(x,deg_freedom)

The CHISQ.DIST.RT function syntax has the following arguments:

• X     Required. The value at which you want to evaluate the distribution.

• Deg_freedom     Required. The number of degrees of freedom.