Question

A city in a particular county has a population of 459,649; the population of the county is 9,519,338. Conduct a goodness of fit test at the 5% level to determine if the racial demographics of the city fit that of the county. Round expected frequency to two decimal places.

Race	Percent, county	Expected #, city	Actual #, city
American Indian and Alaska Native	0.8	3677.19	3,865
Asian	11.9	54698.23	55,595
Black or African American	9.8	45045.60	68,641
Native Hawaiian and other Pacific Islander	0.3	1378.95	5,615
White, including Hispanic/Latino	48.7	223849.06	206,526
Other	23.5	108017.52	95,031
Two or more races	5.0	22982.45	24,376

(1) What are the degrees of freedom = 6

(2) What is the test statistic = 28383

(3) Distribution for this test = X^2₆

(3) What is the p-value = 0.0000

(4) NEED HELP WITH #4

Sketch a picture of this situation. Label and scale the horizontal axis, and shade the region(s) corresponding to the p-value. (Upload your file below.)

(5) Alpha a= 0.05

Answer 1

a) The degrees of freedom here is computed as:

Df = num of categories - 1 = k - 1 = 7 - 1 = 6
Therefore 6 is the degrees of freedom here.

b) The test statistic here is computed as:

The computations here are made as:

Race	Percent, county	Expected #, city	Actual #, city	(O_i - E_i)^2/E_i
American Indian and Alaska Native	0.8	3677.19	3,865	9.592269124
Asian	11.9	54698.23	55,595	14.7024215
Black or African American	9.8	45045.6	68,641	12359.54014
Native Hawaiian and other Pacific Islander	0.3	1378.95	5,615	13012.88633
White, including Hispanic/Latino	48.7	223849.06	2,06,526	1340.58373
Other	23.5	108017.52	95,031	1561.318032
Two or more races	5	22982.45	24,376	84.49845872
				28383.12138

Therefore 28383.12138 is the test statistic value here.

c) For 6 degrees of freedom, as we are doing a goodness of fit test here, the distribution of the test statistic here is given as:

d) Note that as the test statistic value here is so high, the p-value here is obtained as:

As the p-value here is 0, there will be no area shaded here in the chi square graph. This is shown here as:

Note that there is no shaded area here, as the p-value here is approx 0 here.

Q5) As the p-value here is 0 < 0.05 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 racial demographics of the city do not fit that of the county