a) Perform a? chi-square test of homogeneity to determine if there is evidence t
ID: 3370905 • Letter: A
Question
a) Perform a? chi-square test of homogeneity to determine if there is evidence that blacks are less likely to rent in section A than in section B.
What is the value of the test statistic chi squared ?2?
b) Check the data for evidence of racial bias using? two-proportion z procedures. (solve for z)
c) Compare the? z-value to the value of chi squared ?2 calculated in part a.
New Renters A complex is divided into two parts: Section A and Section B. The plaintiffs in a lawsuit claimed that white potential renters were steered to section A, while black renters were steered to section B. The table displays the locations of recently rented apartments. Do you think there is evidence of a racial bias? Assume the conditions for inference are satisfied. Complete parts a through c. Black Section A Section B Total 90 87 29 40 101 115Explanation / Answer
A)
Ho:section and new renter are independent of each other
H1: section and New renter are dependent on each other
Degrees of freedom = (r-1)*(c-1) = 1
With chisq=7.14, p<5%, I reject the null hypothesis at 5% level of significance. And conclude that section and New renter are dependent on each other
B)
Ho: there is no significant difference in the proportion of black renters between section a and b. Pa = Pb
H1: the proportion of black renters are less likely in section A in as compared to Section B. Pa < Pb
P = (p1 * n1 + p2 * n2)/(n1 + n2)
=(0.275*40+0.725*40)/(40+40)
0.50000
SE = sqrt{ p * (1-p) * [ (1/n1) + (1/n2) ] }
=SQRT(0.5*(1-0.5)*(1/40+1/40))
0.111803
z = (p1 -p2)/SE
=(0.275-0.725)/0.111803
-4.024937
P(Z<z)
P(z<-4.024937)
=NORMSDIST(-4.024937)
2.84953E-05
With z=-4.02, p<5%I reject the null hypothesis at 5% level of significance and conclude that the proportion of black renters are less likely in section A in as compared to Section B. Pa < Pb
C)
Chi square value in part a = 7.148136
Z value in part B = -4.024937
The same result is obtained using chi-square test of virginity and Z test for difference in proportions
Observed Frequencies Renters section white black Total A 90 11 101 B 87 29 116 Total 177 40 217Related Questions
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