Question

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. New Renters Black White Total Section A Section B Total NOTE: With respect, this data can be considered sensitive to some individuals, but the intent of this question is to better understand some aspects of a relevant social issue in society through the use of science; in this case, statistical analysis. Please keep in this mind as you complete the question. 97 128 224 182 42 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 x?? x2 = (Round to three decimal places as needed.) What is the P-value of the chi-square test of homogeneity? P-value = (Round to three decimal places as needed.) b) You can also check the data for evidence of racial bias using two-proportion z procedures Find the z-value for this approach. z= (Round to two decimal places as needed.) What is the P-value of the two-proportion z-test? P-value=N (Round to three decimal places as needed.)

Answer 1

PART A.
Given table data is as below

MATRIX	col1	col2	TOTALS
row 1	85	11	96
row 2	97	31	128
TOTALS	182	42	N = 224

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calculation formula for E table matrix
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expected frequencies calculated by applying E - table matrix formula

E-TABLE	col1	col2
row 1	row1*col1/N	row1*col2/N
row 2	row2*col1/N	row2*col2/N

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calculate chi-square test statistic using given observed frequencies, calculated expected frequencies from above

E-TABLE	col1	col2
row 1	78	18
row 2	104	24

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Oi	Ei	Oi-Ei	(Oi-Ei)^2	(Oi-Ei)^2/Ei
85	78	7	49	0.6282
11	18	-7	49	2.7222
97	104	-7	49	0.4712
31	24	7	49	2.0417
				ᴪ^2 o = 5.8633

set up null vs alternative as
level of significance, alpha = 0.05
from standard normal table, chi square value at right tailed, ᴪ^2 alpha/2 =3.8415
since our test is right tailed,reject Ho when ᴪ^2 o > 3.8415
we use test statistic ᴪ^2 o = Σ(Oi-Ei)^2/Ei
from the table , ᴪ^2 o = 5.8633
critical value
the value of |ᴪ^2 alpha| at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 2 - 1 ) = 1 * 1 = 1 is 3.8415
we got | ᴪ^2| =5.8633 & | ᴪ^2 alpha | =3.8415
make decision
hence value of | ᴪ^2 o | > | ᴪ^2 alpha| and here we reject Ho
ᴪ^2 p_value =0.0155
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ᴪ^2 test statistic: 5.8633
p-value:0.0155  

PART B.

Given that,
sample one, x1 =11, n1 =96, p1= x1/n1=0.115
sample two, x2 =31, n2 =128, p2= x2/n2=0.242
finding a p^ value for proportion p^=(x1 + x2 ) / (n1+n2)
p^=0.188
q^ Value For Proportion= 1-p^=0.813
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.115-0.242)/sqrt((0.188*0.813(1/96+1/128))
zo =-2.421
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.4214 ) = 0.0155
hence value of p0.05 > 0.0155, here we reject Ho
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test statistic: -2.421
p-value: 0.0155