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 White 92 78 170 Total 103 109 212 Black 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 Section A SectionB Total 31 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 ? (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. (Round to two decimal places as needed.) What is the P-value of th P-value- e two-proportion z-test? (Round to three decimal places as needed.)

Answer 1

using excel add ins tool PHStat, folllowing o/p is obtained,

Chi-Square Test
Observed Frequencies
	Column variable				Calculations
Row variable	white	Black	Total		fo-fe
section A	92	11	103		9.40566	-9.40566
section B	78	31	109		-9.40566	9.40566
Total	170	42	212
Expected Frequencies
	Column variable
Row variable	white	Black	Total		(fo-fe)^2/fe
section A	82.59434	20.40566	103		1.071096	4.335388
section B	87.40566	21.59434	109		1.012136	4.096742
Total	170	42	212

a) chi square stat = Σ(fo-fe)^2/fe = 10.515

df=(row-1)(column-1) = 1

p-value = 0.001

b)

first sample size,     n1=   103
number of successes, sample 1 =     x1=   92
proportion success of sample 1 , p̂1=   x1/n1=   0.893203883
      
second sample size,     n2 =    109
number of successes, sample 2 =     x2 =    78
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.71559633
      
difference in sample proportions,    p̂1 - p̂2 =     0.177607553
      
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.801886792
      
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)   0.054770829
      
Z-statistic =    (p̂1 - p̂2)/SE =     3.24
p-value=0.001

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c)

we can observe that

Z stat ^2 = chi square stat

3.24² = 10.515

so, answer is option c)

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d)

we can observe that, two p-values are equal.

answer: option b