In 1960 sociologists studied a random sample of 1,018 families that consisted of a husband, a wife, and at least one child. Of those families, 5.8 percent reported that the wife was the primary wage earner of the family. In 2011 the study was replicated with a random sample of 1,013 families that consisted of a husband, a wife, and at least one child. Of those families, 22.3 percent reported that the wife was the primary wage earner of the family. Which of the following represents a 99 percent confidence interval for the difference between the proportions of families that consisted of a husband, a wife, and at least one child from 1960 to 2011 that would have reported the wife as the primary wage carner? (0.223)(0.777) (0.058) (0.942)
(A) (0.223-0.058) 1.96L03 ) (223-0058) ±196 1.013 1 1.018
(B) (0.223-0058) 226(07,(0580942 1,013 1,018
(C) (0.223-0058)2.576, (02230772)(0.45810942 (0.058)(0.942) 1,018
(D) (0.223-0.058) t2.326, 2,031 1,013 1.018
(E) (0.223-0058)?2.57612031(1.013 +1.18

Answer:

(0.223 - 0.058) ± 2.576 * sqrt[((0.223)(0.777))/1018 - ((0.058)(0.942))/1013]

Step-by-step explanation:

Given that :

P1 = 22.3% = 0.223

q1 = 1 - P1 = 1 - 0.223 = 0.777

Sample size, n1 = 1018

P2 = 5.8% = 0.058

q2 = 1 - P2 = 1 - 0.058 = 0.942

Sample size, n2 = 1013

Confidence level, Zcritical at 99% = 2.576

Using the relation :

(P1 - P2) ± Zcritical * Sqrt[(P1q1/n1 + P2q2/n2)

(0.223 - 0.058) ± 2.58 * sqrt[((0.223)(0.777))/1018 - ((0.058)(0.942))/1013]

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