A consumer protection agency is comparing the work of two electrical contractors. The agency plans to inspect residences in which each of these contractors has done the wiring in order toestimate the difference in the proportions of residences that are electrically deficient. Suppose the proportions of residences with deficient work are expected to be about .1 for both contractors. How many homes should be sampled in order to estimate the difference in proportions using a 90% confidence interval of width .1?
A) $n _ { 1 } = n _ { 2 } = 195$
B) $n _ { 1 } = n _ { 2 } = 98$
C) $n _ { 1 } = n _ { 2 } = 217$
D) $n _ { 1 } = \mathrm { n } _ { 2 } = 390$

Question

Quizplus · Accepted Answer

The sample size for each contractor is calculated as:$n = (Z^2 * p * q) / e^2$where:* Z is the z-score for the desired confidence level (90% = 1.645)* p is the expected proportion of residences with deficient work (0.1)* q is the complement of p (1 - p = 0.9)* e is the margin of error (0.1)Plugging in the values, we get:$n = (1.645^2 * 0.1 * 0.9) / 0.1^2 = 195.02$Rounding up to the nearest whole number, we get $n = 195$.

A Consumer Protection Agency Is Comparing the Work of Two n1=n2=195n _ { 1 } = n _ { 2 } = 195n1​=n2​=195

A Consumer Protection Agency Is Comparing the Work of Two $n _ { 1 } = n _ { 2 } = 195$