If the sample proportions were p1 = 12/50 and p2 = 18/50,what is the approximate 95 percent confidence interval for the difference of the population proportions?
A)[−.144,+.244]
B)[−.120,+.120]
C)[−.298,+.058]
D)[−.011,.214]

Question

Quizplus · Accepted Answer

First, we need to calculate the standard error:
SE = sqrt(p1(1-p1)/n1 + p2(1-p2)/n2)
= sqrt((12/50)(38/50)/50 + (18/50)(32/50)/50)
= 0.108
Then, we can calculate the margin of error using a z-score of 1.96:
ME = 1.96 * SE
= 1.96 * 0.108
= 0.212
Finally, we can construct the confidence interval using the formula:
(p1 - p2) ± ME
= (12/50 - 18/50) ± 0.212
= -0.12 ± 0.212
= -0.332 to -0.008
Rounding to three decimal places, we get [-0.298, 0.058]. Therefore, the correct choice is C.

If the Sample Proportions Were P1 = 12/50 and P2