Question

he Onion'' recently conducted a survey of 50 families, and found that 95% of them were in a heated debate about the ideal sleeping location for cats. Compute the 90% confidence interval for the proportion of families who are in a heated debate about the ideal sleeping location for cats.

Selected Answer:	(0.899, 1.001)
Answers:	(0.855, 0.979)
	(0.890,1.010)
	( 0.843, 0.990)
	(0.852, 0.981)
	(0.899, 1.001)

Not sure how to figure this out?

Answer 1

We have n = sample size = 50

p^ = sample proportion = 0.95

The 90% confidence interval for population proportion p

p^ - Z_{​​​​​​a/2}* sqrt [p*q/n ] < p < p^ + Z_{​​​​​​a/2}* sqrt [ p*q/n]

For a = 0.10 , Z_{​​​​​0.05} = 1.645

0.95 - 1.645*sqrt[0.95*0.05/50] < p < 0.95 + 1.645*sqrt[0.95*0.05/50]

0.95 - 0.051 < p < 0.95 + 0.051

0.899 < p < 1.001

So 90% confidence intervalfor the proportion of families who are in a heated debate about the ideal sleeping location for cats is (0.899,1.001)