Of 88 adults selected randomly from one town,60 have health insurance.Construct a 90% confidence interval for the percentage of all adults in the town who have health insurance.
A)(58.5%,77.9%)
B)(61.0%,75.4%)
C)(56.6%,79.8%)
D)(55.4%,81.0%)
E)(60.0%,76.3%)

Question

Quizplus · Accepted Answer

The sample size is 88, and 60 have health insurance. So the sample proportion is 60/88 = 0.682. The standard error of the proportion is sqrt(0.682 * 0.318 / 88) = 0.047. The 90% confidence interval is 0.682 +/- 1.645 * 0.047 = (0.600, 0.763).

Of 88 Adults Selected Randomly from One Town,60 Have Health