Question

Use the one-proportion z-interval procedure to find the required confidence interval.

Of 98 adults selected randomly from one town, 64 have health insurance. Find a 90% confidence interval for the proportion of all adults in the town who have health insurance.

Group of answer choices

0.541 to 0.765

0.574 to 0.732

0.559 to 0.747

0.529 to 0.777

Answer 1

n = 98

x = number of adults having health insurance 64

Sample proportion :

(Round to 4 decimal)

Confidence level = c = 0.90

90% Confidence interval for proportion of all adults in the town who have health insurance is

where zc is z critical value for (1+c)/2 = (1+0.90)/2 = 0.95

zc = 1.645 (From statistical table of z values, average of (1.64+1.65)/2 = 1.645

(Round to 3 decimal)

90% Confidence interval for proportion of all adults in the town who have health insurance is 0.574 to 0.732