Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
-Of 92 adults selected randomly from one town, 61 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance.
A) 0.582

Question

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.
-Of 92 adults selected randomly from one town, 61 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance.
A) 0.582 < p < 0.744
B) 0.566 < p < 0.760
C) 0.548 < p < 0.778
D) 0.536 < p < 0.790

Quizplus · Accepted Answer

We use the formula:
sample proportion ± margin of error 
where sample proportion = 61/92 ≈ 0.663,
margin of error = z* (standard error)
with z* corresponding to the 90% confidence level, which is 1.645,
and standard error = sqrt[pq/n], where q = 1-p.
Thus, we have:
0.663 ± 1.645*sqrt[(0.663*0.337)/92]
= 0.663 ± 0.081
= (0.582, 0.744)
Therefore, the 90% confidence interval for p is (0.582, 0.744), which is option A.

Use the Given Degree of Confidence and Sample Data to Construct