To budget for the coming year, a community hospital must estimate the proportion of treated patients who will not be able to make payment in full.In a random sample of 2568 patients treated in the previous year, 381 failed to make payment in full.What is an 80% confidence interval for the proportion of patients who will not make full payment to the hospital?
A) (0.137, 0.160)
B) (0.125, 0.162)
C) (0.139, 0.157)
D) (0.132, 0.165)

Question

Quizplus · Accepted Answer

We can use a confidence interval formula for a proportion:
p̂ ± z*sqrt((p̂(1-p̂))/n)
where p̂ is the sample proportion, z is the z-score corresponding to the desired confidence level (80% in this case, which corresponds to a z-score of 1.28), and n is the sample size. 
Plugging in the values, we get:
381/2568 ± 1.28*sqrt(((381/2568)(1-381/2568))/2568)
0.148 ± 0.010
The 80% confidence interval is therefore (0.138, 0.158), which is closest to choice C.

To Budget for the Coming Year, a Community Hospital Must