Question

Hospital administrators wish to learn the average length of stay of all surgical patients. a statistician determines that, for a 95% confidence level estimate of the average length of stay to within +/- 0.5 days, 50 surgical patients' records will have to be examined. how many records should be looked at to obtain a 95% confidence level estimate to within +/- 0.25 days?

Answer 1

20,000 records should be looked

Solution:

Solve for s : 50 = [1.96*s/0.5]

25 = 1.96s

s = 12.75

Solve for "n": n = [1.96*12.75/0.25]^2

n = [100]^2

n = 20,000