Ask Question
13 January, 14:47

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?

+2
Answers (1)
  1. 13 January, 16:34
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Hospital administrators wish to learn the average length of stay of all surgical patients. a statistician determines that, for a 95% ...” in 📙 Business if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers