SOLUTION: Hospital administrators wish to determine the average length of stay for all surgical patients. A statistician determines that a 95% confidence level estimate of the average lengt

Question 301945: Hospital administrators wish to determine the average length of stay for all surgical patients. A statistician determines that a 95% confidence level estimate of the average length of stay to within 0.50 days, 100 surgical patients' records would have to be examined. How many records should be looked at for a 95% confidence level within 0.25 days?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Hospital administrators wish to determine the average length of stay for all surgical patients. A statistician determines that a 95% confidence level estimate of the average length of stay to within 0.50 days, 100 surgical patients' records would have to be examined. How many records should be looked at for a 95% confidence level within 0.25 days?
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Solve for s :
100 = [1.96*s/0.5]
50 = 1.96s
s = 25.51
------
Solve for "n":
n = [1.96*25.51/0.25]^2
---
n = [200]^2
n = 40,000
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Cheers,
Stan H.
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