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19 March, 21:11

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?

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  1. 19 March, 21:32
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    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
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