SOLUTION: The length of stay for patients in some 20 hospitals was 3.8 days with a standard deviation of 1.2 days. Can you determine the confidence interval for the length of stay?
a=0.1
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Question 894905: The length of stay for patients in some 20 hospitals was 3.8 days with a standard deviation of 1.2 days. Can you determine the confidence interval for the length of stay?
a=0.1 Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The length of stay for patients in some 20 hospitals was 3.8 days with a standard deviation of 1.2 days. Can you determine the confidence interval for the length of stay?
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x-bar = 3.8
ME = z*s/sqrt(n) = 2.5758*1.2/sqrt(20) = 0.6912
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99% CI:: 3.8-0.6912 < u < 3.8+0.6912
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Cheers,
Stan H.
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