document.write( "Question 508913: The length of stay (in hours) for a sample of 12 pneumonia patients at Santa Theresa Memorial Hospital was found to have a mean of 126 hours and a standard deviation of 21.663 hours. Find the 90 percent confidence interval on the standard deviation of the length of stay. \n" ); document.write( "
Algebra.Com's Answer #341274 by stanbon(75887)\"\" \"About 
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The length of stay (in hours) for a sample of 12 pneumonia patients at Santa Theresa Memorial Hospital was found to have a mean of 126 hours and a standard deviation of 21.663 hours. Find the 90 percent confidence interval on the standard deviation of the length of stay.
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\n" ); document.write( "sample mean = 126
\n" ); document.write( "margin of error = 1.645*21.663/sqrt(12) = 10.29
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\n" ); document.write( "90%CI: 126-10.29 < u < 126+10.29
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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