document.write( "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? \n" ); document.write( "
Algebra.Com's Answer #216533 by stanbon(75887)\"\" \"About 
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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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\n" ); document.write( "Solve for s :
\n" ); document.write( "100 = [1.96*s/0.5]
\n" ); document.write( "50 = 1.96s
\n" ); document.write( "s = 25.51
\n" ); document.write( "------
\n" ); document.write( "Solve for \"n\":
\n" ); document.write( "n = [1.96*25.51/0.25]^2
\n" ); document.write( "---
\n" ); document.write( "n = [200]^2
\n" ); document.write( "n = 40,000
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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