document.write( "Question 847286: A hospital would like to determine the average length of stay for its 
\n" ); document.write( "patients having abdominal surgery. A sample of 15 patients revealed a 
\n" ); document.write( "sample mean of 6.4 days and a sample standard deviation of 1.4 days. 
\n" ); document.write( "Find a 95% confidence interval for the mean stay for patients with 
\n" ); document.write( "abdominal surgery? \n" ); document.write( "
Algebra.Com's Answer #510324 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
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document.write( "Hi,
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document.write( "re TY, for myself like to go with one more decimal point originally..
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document.write( "and then make the decision to go to less accuracy to match parameters
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document.write( "used on the set Up.
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document.write( "ME = \"1.96%2A1.4%2Fsqrt%2815%29+\" = .71
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document.write( "  6.4 - ME < \"mu\" < 6.4 + ME
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document.write( "  6.4 - .71 < \"mu\" < 6.4 + .71
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document.write( "5.69 < \"mu\" < 7.11
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document.write( " 5.7 < \"mu\" < 7.1
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document.write( "Note:
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document.write( "80%	z = 1.28155
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document.write( "90%	z =1.645
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document.write( "92%	z = 1.751
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document.write( "95%	z = 1.96
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document.write( "98%	z = 2.326
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document.write( "99%	z = 2.576
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