SOLUTION: A hospital would like to determine the average length of stay for its  patients having abdominal surgery. A sample of 15 patients revealed a  sample mean of 6.4 days and a sample

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Question 847286: A hospital would like to determine the average length of stay for its 
patients having abdominal surgery. A sample of 15 patients revealed a 
sample mean of 6.4 days and a sample standard deviation of 1.4 days. 
Find a 95% confidence interval for the mean stay for patients with 
abdominal surgery?

Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 
Hi,
re TY, for myself like to go with one more decimal point originally..
and then make the decision to go to less accuracy to match parameters
used on the set Up.
ME = = .71
6.4 - ME < < 6.4 + ME
6.4 - .71 < < 6.4 + .71
5.69 < < 7.11
5.7 < < 7.1
Note:
80% z = 1.28155
90% z =1.645
92% z = 1.751
95% z = 1.96
98% z = 2.326
99% z = 2.576

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