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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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):
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
