Question 1083292: A random sample of health maintenance organizations (HMOs) was selected. For each HMO, the co-payment (in dollars) for a doctor's office visit was recorded. The results are as follows:
9, 8, 12, 5, 9, 12, 7, 5, 12, 6, 8, 11, 11, 11, 6, 6, 10, 5, 10
Under the assumption that co-payment amounts are normally distributed, find a 99% confidence interval for the mean co-payment amount in dollars.
Then complete the table below.
Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place.
(If necessary, consult a list of formulas.)
What is the lower limit of the confidence interval?
What is the upper limit of the confidence interval?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Mean is 8.578 and s=2.567
99%CI is a t df=18, 0.995*s/sqrt(19)
interval is 2.861*2.567/sqrt(19)
=1.685
mean +/- t*SE=
8.578+/-1.685=($6.9, $10.3)
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