SOLUTION: In a hospital, a sample of 8 weeks was selected, and it was found that an average of 438 patients were treated in the emergency room each week. The standard deviation was 16 patie

Algebra ->  Probability-and-statistics -> SOLUTION: In a hospital, a sample of 8 weeks was selected, and it was found that an average of 438 patients were treated in the emergency room each week. The standard deviation was 16 patie      Log On


   

Question 1141170: In a hospital, a sample of 8 weeks was selected, and it was found that an average of 438 patients were treated in the emergency room each week. The standard deviation was 16 patients. Find the 99% confidence interval for the true mean number of patients seen in the emergency room per week.
type of interval
estimate
z or t value
margin of error
interval
summary
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
Estimate is 438 for mean
It is a t-interval.
margin of error is t 0.995 df=7 or 3.499*16/sqrt(8)
the interval is +/- 19.8
the 99%CI is (418.2, 457.8)
There is very high confidence the true mean of number of patients seen in the emergency department will lie in the interval above, if the interval is a true random sampling of the number of patients seen each week.