Answer:
(7.115 ; 13.405)
Explanation:
Given :
Sample size, n = 22
Mean bill, μ = 10.26
Standard deviation, s = 5.21
To obtain the 99% confidence interval for the mean bill of all orders ;
Mean ± margin of error
Margin of Error = Tcritical * s/√n
Tcritical at 99%, df = n-1, 22 - 1 = 21
Tcritical = 2.831
Margin of Error = 2.831 * (5.21/√22) = 3.145
Confidence interval = 10.26 ± 3.145
Lower boundary = 10.26 - 3.145 = 7.115
Upper boundary = 10.26 + 3.145 = 13.405
Confidence interval :
(7.115 ; 13.405)