Question 1159792: A corporation that maintains a large fleet of company cars for the use of its sales staff is interested in the mean distance driven monthly per sales person. The following table gives the monthly distances in miles driven by a random sample of 16 sales persons:
2415, 2221, 2481, 2618, 2660, 1967, 2260, 1938, 2410, 2062, 2466, 2484, 2603, 1994, 2328, 1861
Based on this sample, find a 90% confidence interval for the mean number of miles driven monthly by members of the sales staff, assuming that monthly driving distances are normally distributed.
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 90% confidence interval?
What is the upper limit of the 90% confidence interval?
Found 2 solutions by Boreal, Edwin McCravy:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! mean is 2298
sd is 262.75
half-interval is t(0.95df=15)*s/sqrt(n)
=1.753*262.75/4
=115.15
the 90%CI is (2182.85, 2413.15) units miles
Answer by Edwin McCravy(20086)  (Show Source):
You can put this solution on YOUR website!
On your TI-84, press STAT, choose EDIT, put DATA numbers in L1.
press STAT again, press right arrow key to highlight TESTS choose TInterval
Choose Data, List:L1, Freq:1, C-Level:0.90, Calculate ENTER
TInterval
(2182.8,2313.2)
x̅: 2298
Sx=262.7452505
n:16
2182.8 ≤ x ≤ 2413.2 <--answer for confidence interval
Edwin
|
|
|
