Question 1086581: You are estimating the cost of engine overhauls. A sample of 49 repairs showed an average overhaul of 285 hours with a standard deviation of 60 hours. Calculate a 95% confidence interval for the average overhaul. (Carry intermediate calculations to three decimal places.)
a. 271.00 ≤ μ ≤ 299.00
b. 268.20 ≤ μ ≤ 301.80
c. 281.61 ≤ μ ≤ 288.39
d. 284.52 ≤ μ ≤ 285.48
You are estimating the cost of a personnel shelter and you have collected cost data on 10 shelters. The mean cost was $16,693. The median cost is $17,402. There is a range in the cost of $10,095. The standard deviation is $2,270. If you use the average cost of the shelters as your estimate, then, on average:
a. You could expect to be off by $10,095.
b. You could expect to be off by 20.42%.
c. You could expect to be off by 13.60%.
d. You could expect to be off by $340.90.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! I'll do the first one to get you started
z = 1.960 (z critical value at 95% confidence)
n = 49
xbar = 285
s = 60
The lower limit of the confidence interval is
L = xbar - z*(s/sqrt(n))
L = 285 - 1.96*(60/sqrt(49))
L = 285 - 1.96*(60/7)
L = 285 - 1.96*(8.57142857142857)
L = 285 - 16.8
L = 268.2
The upper limit of the confidence interval is
U = xbar + z*(s/sqrt(n))
U = 285 + 1.96*(60/sqrt(49))
U = 285 + 1.96*(60/7)
U = 285 + 1.96*(8.57142857142857)
U = 285 + 16.8
U = 301.8
(L,U) = (268.2,301.8)
The final answer is choice B) 268.20 < µ < 301.80
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