Question

Suppose you are trying to determine the capacity (in gallons) of the gas tank needed on an airplane you are constructing. You want to be able to travel 3200 nautical miles without stopping, and have gathered data on the amount of fuel similar planes used during flights of comparable length. Show complete calculation and your steps, also interpetation and explanation as asked.

Consider a sample with the following properties: x̅ = 261.5, s = 18.73, n = 26

A) Calculate a confidence interval with α = 0.10

B) Calculate a confidence interval with α = 0.01

C) How would you interpret the results for the confidence interval from part B?

Answer 1

a)

sample mean, xbar = 261.5
sample standard deviation, s = 18.73
sample size, n = 26
degrees of freedom, df = n - 1 = 25

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.708

ME = tc * s/sqrt(n)
ME = 1.708 * 18.73/sqrt(26)
ME = 6.274

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (261.5 - 1.708 * 18.73/sqrt(26) , 261.5 + 1.708 * 18.73/sqrt(26))
CI = (255.23 , 267.77)

b)

sample mean, xbar = 261.5
sample standard deviation, s = 18.73
sample size, n = 26
degrees of freedom, df = n - 1 = 25

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.787

ME = tc * s/sqrt(n)
ME = 2.787 * 18.73/sqrt(26)
ME = 10.237

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (261.5 - 2.787 * 18.73/sqrt(26) , 261.5 + 2.787 * 18.73/sqrt(26))
CI = (251.26 , 271.74)

c)

Therefore, based on the data provided, the 99% confidence interval for the population mean is 251.26 < μ < 271.74 which indicates that we are 99% confident that the true population mean μ is contained by the interval (251.26 , 271.74)