Question

Bus travel. The length of time a bus takes to travel from two locations in a big city is normally distributed. A random sample of 6 trips took 23, 19, 25, 34, 24, and 28 minutes, respectively. Assume that the sample standard deviation is 5.09. a) Determine the margin of error (i.e. error of the estimate) at 90% confidence. b) Determine and interpret the 90% confidence interval for the true mean travel time. c) Suppose that the bus company claims that the average bus travel time is 20 minutes. Would you say this is consistent with the results? Why or why not?

Answer 1

sample mean, xbar = 25.5
sample standard deviation, s = 5.09
sample size, n = 6
degrees of freedom, df = n - 1 = 5

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 2.015

a)
Margin of error,
ME = tc * s/sqrt(n)
ME = 2.015 * 5.09/sqrt(6)
ME = 4.187

b)
CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (25.5 - 2.015 * 5.09/sqrt(6) , 25.5 + 2.015 * 5.09/sqrt(6))
CI = (21.31 , 29.69)

One can be 90% confident that the mean of a sample selected from the population of size 6, the mean will lie within (21.31 , 29.69)

c)
No, the results are not consistent with the claim as 20 is not included in the CI