Question

The accompanying data represent the total travel tax (in dollars) for a 3-day business trip in 8 randomly selected cities. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts (a) through (c) below. 68.55 80.03 69.12 84.74 81.82 88.55 101.67 98.76 (a) Determine a point estimate for the population mean travel tax. A point estimate for the population mean travel tax is $ 84.16. (Round to two decimal places as needed.) (b) Construct and interpret a 90% confidence interval for the mean tax paid for a three-day business trip. Select the correct choice below and fill in the answer boxes to complete your choice. (Round to two decimal places as needed.) O A. The travel tax is between $ and $ for % of all cities. OB. One can be % confident that all cities have a travel tax between $ and S OC. One can be % confident that the mean travel tax for all cities is between $ and $ OD. There is a % probability that the mean travel tax for all cities is between $ and $

Answer 1

sample mean, xbar = 84.16
sample standard deviation, s = 12.1407
sample size, n = 8
degrees of freedom, df = n - 1 = 7

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.895

ME = tc * s/sqrt(n)
ME = 1.895 * 12.1407/sqrt(8)
ME = 8.1

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (84.16 - 1.895 * 12.1407/sqrt(8) , 84.16 + 1.895 * 12.1407/sqrt(8))
CI = (76.03 , 92.29)

One can be 95% confident that the mean travel for all cities is between 6.03 and 92.29