Question

The Traveler Rent-A-Car Company is interested in estimating the mean

number of miles its cars are driven on a particular holiday. It has 23,000 cars

nationwide and samples 200 cars on the holiday in question. The mileage for

each car was recorded. The following data were computed from the sample data:

x bar = 54.5

s = 14.0

a. Produce a 95% confidence interval estimate for the mean number of miles

driven by each car. (8 pts)

b. Traveler’s vice-president of operations received a report from a city in the

company’s southwestern region. It indicated the location had rented 200 cars

during the holiday and had received $2,500 in fees charged for mileage. Assume

the Traveler charges $0.25 per mile as a mileage fee. On the basis of the

confidence interval, would you say that the vice president should investigate the

billing practices of the location? Support your opinion with statistical reasoning

and logic. (5 pts)

Answer 1

a)

Level of Significance ,    α =    0.05          
'   '   '          
z value=   z α/2=   1.9600   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   14.0000   / √   200   =   0.989949
margin of error, E=Z*SE =   1.9600   *   0.98995   =   1.940265
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    54.50   -   1.940265   =   52.5597
Interval Upper Limit = x̅ + E =    54.50   -   1.940265   =   56.4403
95%   confidence interval is (   52.56   < µ <   56.44   )

b)

2500$ for 200 car

so for 1 car=2500/200=12.5$

so average distance by each car=12.5/0.25=50

as average distance is less than confidence interval it is profitable to give car on rent