Question

The Toyota Camry is one of the best-selling cars in North America. The cost of a previously owned Camry depends upon many factors, including the model year, mileage, and condition. Our interest is to estimate the sales price using the car’s mileage. The file “ToyotaCamry” contains data for sales price and the car’s mileage for a 2007 model year Camry for 19 sales (PriceHub website, February 24, 2012). As demonstrated in the lecture, please create a subset data of size 15 and perform your statistical analysis for the subset data. Please note that the subset data should be a random sample of the given data.

Car       price        mileage

4            20400   36504

5            19495   23781

6            18700   33796

7            15995   51706

8            15000   78251

10          13900   59060

12          13500   86137

13          12999   53969

14          12500   54718

15          11600   50366

16          11400   64567

17          11295   69953

18          9995      92367

19          9995      78880

20          9600      55512

(c) State the estimated regression equation that can be used to predict the sales price given the miles.

(d) Suppose that you are considering purchasing a previously owned 2007 Camry that has been driven 50,000 miles. Using the estimated regression equation developed in part (c), predict the price for this car. Is this interpolation or extrapolation? Explain.
(e) Provide an interpretation for the slope of the estimated regression equation.
(f) Construct and interpret a 97% confidence interval for the coefficient of slope.

Answer 1

let price=y, milege=x

y=a+bx

b=sum((xi=xbar)*(yi-ybar))/sum(xi-xbar)^2=-0.126

a=ybar-b*xbar=21252
The regression equation is
Price = 21252 - 0.126 mileage

for 50000miles predicted price of car is (i.e for mileage=50000)

Price = 21252 - 0.126(50000)=14952

predicted price of the car is interpolation because this price lies in the previous prices of the car.

e) interpretation of slope:

as unit change in mileage of car, price of car decreased by 0.126 units.

f)97% cinfidence interval for slope

n=15,

ttab=t(0.015,n-2)=-2.06

MSE=(yi-ypre)^2/n-2=6694241

se(b)=sqrt(MSE/sum(xi-xbar)^2)=0.03544

lb=b-Ztab*se(b)=-0.126-(-2.06*0.03544)=-0.053

ub=b-Ztab*se(b)=-0.126+(-2.06*0.03544)=-0.199

(-0.053,-0.199)