In: Statistics and Probability
Car manufacturers produced a variety of classic cars that continue to increase in value. Suppose the following data is based upon the Martin Rating System for Collectible Cars, and shows the rarity rating (1–20) and the high price ($1,000) for 15 classic cars.
Model | Rating | Price ($1,000) |
---|---|---|
A | 16 | 125.0 |
B | 13 | 70.0 |
C | 17 | 140.0 |
D | 19 | 1,325.0 |
E | 19 | 3,975.0 |
F | 18 | 1,025.0 |
G | 14 | 87.0 |
H | 17 | 425.0 |
I | 19 | 2,675.0 |
J | 16 | 350.0 |
K | 16 | 250.0 |
L | 18 | 350.0 |
M | 17 | 425.0 |
N | 18 | 1,625.0 |
O | 15 | 77.5 |
b) Develop an estimated multiple regression equation with x = rarity rating and x2 as the two independent variables. (Round b0 and b1 to the nearest integer and b2 to one decimal place.)
ŷ =
c) Consider the nonlinear relationship shown by equation (16.7):E(y) = β0β1xUse logarithms to develop an estimated regression equation for this model. (Round b0 to three decimal places and b1to four decimal places.)
log(ŷ) =
Solution
we will solve it by using excel and the steps are
Enter the Data into excel
Model | Rating | Rating^2 | Price |
A | 16 | 256 | 125 |
B | 13 | 169 | 70 |
C | 17 | 289 | 140 |
D | 19 | 361 | 1,325.00 |
E | 19 | 361 | 3,975.00 |
F | 18 | 324 | 1,025.00 |
G | 14 | 196 | 87 |
H | 17 | 289 | 425 |
I | 19 | 361 | 2,675.00 |
J | 16 | 256 | 350 |
K | 16 | 256 | 250 |
L | 18 | 324 | 350 |
M | 17 | 289 | 425 |
N | 18 | 324 | 1,625.00 |
O | 15 | 225 | 77.5 |
Click on Data tab
Click on Data Analysis
Select Regression
Select input Y Range as Range of dependent variable.
Select Input X Range as Range of independent variable
click on labels if your selecting data with labels
click on ok.
So this is the output of Regression in Excel.
SUMMARY OUTPUT | ||||||||
Regression Statistics | ||||||||
Multiple R | 0.8 | |||||||
R Square | 0.7 | |||||||
Adjusted R Square | 0.7 | |||||||
Standard Error | 664.0 | |||||||
Observations | 15.0 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 2.0 | 12698545.4 | 6349272.7 | 14.4 | 0.0 | |||
Residual | 12.0 | 5290849.8 | 440904.2 | |||||
Total | 14.0 | 17989395.2 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | 34041.9 | 13568.8 | 2.5 | 0.0 | 4477.9 | 63605.8 | 4477.9 | 63605.8 |
Rating | -4596.9 | 1677.2 | -2.7 | 0.0 | -8251.3 | -942.5 | -8251.3 | -942.5 |
Rating^2 | 154.4 | 51.4 | 3.0 | 0.0 | 42.4 | 266.3 | 42.4 | 266.3 |
Price =34042 -4597*Rating+154.4*Rating^2
c) Consider the nonlinear relationship shown by equation (16.7):E(y) = β0β1xUse logarithms to develop an estimated regression equation for this model.
take log(price) and log(Rating)
SUMMARY OUTPUT | ||||||||
Regression Statistics | ||||||||
Multiple R | 0.8716 | |||||||
R Square | 0.7596 | |||||||
Adjusted R Square | 0.7411 | |||||||
Standard Error | 0.6654 | |||||||
Observations | 15.0000 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 1.0000 | 18.1908 | 18.1908 | 41.0854 | 0.0000 | |||
Residual | 13.0000 | 5.7558 | 0.4428 | |||||
Total | 14.0000 | 23.9467 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | -22.4232 | 4.4363 | -5.0545 | 0.0002 | -32.0073 | -12.8391 | -32.0073 | -12.8391 |
log(Rating) | 10.0919 | 1.5744 | 6.4098 | 0.0000 | 6.6905 | 13.4933 | 6.6905 | 13.4933 |
Log(Price ) = -22.423+10.0919*log(Rating)