Question

A statistical program is recommended.

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	18	1,575.0
B	14	37.0
C	15	52.5
D	18	1,000.0
E	13	95.0
F	19	1,275.0
G	19	2,675.0
H	17	375.0
I	16	325.0
J	17	165.0
K	17	425.0
L	16	225.0
M	16	150.0
N	18	400.0
O	19	4,025.0

(a)

Develop a scatter diagram of the data using the rarity rating as the independent variable and price as the independent variable.

Does a simple linear regression model appear to be appropriate?

No, there appears to be a curvilinear relationship between the two variables.

No, there doesn't appear to be a relationship between the two variables.   

Yes, there appears to be a linear relationship between the two variables.

(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) = β₀β₁^x

Use logarithms to develop an estimated regression equation for this model. (Round b₀ to three decimal places and b₁ to four decimal places.)

log(ŷ) =

(d)

Do you prefer the estimated regression equation developed in part (b) or part (c)? Explain.

The model in part (b) is preferred because r² is higher and the p-value is lower.

The model in part (b) is preferred because r² is lower and the p-value is lower.   

The model in part (c) is preferred because r² is lower and the p-value is lower.

The model in part (c) is preferred because r² is higher and the p-value is lower.

Answer 1

a)

Yes, there appears to be a linear relationship between the two variables.

b)

We will be applying the Linear regression model here, it can be done by using the function =LINEST(y_value, x_value, TRUE, TRUE) where y_values contain values of Price(y) here and x_values have Rating(x) and Rating square(x**2) values.

Select 5 rows and 3 columns and then write the formula in the first cell and after that, press Shift + Ctrl + Enter.    

The equation comes out to be -
y = 34473.67 - 4653.34*x + 156.15*x**2  

c)

We will be applying the Linear regression model here, it can be done by using the function =LINEST(y_value, x_value, TRUE, TRUE) where y_values contain values of log of price here and x_values have ratings values.

Select 5 rows and 2 columns and then write the formula in the first cell and after that, press Shift + Ctrl + Enter.    

The equation comes out to be -
log(y) = -5.5 + 0.68*x  

d)

The model in part c have the r**2 value of 0.79, which is higher than r**2 value of 0.70 in part b. Hence, it will be used.

The model in part (c) is preferred because r² is higher and the p-value is lower.