Question

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	17	400.0
B	16	325.0
C	15	102.5
D	13	70.0
E	14	37.0
F	16	225.0
G	18	350.0
H	17	140.0
I	18	1,575.0
J	16	150.0
K	18	1,000.0
L	19	2,675.0
M	19	1,275.0
N	19	4,025.0
O	17	450.0

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

(a1)Does a simple linear regression model appear to be appropriate?

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

-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.

(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 (c) is preferred because r² is higher 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 (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.

Answer 1

Solution

a

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

b)Develop an estimated multiple regression equation with x = rarity rating and x² as the two independent variables.

we will solve it by using excel and the steps are

Enter the Data into excel

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	0.835
R Square	0.697
Adjusted R Square	0.646
Standard Error	678.704
Observations	15.000
ANOVA
	df	SS	MS	F	Significance F
Regression	2.000	12713138.982	6356569.491	13.799	0.001
Residual	12.000	5527672.918	460639.410
Total	14.000	18240811.900
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	33988.403	13869.174	2.451	0.031	3770.069	64206.737	3770.069	64206.737
Rating	-4591.840	1714.375	-2.678	0.020	-8327.142	-856.538	-8327.142	-856.538
Rating^2	154.233	52.522	2.937	0.012	39.798	268.668	39.798	268.668

Price = 33988 - 4592*Ratin + 154.2*Rating^2

(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.

Log(Price) = -24.251 +10.7254*log(Rating)

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.8889
R Square	0.7901
Adjusted R Square	0.7740
Standard Error	0.6479
Observations	15.0000
ANOVA
	df	SS	MS	F	Significance F
Regression	1.0000	20.5464	20.5464	48.9483	0.0000
Residual	13.0000	5.4569	0.4198
Total	14.0000	26.0033
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	-24.2509	4.3196	-5.6142	0.0001	-33.5827	-14.9190	-33.5827	-14.9190
log(Rating)	10.7254	1.5330	6.9963	0.0000	7.4135	14.0373	7.4135	14.0373

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

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