Question

(Round all intermediate calculations to at least 4 decimal places.) Consider the following sample regressions for the linear, the quadratic, and the cubic models along with their respective R2 and adjusted R2. Linear Quadratic Cubic Intercept 25.97 20.73 16.20 x 0.47 2.82 6.43 x2 NA −0.20 −0.92 x3 NA NA 0.04 R2 0.060 0.138 0.163 Adjusted R2 0.035 0.091 0.093 pictureClick here for the Excel Data File a. Predict y for x = 3 and 5 with each of the estimated models. (Round your answers to 2 decimal places.) Linear yˆ Quadratic yˆ Cubic yˆ x = 3 x = 5 b. Select the most appropriate model. Cubic model Quadratic model Linear model

	Linear	Quadratic	Cubic
Intercept	25.97	20.73	16.20
x	0.47	2.82	6.43
x2	NA	-0.20	-0.92
x3	NA	NA	0.04
R2	0.060	0.138	0.163
Adjusted R2	0.035	0.091	0.093

Answer 1

Answer

(A)Linear model equation is y = 25.97 + 0.47*x

for x = 3, we get

= 25.97+0.47*3

= 27.38

and

for x = 5, we get

= 25.97+0.47*5

= 28.32

Quadratic model equation is y = 20.73 + 2.82*(x1)^2 -0.20*(x2)

for x = 3, we get

= 20.73 + 2.82*(3)^2 -0.20*(3)

=45.51

and

for x = 5, we get

= 20.73 + 2.82*(5)^2 -0.20*(5)

= 90.23

Cubic model equation is y = 16.20 + 6.43*(x1)^3 -0.92*(x2^2)+0.04(x3)

for x = 3, we get

= 16.20 + 6.43*(3)^3 -0.92*(3^2)+0.04(3)

= 181.65

and

for x = 5, we get

= 16.20 + 6.43*(5)^3 -0.92*(5^2)+0.04(5)

= 797.15

(b)We have to compare the value of coefficient of determination or R^2 and the adjusted R squared values in order to determine which model is the best

Using the given data table, it is clear that the value of coefficient of determination is highest for cubic model, i.e. R^2 = 0.163

and the value of adjusted R squared coefficient is also highest for the cubic model, i.e. adj R^2 = 0.093

cubic model is also giving highest estimates  

therefore, we can say that the cubic model is best suited for the given data set as it has largest R^2 as well as largest adj R^2