85E Consider the model:E(y)=尾0+尾1x... [FREE SOLUTION]

91影视

Statistics For Business And Economics

James T. McClave, P. George Benson, Terry Sincich

$Math Studyset 91影视 Explanations$ Math

13th Edition 路 888 pages

ISBN: 9780134506593

Chapter 12: 85E (page 764)

Consider the model:
$E (y) = 尾_{0} + 尾_{1} x_{1} + 尾_{2} x_{2} + 尾_{3} {x_{2}}^{2} + 尾_{4} x_{3} + 尾_{5} x_{1} {x_{2}}^{2}$
where x₂ is a quantitative model and
$x_{1} = (\begin{matrix} 1 r e c e i v e d t r e a t m e n t \\ 0 d i d n o t r e c e i v e t r e a t m e n t \end{matrix})$
The resulting least squares prediction equation is
localid="1649802968695" $\overset{铃�}{y} = 2 + x_{1} - 5 x_{2} + 3 {x_{2}}^{2} - 4 x_{3} + x_{1} {x_{2}}^{2}$
a. Substitute the values for the dummy variables to determine the curves relating to the mean value E(y) in general form.
b. On the same graph, plot the curves obtained in part a for the independent variable between 0 and 3. Use the least squares prediction equation.

Short Answer

Expert verified

The mean value of E(y)for x₁ = 0 is $E (y) = 尾_{0} + 尾_{2} x_{2} + 尾_{3} {x_{2}}^{2} + 尾_{4} x_{3}$ andthe mean value of E(y)for x₁= 1 is . $E (y) = (尾_{0} + 尾_{1}) + 尾_{2} x_{2} + 尾_{3} {x_{2}}^{2} + 尾_{4} x_{3} + 尾_{5} {x_{2}}^{2}$

Step by step solution

Dummy variable equation in general form

The mean value of E(y) in general form can be written for the value of x₁ = 1 and x₁= 0

For x₁ = 0, $E (y) = 尾_{0} + 尾_{1} x_{1} + 尾_{2} x_{2} + 尾_{3} {x_{2}}^{2} + 尾_{4} x_{3} + 尾_{5} x_{1} {x_{2}}^{2} E (y) = 尾_{0} + 尾_{1} 脳 0 + 尾_{2} x_{2} + 尾_{3} {x_{2}}^{2} + 尾_{4} x_{3} + 尾_{5} 脳 0 脳 {x_{2}}^{2} E (y) = 尾_{0} + 尾_{2} x_{2} + 尾_{3} {x_{2}}^{2} + 尾_{4} x_{3}$

For x₁= 1, $E (y) = 尾_{0} + 尾_{1} x_{1} + 尾_{2} x_{2} + 尾_{3} {x_{2}}^{2} + 尾_{4} x_{3} + 尾_{5} x_{1} {x_{2}}^{2} E (y) = 尾_{0} + 尾_{1} 脳 1 + 尾_{2} x_{2} + 尾_{3} {x_{2}}^{2} + 尾_{4} x_{3} + 尾_{5} 脳 1 脳 {x_{2}}^{2} E (y) = 尾_{0} + 尾_{1} + 尾_{2} x_{2} + 尾_{3} {x_{2}}^{2} + 尾_{4} x_{3} + 尾_{5} {x_{2}}^{2}$

Graph

For the value of x₂ = 0, the mean value E(y) least squares prediction equation will look like

$E (y) = 2 + x_{1} - 5 x_{2} + 3 {x_{2}}^{2} - 4 x_{3} + x_{1} {x_{2}}^{2} E (y) = 2 + 0 - 5 脳 0 + 3 脳 0^{2} - 4 x_{3} + 0 脳 0^{2} f o r x_{1} = 0 a n d x_{2} = 0 E (y) = 2 - 4 x_{3}$

For the value of x₂ = 0, the mean value E(y) least squares prediction equation will look like

$E (y) = 2 + x_{1} - 5 x_{2} + 3 {x_{2}}^{2} - 4 x_{3} + x_{1} {x_{2}}^{2} E (y) = 2 + 1 - 5 脳 0 + 3 脳 0^{2} - 4 x_{3} + 1 脳 0^{2} f o r x_{1} = 1 a n d x_{2} = 0 E (y) = 3 - 4 x_{3}$