Question

Consider the model:

$\mathit{E}\left(y\right)\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}{\mathit{x}}_{\mathbf{1}}\mathbf{+}{\mathit{\beta }}_{\mathbf{2}}{\mathit{x}}_{\mathbf{2}}\mathbf{+}{\mathit{\beta }}_{\mathbf{3}}{{\mathbf{x}}_{\mathbf{2}}}^{\mathbf{2}}\mathbf{+}{\mathit{\beta }}_{\mathbf{4}}{\mathit{x}}_{\mathbf{3}}\mathbf{+}{\mathit{\beta }}_{\mathbf{5}}{\mathit{x}}_{\mathbf{1}}{{\mathbf{x}}_{\mathbf{2}}}^{\mathbf{2}}$

where x₂ is a quantitative model and

${\mathit{x}}_{\mathbf{1}}\mathbf{=}\left(\begin{array}{c}1receivedtreatment\\ 0didnotreceivetreatment\end{array}\right)$

The resulting least squares prediction equation is

localid="1649802968695" $\stackrel{\mathbf{⏜}}{\mathbf{y}}\mathbf{=}\mathbf{2}\mathbf{+}{\mathit{x}}_{\mathbf{1}}\mathbf{-}\mathbf{5}{\mathit{x}}_{\mathbf{2}}\mathbf{+}\mathbf{3}{{\mathbf{x}}_{\mathbf{2}}}^{\mathbf{2}}\mathbf{-}\mathbf{4}{\mathit{x}}_{\mathbf{3}}\mathbf{+}{\mathit{x}}_{\mathbf{1}}{{\mathbf{x}}_{\mathbf{2}}}^{\mathbf{2}}$

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

The mean value of E(y)for x₁ = 0 is$E\left(y\right)={\beta }_0+{\beta }_2{x}_2+{\beta }_3{x_2}^2+{\beta }_4{x}_3$ andthe mean value of E(y)for x₁= 1 is .$E\left(y\right)=\left({\beta }_0+{\beta }_1\right)+{\beta }_2{x}_2+{\beta }_3{x_2}^2+{\beta }_4{x}_3+{\beta }_5{x_2}^2$

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,$$\begin{gathered} E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+{\beta }_3{x_2}^2+{\beta }_4{x}_3+{\beta }_5{x}_1{x_2}^2 \\ E\left(y\right)={\beta }_0+{\beta }_1\times 0+{\beta }_2{x}_2+{\beta }_3{x_2}^2+{\beta }_4{x}_3+{\beta }_5\times 0\times {x_2}^2 \\ E\left(y\right)={\beta }_0+{\beta }_2{x}_2+{\beta }_3{x_2}^2+{\beta }_4{x}_3 \end{gathered}$$

For x₁= 1, $$\begin{gathered} E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+{\beta }_3{x_2}^2+{\beta }_4{x}_3+{\beta }_5{x}_1{x_2}^2 \\ E\left(y\right)={\beta }_0+{\beta }_1\times 1+{\beta }_2{x}_2+{\beta }_3{x_2}^2+{\beta }_4{x}_3+{\beta }_5\times 1\times {x_2}^2 \\ E\left(y\right)={\beta }_0+{\beta }_1+{\beta }_2{x}_2+{\beta }_3{x_2}^2+{\beta }_4{x}_3+{\beta }_5{x_2}^2 \end{gathered}$$

Graph

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

$$\begin{gathered} E\left(\mathrm{y}\right)=2+x_1-5x_2+3{{\mathrm{x}}_2}^2-4x_3+x_1{{\mathrm{x}}_2}^2 \\ E\left(\mathrm{y}\right)=2+0-5\times 0+3\times 0^2-4x_3+0\times 0^{2}for{x}_1=0and{x}_2=0 \\ E\left(\mathrm{y}\right)=2-4x_3 \end{gathered}$$

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

$$\begin{gathered} E\left(\mathrm{y}\right)=2+x_1-5x_2+3{{\mathrm{x}}_2}^2-4x_3+x_1{{\mathrm{x}}_2}^2 \\ E\left(\mathrm{y}\right)=2+1-5\times 0+3\times 0^2-4x_3+1\times 0^{2}for{x}_1=1and{x}_2=0 \\ E\left(\mathrm{y}\right)=3-4x_3 \end{gathered}$$