The Minitab printout below was obtained from fitting the model$\mathit{y}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}{\mathit{x}}_{\mathbf{1}}\mathbf{+}{\mathit{\beta }}_{\mathbf{2}}{\mathit{x}}_{\mathbf{2}}\mathbf{+}{\mathit{\beta }}_{\mathbf{3}}{\mathit{x}}_{\mathbf{1}}{\mathit{x}}_{\mathbf{2}}\mathbf{+}\mathit{\epsilon }$to n = 15 data points.

a) What is the prediction equation?

b) Give an estimate of the slope of the line relating y to x₁ when x₂ =10 .

c) Plot the prediction equation for the case when x₂ =1 . Do this twice more on the same graph for the cases when x₂ =3 and x₂ =5 .

d) Explain what it means to say that x₁and x₂interact. Explain why your graph of part c suggests that x₁and x₂interact.

e) Specify the null and alternative hypotheses you would use to test whetherx₁andx₂interact.

f)Conduct the hypothesis test of part e using $\mathit{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$.

The prediction equation is the same as the regression equation which is calculated and given in the image.

The prediction equation here will be $\mathit{y}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{550}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{815}{\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{63}{\mathit{x}}_{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{285}{\mathit{x}}_{\mathbf{1}}{\mathit{x}}_{\mathbf{2}}$.

Given,

$$\begin{gathered} \mathit{E}\left(y\right)\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{550}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{815}{\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{63}{\mathit{x}}_{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{285}{\mathit{x}}_{\mathbf{1}}{\mathit{x}}_{\mathbf{2}} \\ \mathit{y}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{550}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{815}{\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{63}\left(10\right)\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{285}{\mathit{x}}_{\mathbf{1}}\left(10\right)\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}{\mathit{x}}_{\mathbf{2}}\mathbf{=}\mathbf{10} \\ \mathit{y}\mathbf{=}\mathbf{17}\mathbf{.}\mathbf{565}\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{185}{\mathit{x}}_{\mathbf{1}} \end{gathered}$$

The slope of the line relating y to x₁ when x₂=10is -6.185.

Given,

$$\begin{gathered} \mathit{E}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{550}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{815}{\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{63}{\mathit{x}}_{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{285}{\mathit{x}}_{\mathbf{1}}{\mathit{x}}_{\mathbf{2}} \\ \mathit{y}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{550}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{815}{\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{63}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{285}{\mathit{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}{\mathit{x}}_{\mathbf{2}}\mathbf{=}\mathbf{1} \\ \mathit{y}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{08}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{53}{\mathit{x}}_{\mathbf{1}} \end{gathered}$$

Now to plot this equation, make a table

Given,

$$\begin{gathered} \mathit{E}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{550}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{815}{\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{63}{\mathit{x}}_{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{285}{\mathit{x}}_{\mathbf{1}}{\mathit{x}}_{\mathbf{2}} \\ \mathit{y}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{550}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{815}{\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{63}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{285}{\mathit{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}{\mathit{x}}_{\mathbf{2}}\mathbf{=}\mathbf{3} \\ \mathit{y}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{34}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{04}{\mathit{x}}_{\mathbf{1}} \end{gathered}$$

Now to plot this equation, make a table

Given,

$$\begin{gathered} \mathit{E}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{550}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{815}{\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{63}{\mathit{x}}_{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{285}{\mathit{x}}_{\mathbf{1}}{\mathit{x}}_{\mathbf{2}} \\ \mathit{y}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{550}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{815}{\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{63}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{285}{\mathit{x}}_{\mathbf{1}}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}{\mathit{x}}_{\mathbf{2}}\mathbf{=}\mathbf{5} \\ \mathit{y}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{6}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{61}{\mathit{x}}_{\mathbf{1}} \end{gathered}$$

Now to plot this equation, make a table

Interaction between the variables x₁ and x₂ indicates that the two variables are not entirely independent of each other and that they are dependent to some extent on each other. Hence for this reason, a new variable is added in the model where there is interaction amongst two variables like ‘x₁ x₂’ to represent the dependency in the model.

To test whether x₁ and x₂ variables interact in the model, the presence of the model parameter${\beta }_3$is tested.

Hence the null hypothesis would be the absence of the model parameter ${\beta }_3$ and the alternate hypothesis would be the presence of ${\beta }_3$

Mathematically, $H_0:{\beta }_3=0,H_a:{\beta }_3\ne 0$

$$\begin{gathered} H_0:{\beta }_3=0 \\ {H}_q:{\beta }_3\ne 0 \end{gathered}$$

Here, t-test statistic$=\frac{{\beta }_3}{s{\beta }_3}=\frac{-1.285}{0.159}=-8.082$

Value of$t_{0.05,14}$is 1.761

is rejected if t statistic >$t_{0.05,14}$ . For $\alpha =0.05$, since t < $t_{0.05,31}$

Not sufficient evidence to reject $H_0$at a 95% confidence interval.

Therefore,role="math" localid="1649963401754" ${\mathit{\beta }}_{\mathbf{3}}\mathbf{=}\mathbf{0}$. Hence it can be concluded with enough evidence that x₁ and x₂ do not interact in the model.