Question

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 $\mathit{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$.

Short answer

a) The prediction equation here will be $y=-2.550+3.815x_1+2.63x_2-1.285x_1{x}_2$ .

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

c) Graph

d) 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.

e) The null hypothesis and alternate hypothesis will be $H_0:{\beta }_3=0$and $H_a:{\beta }_3\ne 0$

f) At 5% significance level ${\beta }_3=0$,. Hence, it can be concluded with enough evidence that x₁ and x₂ do not interact in the model.

Step-by-step solution

Prediction equation

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

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

Slope of the line

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.

Graph

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

Y	0.08	-2.45
X1	0	1

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

Y	5.34	5.3
X1	0	1

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

Y	10.6	7.99
X1	0	1

Interpretation of Graph

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.

Significance of  β3

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

Substance of β3 

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

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.