Question

The Excel printout below resulted from fitting the following model to n = 15 data points:

$\mathit{y}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}{\mathit{x}}_{\mathbf{1}}\mathbf{+}{\mathit{\beta }}_{\mathbf{2}}{\mathit{x}}_{\mathbf{2}}\mathbf{+}\mathit{\epsilon }$

Where,

${\mathit{x}}_{\mathbf{1}}\mathbf{=}\left(\begin{array}{c}1iflevel2\\ 0ifnot\end{array}\right)$ role="math" localid="1651883353071" ${\mathit{x}}_{\mathbf{2}}\mathbf{=}\left(\begin{array}{c}1iflevel3\\ 0ifnot\end{array}\right)$

a) Report the least squares prediction equation.

b) Interpret the values of β₁ and β₂.

c) Interpret the following hypotheses in terms of µ₁, µ₂,and µ₃:

d) At least one of parameters and differs from 0

Conduct the hypothesis test of part c.

Short answer

a) From the excel printout, the coefficient values can be used to write the least square prediction equation for the model. Here, $\widehat{y=80+16.8x_1+40.4x_2+\epsilon }$

b) ${\beta }_1$and ${\beta }_2$denotes the difference between the mean levels for different dummy variables. This means that ${\beta }_1={\mu }_2-{\mu }_1$ while ${\beta }_3={\mu }_3-{\mu }_1$

c) Here, the null hypothesis becomes that the means for the three groups are equal meaning ${\mu }_1={\mu }_2={\mu }_3$while the alternate hypothesis implies that at least two of the three means ${\beta }_1\ne {\beta }_2\ne {\beta }_3$ differ

d) At 95% confidence level,${\beta }_1\ne {\beta }_2\ne 0$ Hence two of the three means differ in the model.

Step-by-step solution

Least squares prediction equation

From the excel printout, the values of the coefficients can be used to write the least square prediction equation for the model

Here,

$y=80+16.8x_1+40.4x_2+\epsilon$

Interpretation of β1and β2

${\beta }_1$and${\beta }_2$denotes difference between the mean levels for different dummy variables.

This means ${\beta }_1={\mu }_2-{\mu }_1$while${\beta }_3={\mu }_3-{\mu }_1$

Simplification of hypothesis 

$H_0:{\beta }_1={\beta }_2=0$

$H_a:$At least one of parameters${\beta }_1$ and ${\beta }_2$differs from 0

Here, the null hypothesis becomes that the means for the three groups are equal meaning${\mu }_1={\mu }_2={\mu }_3$ while the alternate hypothesis implies that at least two of the three means$\left({\mu }_1,{\mu }_2,{\mu }_3\right)$ differ.

Hypothesis testing

$H_0:{\beta }_1={\beta }_2=0$

$H_a$At least one of parameters${\beta }_1$or role="math" localid="1651888100746" ${\beta }_2$is non zero

Here, F test statistic$=\frac{SSE}{\left(n-k\right)}+1$and the p-value is 0

$H_0$is rejected if p- value$<\alpha$,For since $\alpha =0.05$then $p$is less than 0.05

Sufficient evidence to reject$H_0$ at 95% confidence interval.

Therefore,Hence ${\beta }_1\ne {\beta }_2$two of the three means differ in the model