Question

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,

${\mathbf{\text{x}}}_{\mathbf{1}}\mathbf{=}\left(\begin{array}{cc}1& iflevel2\\ 0& ifnot\end{array}\right)$${\mathbf{\text{x}}}_{\mathbf{2}}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{1}& \mathbf{if}\mathbf{}\mathbf{level}\mathbf{}\mathbf{3}\\ \mathbf{0}& \mathbf{if}\mathbf{}\mathbf{not}\end{array}\mathbf{\right)}$

Short answer

Answer:

1. From the excel printout, the coefficient values can be used to write the least square prediction equation for the model. Here,$\hat{\mathbf{y}}\mathbf{=}\mathbf{80}\mathbf{+}\mathbf{16}\mathbf{.}\mathbf{8}{\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathbf{40}\mathbf{.}\mathbf{4}{\mathit{x}}_{\mathbf{2}}\mathbf{+}\mathit{\epsilon }$
2. ${\mathit{\beta }}_{\mathbf{1}}$and${\mathit{\beta }}_2$denotes the difference between the mean levels for different dummy variables. This means that${\mathit{\beta }}_{\mathbf{1}}\mathbf{=}{\mathit{\mu }}_{\mathbf{2}}\mathbf{-}{\mathit{\mu }}_{\mathbf{1}}$while${\mathit{\beta }}_{\mathbf{2}}\mathbf{=}{\mathit{\mu }}_{\mathbf{3}}\mathbf{-}{\mathit{\mu }}_{\mathbf{1}}$
3. Here, the null hypothesis becomes that the means for the three groups are equal meaning${\mathit{\mu }}_{\mathbf{1}}\mathbf{=}{\mathit{\mu }}_{\mathbf{2}}\mathbf{=}{\mathit{\mu }}_{\mathbf{3}}$while the alternate hypothesis implies that at least two of the three means ${\mathit{\beta }}_{\mathbf{1}}\mathbf{\ne }{\mathit{\beta }}_{\mathbf{2}}\mathbf{\ne }{\mathit{\beta }}_{\mathbf{3}}$differ
4. At 95% confidence level,${\mathit{\beta }}_{\mathbf{1}}\mathbf{\ne }{\mathit{\beta }}_{\mathbf{2}}\mathbf{\ne }\mathbf{0}$ 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,$\hat{\mathbf{y}}\mathbf{=}\mathbf{80}\mathbf{+}\mathbf{16}\mathbf{.}\mathbf{8}{\mathit{x}}_{\mathbf{1}}\mathbf{+}\mathbf{40}\mathbf{.}\mathbf{4}{\mathit{x}}_{\mathbf{2}}\mathbf{+}\mathit{\epsilon }$

Interpretation of   β1 and β2

${\mathit{\beta }}_{\mathbf{1}}$and${\mathit{\beta }}_{\mathbf{2}}$ denotes difference between the mean levels for different dummy variables.

This means ${\mathit{\beta }}_{\mathbf{1}}\mathbf{=}{\mathit{\mu }}_{\mathbf{2}}\mathbf{-}{\mathit{\mu }}_{\mathbf{1}}$while${\mathit{\beta }}_{\mathbf{2}}\mathbf{=}{\mathit{\mu }}_{\mathbf{3}}\mathbf{-}{\mathit{\mu }}_{\mathbf{1}}$

Simplification of hypothesis

$\mathbf{}{\mathit{H}}_{\mathbf{0}}\mathbf{}\mathbf{:}{\mathit{\beta }}_{\mathbf{1}}\mathbf{=}{\mathit{\beta }}_{\mathbf{2}}\mathbf{=}\mathbf{0}$

${\mathit{H}}_{\mathbf{a}}\mathbf{}\mathbf{:}$At least one of parameters${\mathit{\beta }}_{\mathbf{1}}$ and${\mathit{\beta }}_{\mathbf{2}}$ differs from 0

Here, the null hypothesis becomes that the means for the three groups are equal meaning ${\mathit{\mu }}_{\mathbf{1}}\mathbf{=}{\mathit{\mu }}_{\mathbf{2}}\mathbf{=}{\mathit{\mu }}_{\mathbf{3}}$while the alternate hypothesis implies that at least two of the three means role="math" localid="1649851966245" ${\mathbf{\text{(μ}}}_{\mathbf{1}}\mathbf{,}{\mathit{\mu }}_{\mathbf{2}}\mathit{a}\mathit{n}\mathit{d}{\mathit{\mu }}_{\mathbf{3}}\mathbf{)}$differ

Hypothesis testing

$\mathbf{}{\mathit{H}}_{\mathbf{0}}\mathbf{}\mathbf{:}{\mathit{\beta }}_{\mathbf{1}}\mathbf{=}{\mathit{\beta }}_{\mathbf{2}}\mathbf{=}\mathbf{0}$

$\mathbf{}{\mathit{H}}_{\mathit{a}}\mathbf{}\mathbf{:}$At least one of parameters${\mathit{\beta }}_{\mathbf{1}}$or${\mathit{\beta }}_{\mathbf{2}}$is non zero

Here, F test statistic$\mathbf{=}\frac{\mathbf{S}\mathbf{S}\mathbf{E}}{\mathbf{n}\mathbf{-}\mathbf{k}\mathbf{+}\mathbf{1}}\mathbf{=}\mathbf{24}\mathbf{.}\mathbf{72}$and the p-value is 0

$\mathbf{}{\mathit{H}}_{\mathbf{0}}$is rejected if$\mathit{P}\mathbf{-}\mathit{v}\mathit{a}\mathit{l}\mathit{u}\mathit{e}\mathbf{}\mathbf{<}\mathit{a}$For$\mathit{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$since p- value is less than 0.05

Sufficient evidence to reject$\mathbf{}{\mathit{H}}_{\mathbf{0}}\mathbf{}$at 95% confidence interval.

Therefore, ${\mathit{\beta }}_{\mathbf{1}}\mathbf{\ne }{\mathit{\beta }}_{\mathbf{2}}\mathbf{\ne }\mathbf{0}$Hence two of the three means differ in the model