Question

The following model was used to relate E(y) to a single qualitative variable with four levels:$\mathbf{E}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}{\mathbf{\beta }}_{\mathbf{0}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{2}}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{3}}{\mathbf{x}}_{\mathbf{3}}$where

${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\left(\begin{array}{l}\mathbf{1}\mathbf{if}\mathbf{level}\mathbf{2}\\ \mathbf{0}\mathbf{if}\mathbf{not}\end{array}\right){\mathbf{x}}_{\mathbf{2}}\mathbf{=}\left(\begin{array}{l}\mathbf{1}\mathbf{if}\mathbf{level}\mathbf{3}\\ \mathbf{0}\mathbf{if}\mathbf{not}\end{array}\right){\mathbf{x}}_{\mathbf{3}}\mathbf{=}\left(\begin{array}{l}\mathbf{1}\mathbf{if}\mathbf{level}\mathbf{4}\\ \mathbf{0}\mathbf{if}\mathbf{not}\end{array}\right)$

This model was fit to n = 30 data points, and the following result was obtained:

$\hat{\mathbf{y}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{2}\mathbf{-}\mathbf{4}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}\mathbf{12}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{x}}_{\mathbf{3}}$

1. Use the least-squares prediction equation to find the estimate of E(y) for each level of the qualitative independent variable.
2. Specify the null and alternative hypotheses you would use to test whether E(y) is the same for all four levels of the independent variable.

Short answer

a.For ${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{1}$,${\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{0}$and ${\mathbf{x}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{;}$

$\hat{\mathbf{y}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{2}\mathbf{-}\mathbf{4}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{+}\mathbf{12}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{2}$

For ${\mathit{x}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}{\mathit{x}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$and ${\mathit{x}}_{\mathbf{3}}\mathbf{=}\mathbf{0}$

$\hat{\mathbf{y}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{2}\mathbf{-}\mathbf{4}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{+}\mathbf{12}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{8}$

. $\text{For}\mathbf{}\mathit{x}{\mathbf{3}}_{}\mathbf{=}\mathbf{1}\mathbf{,}{\mathit{x}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\mathbf{}\text{and}\mathbf{}{\mathit{x}}_{\mathbf{2}}\mathbf{=}\mathbf{0}$

$\hat{\mathbf{y}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{2}\mathbf{-}\mathbf{4}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{+}\mathbf{12}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{2}$

b. The hypothesis to be tested can be written as${\mathbf{H}}_{\mathbf{0}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{3}}\mathbf{=}\mathbf{0}$

${\mathit{H}}_{\mathbf{a}}\mathbf{:}$At least one of the parameters ${\mathit{\beta }}_{\mathbf{1}}\mathbf{,}{\mathit{\beta }}_{\mathbf{2}}\mathbf{}\text{and}\mathbf{}{\mathit{\beta }}_{\mathbf{3}}\mathbf{}\text{differfrom}\mathbf{}\mathbf{0}$

Step-by-step solution

Mean values for each level of the qualitative independent variable

The value E(Y) for a different level of the qualitative independent variable

Can be found by substituting in place of other two variables

$\text{For}{\mathit{x}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathit{x}}_{\mathbf{2}}\mathbf{=}\mathbf{0}and\mathbf{}{\mathit{x}}_{\mathbf{3}}\mathbf{=}\mathbf{0}\mathbf{;}$

$\hat{\mathbf{y}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{2}\mathbf{-}\mathbf{4}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{+}\mathbf{12}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{2}$

$\text{For}{\mathit{x}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathit{x}}_{\mathbf{1}}\mathbf{=}\mathbf{0},\text{and}{\mathit{x}}_{\mathbf{3}}\mathbf{=}\mathbf{0};$

$\hat{\mathbf{y}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{2}\mathbf{-}\mathbf{4}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{+}\mathbf{12}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{8}$

$\text{For}\mathbf{}{\mathit{x}}_{\mathbf{3}}\mathbf{=}\mathbf{1},{\mathit{x}}_{\mathbf{1}}\mathbf{=}\mathbf{0},\text{and}{\mathit{x}}_{\mathbf{2}}\mathbf{=}\mathbf{0};$

$\hat{\mathbf{y}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{2}\mathbf{-}\mathbf{4}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{+}\mathbf{12}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{2}$

Hypothesis testing

Here, the null hypothesis becomes that the means for the three groups are equal meaning _{${\mathbf{\mu }}_{\mathbf{1}}\mathbf{=}{\mathbf{\mu }}_{\mathbf{2}}\mathbf{=}{\mathbf{\mu }}_{\mathbf{3}}$}while the alternate hypothesis implies that at least two of the three means (${\mathit{\mu }}_{\mathbf{1}}$,${\mathit{\mu }}_{\mathbf{2}}$_,and ${\mathit{\mu }}_{\mathbf{3}}$) differ.