Minitab was used to fit the complete second-order mode$\mathbf{E}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}{\mathbf{\beta }}_{\mathbf{0}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{2}}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{3}}{\mathbf{x}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{4}}{\mathbf{x}}_{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{5}}{\mathbf{x}}_{\mathbf{2}}^{\mathbf{2}}$to n = 39 data points. The printout is shown on the next page.

a. Is there sufficient evidence to indicate that at least one of the parameters—${\mathbf{\beta }}_{\mathbf{1}}\mathbf{,}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{,}{\mathbf{\beta }}_{\mathbf{3}}\mathbf{,}{\mathbf{\beta }}_{\mathbf{4}}$, and${\mathbf{\beta }}_{\mathbf{1}}\mathbf{,}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{,}{\mathbf{\beta }}_{\mathbf{3}}\mathbf{,}{\mathbf{\beta }}_{\mathbf{4}}$—is nonzero? Test using$\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$.

b. Test${\mathbf{H}}_{\mathbf{0}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{4}}\mathbf{=}\mathbf{0}\mathbf{against}{\mathbf{H}}_{\mathbf{a}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{4}}\ne \mathbf{0}$. Use$\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$.

c. Test${\mathbf{H}}_{\mathbf{0}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{5}}\mathbf{=}\mathbf{0}\mathbf{against}{\mathbf{H}}_{\mathbf{a}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{5}}\ne \mathbf{0}$. Use$\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$.

d. Use graphs to explain the consequences of the tests in parts b and c.

${\mathbf{H}}_{\mathbf{0}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{3}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{4}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{5}}\mathbf{=}\mathbf{0}$

${\mathbf{H}}_{\mathbf{a}}\mathbf{:}$at least one of the parameters ${\mathbf{\beta }}_{\mathbf{1}}$, ${\mathbf{\beta }}_{\mathbf{2}}$, ${\mathbf{\beta }}_{\mathbf{3}}$,${\mathbf{\beta }}_{\mathbf{4}}$ and${\mathbf{\beta }}_{\mathbf{5}}$ are non-zero.

Here, F test statistic =$\frac{\mathbf{SSE}}{\mathbf{n}\mathbf{-}\left(\mathbf{K}\mathbf{+}\mathbf{1}\right)}\mathbf{=}\frac{\mathbf{251}\mathbf{.}\mathbf{81}}{\mathbf{34}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{406}$

${\mathbf{H}}_{\mathbf{0}}$is rejected if F – statistics < ${\mathbf{F}}_{\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{,}\mathbf{34}\mathbf{,}\mathbf{34}}$. For $\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$, since 7.406 > 1.843.

Not sufficient evidence to reject${\mathbf{H}}_{\mathbf{0}}$ at 95% confidence interval.

Therefore,${\mathbf{\beta }}_{\mathbf{1}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{3}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{4}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{5}}\mathbf{=}\mathbf{0}$

$$\begin{gathered} {\mathbf{H}}_{\mathbf{0}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{4}}\mathbf{=}\mathbf{0} \\ {\mathbf{H}}_{\mathbf{a}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{4}}\ne \mathbf{0} \end{gathered}$$

Here, t-test statistic$\mathbf{=}\frac{{\hat{\mathbf{\beta }}}_{\mathbf{4}}}{\mathbf{s}{\hat{\mathbf{\beta }}}_{\mathbf{4}}}\mathbf{=}\frac{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{0043}}{\mathbf{0}\mathbf{.}\mathbf{0004}}\mathbf{=}\mathbf{-}\mathbf{10}\mathbf{.}\mathbf{75}$

Value of${\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{005}\mathbf{,}\mathbf{34}}$is 2.728

${\mathbf{H}}_{\mathbf{0}}$is rejected if t statistic > ${\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{005}\mathbf{,}\mathbf{34}}$. For $\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$, since t <${\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{,}\mathbf{199}}$

Not sufficient evidence to reject${\mathbf{H}}_{\mathbf{0}}$ at a 95% confidence interval.

Thus,${\mathbf{\beta }}_{\mathbf{4}}\mathbf{=}\mathbf{0}$

$$\begin{gathered} {\mathbf{H}}_{\mathbf{0}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{5}}\mathbf{=}\mathbf{0} \\ {\mathbf{H}}_{\mathbf{a}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{5}}\ne \mathbf{0} \end{gathered}$$

Here, t-test statistic$\mathbf{=}\frac{{\hat{\mathbf{\beta }}}_{\mathbf{5}}}{\mathbf{s}{\hat{\mathbf{\beta }}}_{\mathbf{5}}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{0020}}{\mathbf{0}\mathbf{.}\mathbf{0033}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6060}$

Value of${\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{,}\mathbf{34}}$is 2.728

${\mathbf{H}}_{\mathbf{0}}$is rejected if t statistic > ${\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{005}\mathbf{,}\mathbf{34}}$. For $\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$, since t <${\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{005}\mathbf{,}\mathbf{34}}$

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

So, ${\mathbf{\beta }}_{\mathbf{5}}\mathbf{=}\mathbf{0}$.

Since, the value of${\mathbf{\beta }}_{\mathbf{4}}\mathbf{=}\mathbf{0}$and${\mathbf{\beta }}_{\mathbf{5}}\mathbf{=}\mathbf{0}$at 99% confidence level, this means that the parabola does not have a curvature and it essentially is a straight line.

Here, a straight line can be drawn relating y toholding${\mathbf{x}}_{\mathbf{2}}$constant or the other way round.