Question

Question:Suppose you fit the first-order model $\mathbf{y}\mathbf{=}{\mathbf{\beta }}_{\mathbf{0}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{2}}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{3}}{\mathbf{x}}_{\mathbf{3}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{4}}{\mathbf{x}}_{\mathbf{4}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{5}}{\mathbf{x}}_{\mathbf{5}}\mathbf{+}\mathbf{\epsilon }$to n=30 data points and obtain SSE = 0.33 and ${\mathbf{R}}^{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{92}$

(A) Do the values of SSE and ${\mathbf{R}}^{\mathbf{2}}$suggest that the model provides a good fit to the data? Explain.

(B) Is the model of any use in predicting Y ? Test the null hypothesis ${\mathbf{H}}_{\mathbf{0}}\mathbf{:}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{3}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{4}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{5}}\mathbf{=}\mathbf{0}$ against the alternative hypothesis $\left\{H\right\}$at least one of the parameters ${\mathbf{\beta }}_{\mathbf{1}},{\mathbf{\beta }}_{\mathbf{2}},...,{\mathbf{\beta }}_5$ is non zero.Use$\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$ .

Short answer

(A) Low value of SSE and high value of ${\mathbf{R}}^{\mathbf{2}}$ indicate that the model is a good fit to the data.

(B) At 95% confidence interval, it can be concluded that ${\mathbf{H}}_{\mathbf{\alpha }}\mathbf{:}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{3}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{4}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{5}}\mathbf{=}\mathbf{0}$

Step-by-step solution

Step-by-Step Solution Step 1: Good fit to the data

A low sum of squares of error values like 0.33 indicate that the there is low variability in the set of observations. It means that the data points are close to the model fitted. Also, high value like 0.92 for${\mathbf{R}}^{\mathbf{2}}$indicates that around 92% of the variation in the data is explained by the model which is a good thing.

Hence, we can say that SSE = 0.33 andis a good measure of good fit to the data.

Overall significance of the model

${\mathbf{H}}_0\mathbf{:}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{3}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{4}}\mathbf{=}{\mathbf{\beta }}_{\mathbf{5}}\mathbf{=}\mathbf{0}$

${\text{H}}_{\text{a}}\text{:}$At least one of the parameters ${\mathbf{\beta }}_{\mathbf{1}},{\mathbf{\beta }}_{\mathbf{2}},...,{\mathbf{\beta }}_5$is non zero

Here, F test statistic $\mathbf{=}\frac{\mathbf{SSE}}{\mathbf{n}\mathbf{-}\mathbf{(}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{)}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{33}}{\mathbf{30}\mathbf{-}\mathbf{6}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01375}$

Value of${\mathbf{F}}_{\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{,}\mathbf{24}\mathbf{,}\mathbf{24}}$is 1.984

${\text{H}}_0$is rejected if F statistic > ${\mathbf{F}}_{\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{,}\mathbf{24}\mathbf{,}\mathbf{24}}$For $\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$, since F <${\mathbf{F}}_{\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{,}\mathbf{24}\mathbf{,}\mathbf{24}}$

Do not have sufficient evidence to reject ${\text{H}}_0$ at 95% confidence interval.

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