Question

Consider the following data that fit the quadratic model$\mathbf{E}\left(y\right)\mathbf{=}{\mathbf{\beta }}_{\mathbf{0}}\mathbf{+}{\mathbf{\beta }}_{\mathbf{1}}\mathbf{x}\mathbf{+}{\mathbf{\beta }}_{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}$:

a. Construct a scatterplot for this data. Give the prediction equation and calculate ${\mathbf{R}}^{\mathbf{2}}$based on the model above.

b. Interpret the value of${\mathbf{R}}^{\mathbf{2}}$.

c. Justify whether the overall model is significant at the 1% significance level if the data result into a p-value of 0.000514.

Short answer

a. Scatter plot, prediction equation is $\mathbf{y}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{541667}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{25357}\mathbf{x}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{386905}{\mathbf{x}}^{\mathbf{2}}$and value of ${\mathbf{R}}^{\mathbf{2}}$calculated here is 0.9516

b. The value of${\mathbf{R}}^{\mathbf{2}}$here is 0.9516 which is a high value denoting that almost 95% of the variation in the variables is explained by the model. This means that the model is a good fit for the data.

c. At 1% significance level, it can be concluded that ${\mathbf{\beta }}_{\mathbf{1}}\ne {\mathbf{\beta }}_{\mathbf{2}}\ne \mathbf{0}$

Step-by-step solution

Scatterplot for the data

X	Y
0	8
1	7.3
2	5.6
3	5.9
4	5.2
5	6.5
6	8.8
7	12.1

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.975509
R Square	0.951618
Adjusted R Square	0.932265
Standard Error	0.586556
Observations	8
ANOVA
	df	SS	MS	F	Significance F
Regression	2	33.83476	16.91738	49.17163	0.000515
Residual	5	1.720238	0.344048
Total	7	35.555
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	8.541667	0.49366	17.30272	1.18E-05	7.272673	9.810661	7.272673	9.810661
X	-2.25357	0.329452	-6.84036	0.001019	-3.10046	-1.40669	-3.10046	-1.40669
X^2	0.386905	0.045254	8.549672	0.000361	0.270576	0.503233	0.270576	0.503233

On solving the second-order model equation in excel, the output generated is attached here. The prediction equation, therefore is$\mathbf{y}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{541667}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{25357}\mathbf{x}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{386905}{\mathbf{x}}^{\mathbf{2}}$

The value of${\mathbf{R}}^{\mathbf{2}}$ calculated here is 0.9516.

Interpretation of R2

The value of${\mathbf{R}}^{\mathbf{2}}$ here is 0.9516 which is a high value denoting that almost 95% of the variation in the variables is explained by the model. This means that the model is a good fit for the data.

Goodness of fit

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

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

Here, F test statistic$\mathbf{=}\frac{\mathbf{SSE}}{\mathbf{n}\mathbf{-}\mathbf{(}\mathbf{k}\mathbf{+}\mathbf{1}\mathbf{)}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{720238}}{\mathbf{5}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3440}$

${\mathbf{H}}_{\mathbf{0}}$is rejected if p-value < $\mathbf{\alpha }$. For $\mathbf{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$, since 0.000514 < 0.01

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

Therefore,${\mathbf{\beta }}_{\mathbf{1}}\ne {\mathbf{\beta }}_{\mathbf{2}}\ne \mathbf{0}$.