Chapter 15, Problem 1PS

The following is the quadratic regression equation for a sample of $n=25:$

${\hat{Y}}_i=5+3X_{li}+1.5X_{li}^2$

a. Predict $Y$ for $X_1=2.$

b. Suppose that the computed $t_{STAT}$ test statistic for the quadratic regression coefficients is 2.35. At the 0.05 level of significance, is there evidence that the quadratic model is better than the linear model?

c. Suppose that the computer $t_{STAT}$ test statistic for the quadratic regression coefficient is 1.17. At the 0.05 level of significance, is there evidence that the quadratic model is better than the linear model?

d. Suppose the regression coefficient for the linear effect is $-\text{3}.0,$ Predict $Y$ for $X_1=2.$

To determine

Determine the predicted value of Y.

Answer to Problem 1PS

The predicted value of Y is 17.

Explanation of Solution

The quadratic regression equation is given as:

${\widehat{Y}}_i=5+3X_{1i}+1.5X_{1i}^2$

For predicting the value of Y for $X_1=2$ , simply put the value of $X_1$ in quadratic regression equation.

Thus, the predicted value can be calculated as:

$\begin{array}{l}{\widehat{Y}}_i=5+3X_{1i}+1.5X_{1i}^2\\ {\widehat{Y}}_i=5+3\times 2+1.5\times {\left(2\right)}^2\\ {\widehat{Y}}_i=5+6+6\\ {\widehat{Y}}_i=17\end{array}$

Therefore, the predicted value of Y is 17.

To determine

Test whether a quadratic model is better than linear model, if the test statistics for quadratic regression coefficient is 2.35

Answer to Problem 1PS

The quadratic model is better than linear model.

Explanation of Solution

It is given that the value of the test statistics $\left(t_{\text{STAT}}\right)$ for the quadratic regression coefficient is 2.35.

The null and alternative hypotheses can be constructed as:

$\begin{array}{l}{H}_0:\text{The quadratic effect does not significantly improve the model}\text{.}\\ H_1:\text{ The quadratic effect significanly improves the model}\text{.}\end{array}$

The degrees of freedom $\left(df\right)$ for the test can be calculated as:

$\begin{array}{c}df=n-2-1\\ =25-2-1\\ =22\end{array}$

This is a two tailed test. So, the upper critical region will be $2.5\%$ .

The critical value of t statistics from t distribution table, which is given in appendix table E.3 at $2.5\%$ level of significance and 22 degree of freedom is 2.0739.

The decision rule on the basis of the critical value approach:

If, $t_{\text{STAT}}\ge t_{\alpha /2}\text{or}-t_{\text{STAT}}\le -t_{\alpha /2},\text{Reject }{H}_0$

If, $-t_{\alpha /2}\text{< }{t}_{\text{STAT}}<t_{\alpha /2},\text{Do Not Reject }{H}_0$

Since, $t_{STAT}\left(2.35\right)>t_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(2.0739\right)$ . So, reject the null hypothesis.

Therefore, there is a sufficient evidence to conclude that quadratic model is better than linear model.

To determine

Test whether a quadratic model is better than linear model, if the test statistics for quadratic regression coefficient is 1.17

Answer to Problem 1PS

The quadratic model is not better than linear model.

Explanation of Solution

It is given that the value of the test statistics $\left(t_{\text{STAT}}\right)$ for the quadratic regression coefficient is 1.17.

The hypothesis and the critical value of test statistics is same as above.

Since, $-t_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(-2.0739\right)<t_{STAT}\left(1.17\right)<t_{\raisebox{1ex}{$\alpha $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(2.0739\right)$ null hypothesis fails to reject.

Therefore, there is a sufficient evidence to conclude that quadratic model is not better than the linear model.

To determine

Predict Y, when coefficient for the linear effect is $-3.0$ .

Answer to Problem 1PS

The predicted value of Y is 5.

Explanation of Solution

The regression coefficient for the linear model is given as $-3.0$

Thus, the quadratic regression equation for 25 sample will change to:

${\widehat{Y}}_i=5-3X_{1i}+1.5X_{1i}^2$

For predicting the value of Y for $X_1=2$ , simply put the value of $X_1$ in quadratic regression equation.

Thus, the required predicted value is,

$\begin{array}{l}{\widehat{Y}}_i=5-3X_{1i}+1.5X_{1i}^2\\ {\widehat{Y}}_i=5-3\times 2+1.5\times {\left(2\right)}^2\\ {\widehat{Y}}_i=5-6+6\\ {\widehat{Y}}_i=5\end{array}$

Therefore, the predicted value of Y is 5.