Chapter 8.3, Problem 8E

a.

To determine

Construct a linear regression model for the data.

Test whether there is enough evidence to conclude that ${\beta }_0$ is zero.

Test whether there is enough evidence to conclude that ${\beta }_1$ is zero.

Answer to Problem 8E

A simple linear regression model for the data is:

$\underset{\_}{\widehat{y}=2.77818-1.63031x}$.

No, there is not enough evidence to conclude that ${\beta }_0$ is zero.

No, there is not enough evidence to conclude that ${\beta }_1$ is zero.

Explanation of Solution

Given info:

The data represents the values of the variables amount of deflection in mm $\left(y\right)$ and distance in m $\left(x\right)$.

Calculation:

Linear regression model:

A linear regression model is given as $\widehat{y}=b_0+b_{1}x$ where $\widehat{y}$ be the predicted values of response variable and x be the predictor variable. $b_1$ be the slope and $b_0$ be the intercept of the line.

A linear regression model is given as $\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_{1}x$ where $\widehat{y}$ be the predicted values of response variable and x be the predictor variable. The ${\widehat{\beta }}_1$ be the estimate of slope and ${\widehat{\beta }}_0$ be the estimate of intercept of the line.

Regression:

Software procedure:

Step by step procedure to obtain regression using MINITAB software is given as,

• Choose Stat > Regression > General Regression.
• In Response, enter the numeric column containing the response data Y.
• In Model, enter the numeric column containing the predictor variables X.
• Click OK.

Output obtained from MINITAB is given below:

The ‘Coefficient’ column of the regression analysis MINITAB output gives the slopes corresponding to the respective variables stored in the column ‘Term’.

A careful inspection of the output shows that the fitted model is:

$\widehat{y}=2.77818-1.63031x$.

Hence, the linear regression model for the data is:

$\underset{\_}{\widehat{y}=2.77818-1.63031x}$.

Test for slope coefficient ${\beta }_0$:

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_0=0$

That is, intercept of the regression model is not significant.

Alternative hypothesis:

$H_a:{\beta }_0\ne 0$

That is, intercept of the regression model is significant.

Level of significance:

Since, the level of significance is not given. The prior level of significance $\alpha =0.05$ can be used.

The ‘P’ column of the regression analysis MINITAB output gives the P- value corresponding to the respective variables stored in the column ‘Term’.

The P- value corresponding to the coefficient ${\beta }_0$ is 0.000.

Decision criteria based on P-value approach:

If $P\text{-value}\le \alpha$, then reject the null hypothesis $H_0$.

If $P\text{-value}>\alpha$, then fail to reject the null hypothesis $H_0$.

Conclusion:

The P-value is 0.000 and $\alpha$ value is 0.05.

Here, P-value is less than the $\alpha$ value.

That is $0.013\left(=P\right)<0.05\left(=\alpha \right)$.

By the rejection rule, reject the null hypothesis.

Therefore, intercept of the regression model is significant.

Thus, there is not enough evidence to conclude that ${\beta }_0$ is zero.

Test for slope coefficient ${\beta }_1$:

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_1=0$

That is, slope coefficient of the predictor variable distance is not significant.

Alternative hypothesis:

$H_a:{\beta }_1\ne 0$

That is, slope coefficient of the predictor variable distance is significant.

From the MINITAB output, the P- value corresponding to the coefficient ${\beta }_1$ is 0.001.

Conclusion:

The P-value is 0.001 and $\alpha$ value is 0.05.

Here, P-value is less than the $\alpha$ value.

That is $0.001\left(=P\right)<0.05\left(=\alpha \right)$.

By the rejection rule, reject the null hypothesis.

Therefore, the slope coefficient of the predictor variable distance is significant.

Thus, there is not enough evidence to conclude that ${\beta }_1$ is zero.

b.

To determine

Construct a quadratic regression model for the data.

Test whether there is enough evidence to conclude that ${\beta }_0$ is zero.

Test whether there is enough evidence to conclude that ${\beta }_1$ is zero.

Test whether there is enough evidence to conclude that ${\beta }_2$ is zero.

Answer to Problem 8E

The quadratic regression model for the data is $\underset{\_}{\widehat{y}=3.25743-3.51514x+1.03002x^2}$.

No, there is not enough evidence to conclude that ${\beta }_0$ is zero.

No, there is not enough evidence to conclude that ${\beta }_1$ is zero.

Yes, there is enough evidence to conclude that ${\beta }_2$ is zero.

Explanation of Solution

Calculation:

Quadratic model:

The quadratic regression model would be of the form:

$\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_{1}x+{\widehat{\beta }}_2{x}^2$

Regression:

Software procedure:

Step by step procedure to obtain quadratic regression model using MINITAB software is given as,

• Choose Stat > Regression > General Regression.
• In Response, enter the numeric column containing the response data Y.
• In Model, enter the numeric column containing the predictor variables X and X-square.
• Click OK.

Output obtained from MINITAB is given below:

The ‘Coefficient’ column of the regression analysis MINITAB output gives the slopes corresponding to the respective variables stored in the column ‘Predictor’.

A careful inspection of the output shows that the fitted model is:

$\widehat{y}=3.25743-3.51514x+1.03002x^2$.

Hence, the quadratic regression model for the data is:

$\underset{\_}{\widehat{y}=3.25743-3.51514x+1.03002x^2}$.

Test for slope coefficient ${\beta }_0$:

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_0=0$

That is, intercept of the regression model is not significant.

Alternative hypothesis:

$H_a:{\beta }_0\ne 0$

That is, intercept of the regression model is significant.

Level of significance:

Since, the level of significance is not given. The prior level of significance $\alpha =0.05$ can be used.

The ‘P’ column of the regression analysis MINITAB output gives the P- value corresponding to the respective variables stored in the column ‘Term’.

The P- value corresponding to the coefficient ${\beta }_0$ is 0.000.

Decision criteria based on P-value approach:

If $P\text{-value}\le \alpha$, then reject the null hypothesis $H_0$.

If $P\text{-value}>\alpha$, then fail to reject the null hypothesis $H_0$.

Conclusion:

The P-value is 0.000 and $\alpha$ value is 0.05.

Here, P-value is less than the $\alpha$ value.

That is $0.000\left(=P\right)<0.05\left(=\alpha \right)$.

By the rejection rule, fail to reject the null hypothesis.

Therefore, intercept of the regression model is significant.

Thus, there is not enough evidence to conclude that ${\beta }_0$ is zero.

Test for slope coefficient ${\beta }_1$:

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_1=0$

That is, coefficient of the predictor variable distance is not significant.

Alternative hypothesis:

$H_a:{\beta }_1\ne 0$

That is, coefficient of the predictor variable distance is significant.

From the MINITAB output, the P- value corresponding to the coefficient ${\beta }_1$ is 0.000.

Conclusion:

The P-value is 0.000 and $\alpha$ value is 0.05.

Here, P-value is less than the $\alpha$ value.

That is $0.000\left(=P\right)<0.05\left(=\alpha \right)$.

By the rejection rule, fail to reject the null hypothesis.

Therefore, coefficient of the predictor variable distance is significant.

Thus, there is not enough evidence to conclude that ${\beta }_1$ is zero.

Test for slope coefficient ${\beta }_2$:

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_2=0$

That is, coefficient of the predictor variable distance square is not significant.

Alternative hypothesis:

$H_a:{\beta }_2\ne 0$

That is, coefficient of the predictor variable distance square is significant.

From the MINITAB output, the P- value corresponding to the coefficient ${\beta }_2$ is 0.000.

Conclusion:

The P-value is 0.000 and $\alpha$ value is 0.05.

Here, P-value is less than the $\alpha$ value.

That is $0.000\left(=P\right)<0.05\left(=\alpha \right)$.

By the rejection rule, fail to reject the null hypothesis.

Therefore, the coefficient of the predictor variable distance square is significant.

Thus, there is not enough evidence to conclude that ${\beta }_2$ is zero.

c.

To determine

Construct a cubic regression model for the data.

Test whether there is enough evidence to conclude that ${\beta }_0$ is zero.

Test whether there is enough evidence to conclude that ${\beta }_1$ is zero.

Test whether there is enough evidence to conclude that ${\beta }_2$ is zero.

Test whether there is enough evidence to conclude that ${\beta }_3$ is zero.

Answer to Problem 8E

The quadratic regression model for the data is $\underset{\_}{\widehat{y}=3.27390-3.69506x+1.29550x^2-0.09672x^3}$.

No, there is enough evidence to conclude that ${\beta }_0$ is zero.

No, there is enough evidence to conclude that ${\beta }_1$ is zero.

Yes, there is enough evidence to conclude that ${\beta }_2$ is zero.

Yes, there is enough evidence to conclude that ${\beta }_3$ is zero.

Explanation of Solution

Calculation:

Cubic model:

The cubic regression model would be of the form:

$\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_{1}x+{\widehat{\beta }}_2{x}^2+{\widehat{\beta }}_3{x}^3$

Regression:

Software procedure:

Step by step procedure to obtain cubic regression model using MINITAB software is given as,

• Choose Stat > Regression > General Regression.
• In Response, enter the numeric column containing the response data Y.
• In Model, enter the numeric column containing the predictor variables X, X-square and X-cube.
• Click OK.

Output obtained from MINITAB is given below:

The ‘Coefficient’ column of the regression analysis MINITAB output gives the slopes corresponding to the respective variables stored in the column ‘Predictor’.

A careful inspection of the output shows that the fitted model is:

$\widehat{y}=3.27390-3.69506x+1.29550x^2-0.09672x^3$.

Hence, the cubic regression model for the data is:

$\underset{\_}{\widehat{y}=3.27390-3.69506x+1.29550x^2-0.09672x^3}$.

Test for slope coefficient ${\beta }_0$:

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_0=0$

That is, intercept of the regression model is not significant.

Alternative hypothesis:

$H_a:{\beta }_0\ne 0$

That is, intercept of the regression model is significant.

Level of significance:

Since, the level of significance is not given. The prior level of significance $\alpha =0.05$ can be used.

The ‘P’ column of the regression analysis MINITAB output gives the P- value corresponding to the respective variables stored in the column ‘Term’.

The P- value corresponding to the coefficient ${\beta }_0$ is 0.000.

Decision criteria based on P-value approach:

If $P\text{-value}\le \alpha$, then reject the null hypothesis $H_0$.

If $P\text{-value}>\alpha$, then fail to reject the null hypothesis $H_0$.

Conclusion:

The P-value is 0.000 and $\alpha$ value is 0.05.

Here, P-value is less than the $\alpha$ value.

That is $0.000\left(=P\right)<0.05\left(=\alpha \right)$.

By the rejection rule, reject the null hypothesis.

Therefore, intercept of the regression model is significant.

Thus, there is not enough evidence to conclude that ${\beta }_0$ is zero.

Test for slope coefficient ${\beta }_1$:

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_1=0$

That is, coefficient of the predictor variable distance is not significant.

Alternative hypothesis:

$H_a:{\beta }_1\ne 0$

That is, coefficient of the predictor variable distance is significant.

From the MINITAB output, the P- value corresponding to the coefficient ${\beta }_1$ is 0.002.

Conclusion:

The P-value is 0.002 and $\alpha$ value is 0.05.

Here, P-value is less than the $\alpha$ value.

That is $0.002\left(=P\right)<0.05\left(=\alpha \right)$.

By the rejection rule, reject the null hypothesis.

Therefore, coefficient of the predictor variable distance is significant.

Thus, there is not enough evidence to conclude that ${\beta }_1$ is zero.

Test for slope coefficient ${\beta }_2$:

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_2=0$

That is, coefficient of the predictor variable distance square is not significant.

Alternative hypothesis:

$H_a:{\beta }_2\ne 0$

That is, coefficient of the predictor variable distance square is significant.

From the MINITAB output, the P- value corresponding to the coefficient ${\beta }_2$ is 0.081.

Conclusion:

The P-value is 0.081 and $\alpha$ value is 0.05.

Here, P-value is greater than the $\alpha$ value.

That is $0.081\left(=P\right)>0.05\left(=\alpha \right)$.

By the rejection rule, fail to reject the null hypothesis.

Therefore, the coefficient of the predictor variable distance square is not significant.

Thus, there is enough evidence to conclude that ${\beta }_2$ is zero.

Test for slope coefficient ${\beta }_3$:

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_3=0$

That is, coefficient of the predictor variable distance cube is not significant.

Alternative hypothesis:

$H_a:{\beta }_3\ne 0$

That is, coefficient of the predictor variable distance cube is significant.

From the MINITAB output, the P- value corresponding to the coefficient ${\beta }_3$ is 0627.

Conclusion:

The P-value is 0.627 and $\alpha$ value is 0.05.

Here, P-value is greater than the $\alpha$ value.

That is $0.627\left(=P\right)>0.05\left(=\alpha \right)$.

By the rejection rule, fail to reject the null hypothesis.

Therefore, the coefficient of the predictor variable distance cube is not significant.

Thus, there is enough evidence to conclude that ${\beta }_3$ is zero.

d.

To determine

Find the best model among the three models obtained in part (a), part (b) and part (c).

Answer to Problem 8E

The model obtained in part (b) is the best model compared to the other two models in part (a) and part (c).

Explanation of Solution

Calculation:

In the model obtained in part (b), all the coefficients of the model are significantly different from zero.

The coefficient of determination is higher for the model obtained in part (b), than for the model obtained in part (a).

That is, $0.9976>0.8975$.

There is not much difference in the coefficient of determination for the model obtained in part (b) and part (c).

That is, 0.9976% and 0.9978% are not much distinct.

Increasing the number of predictors in an analysis increases the complexity of analysis. An investigator usually does not wish to increase the complications of analysis for such a small increase in $R^2$.

Thus, the model obtained in part (b) is the best model compared to the other two models in part (a) and part (c).

e.

To determine

Estimate the amount of deflection at a distance of 1m using the most appropriate method.

Answer to Problem 8E

The estimate of the amount of deflection at a distance of 1m is 0.77231mm.

Explanation of Solution

Calculation:

The quadratic model is the most appropriate regression model among the obtained three models.

The quadratic regression model is, $\widehat{y}=3.25743-3.51514x+1.03002x^2$

Here, $x=1\text{m}$.

Estimate of amount of deflection:

$\begin{array}{c}\widehat{y}=3.25743-3.51514x+1.03002x^2\\ =3.25743-\left(3.51514\times 1\right)+\left(1.03002\times 1^2\right)\\ =0.77231\end{array}$

Thus, the estimate of the amount of deflection at a distance of 1m is 0.77231mm.