Chapter 13.4, Problem 54E

a.

To determine

Identify the predictors that could be used in the model to incorporate suppliers and lubrication regimens in addition to blank holder pressure.

Answer to Problem 54E

The predictors that could be used in the model to incorporate suppliers and lubrication regimens in addition to blank holder pressure are given below:

$x_2=\{\begin{array}{l}1,\text{ supplier 1}\\ \text{0, otherwise}\end{array}$

$x_3=\{\begin{array}{l}1,\text{ supplier 2}\\ \text{0, otherwise}\end{array}$

$x_4=\{\begin{array}{l}1,\text{ lubricant 1}\\ \text{0, otherwise}\end{array}$

$x_5=\{\begin{array}{l}1,\text{ lubricant 2}\\ \text{0, otherwise}\end{array}$

Explanation of Solution

Given info:

The MINITAB output shows that the springback from the wall opening angle being predicted using blank holder pressure, three types of material suppliers and three types of lubrication regimens.

Calculation:

Dummy or indicator variable:

If there a categorical variable has k levels then k–1 dummy variables would be included in the model.

The material suppliers and lubrication regimens are categorical variables with three levels each.

For material suppliers:

Fix the third level as a base and create two dummy variables as shown below:

$x_2=\{\begin{array}{l}1,\text{ supplier 1}\\ \text{0, otherwise}\end{array}$

$x_3=\{\begin{array}{l}1,\text{ supplier 2}\\ \text{0, otherwise}\end{array}$

For lubrication regimens:

Fix the first level (no lubricant) as a base and create two dummy variables as shown below:

$x_4=\{\begin{array}{l}1,\text{ lubricant 1}\\ \text{0, otherwise}\end{array}$

$x_5=\{\begin{array}{l}1,\text{ lubricant 2}\\ \text{0, otherwise}\end{array}$

b.

To determine

Test the hypothesis to conclude whether the model specifies a useful relationship between springback from the wall opening angle and at least one of the five predictor variables.

Answer to Problem 54E

There is sufficient evidence to conclude that the model specifies a useful relationship between the springback from the wall opening angle and at least one of the five predictor variables

Explanation of Solution

Given info:

The MINTAB output was given.

Calculation:

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_1={\beta }_2={\beta }_3={\beta }_4={\beta }_5=0$

That is, there is no use of linear relationship between springback from the wall opening angle and at least one of the five predictor variables.

Alternative hypothesis:

$H_a:\text{At least one of the }\beta \text{'}\text{s}\ne 0$

That is, there is a use of linear relationship between springback from the wall opening angle and at least one of the five predictor variables.

From the MINITAB output, it can be observed that the P-value corresponding to the F-statistic is 0.000.

Rejection region:

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

If $P\text{-value}<\alpha$, then reject the null hypothesis $\left(H_0\right)$.

Conclusion:

The P- value is 0.000 and the level of significance is 0.001.

The P- value is lesser than the level of significance.

That is, $0.000\left(=P\text{-value}\right)<0.001\left(=\alpha \right)$.

Thus, the null hypothesis is rejected,

Hence, there is sufficient evidence to conclude that there is a use of linear relationship between springback from the wall opening angle and at least one of the five predictor variables.

c.

To determine

Calculate the 95% prediction interval for the springback from the wall opening angle when BHP is 1,000, material from supplier 1 and no lubrication.

Answer to Problem 54E

The 95% prediction interval for the predicted springback from the wall opening angle when BHP is 1,000, material from supplier 1 and no lubrication is $\left(13.80,19.09\right)$

Explanation of Solution

Given info:

The BHP is 1,000, material from supplier 1 and no lubrication. The corresponding standard deviation for prediction is 0.524.

Calculation:

The prediction value of springback from the wall opening angle when BHP is 1,000, material from supplier 1 and no lubrication is calculated as follows:

$\begin{array}{c}\widehat{y}=21.5322-0.0033680\left(1,000\right)-1.7181\left(1\right)-1.4840\left(0\right)-0.3063\left(0\right)-0.8931\left(0\right)\\ =21.5322-3.368-1.7181\\ =16.4461\end{array}$

Thus, the prediction value of springback from the wall opening angle when BHP is 1,000, material from supplier 1 and no lubrication is 16.4461.

95% prediction interval:

The prediction interval is calculated using the formula:

${\widehat{Y}}_i\pm t_{\frac{\alpha }{2},n-\left(k+1\right)}\sqrt{s^2+{\left(s_{\widehat{Y_i}}\right)}^2}$

Where,

$\widehat{Y_i}$ is the estimated value of the dependent variable.

$\alpha$ is the level of significance.

n is the total number of observations.

k is the total number of predictors in the model.

$s_{\widehat{Y_i}}$ is the standard error while calculating the estimated value of the dependent variable.

$s^2$ is the standard deviation obtained after fitting the model.

Critical value:

Software procedure:

Step-by-step procedure to find the critical value is given below:

• Click on Graph, select View Probability and click OK.
• Select t, enter 30 as Degrees of freedom, inShaded Area Tab select Probability under Define Shaded Area By and choose Both tails.
• Enter Probability value as 0.05.
• Click OK.

Output obtained from MINITAB is given below:

The 95% confidence interval for the predicted amount of beta carotene is calculated as follows:

$\begin{array}{c}{\widehat{Y}}_i\pm t_{\frac{\alpha }{2},n-\left(k+1\right)}\sqrt{s^2+{\left(s_{\widehat{Y_i}}\right)}^2}=16.4461\pm t_{\frac{0.05}{2},36-\left(5+1\right)}\sqrt{{\left(1.18413\right)}^2+{\left(0.524\right)}^2}\\ =16.4461\pm t_{0.025,30}\sqrt{1.6767}\\ =16.4461\pm 2.042\left(1.2949\right)\\ =16.4461\pm 2.6442\end{array}$

$=13.80,19.09$

Thus, the 95% prediction interval for the predicted springback from the wall opening angle when BHP is 1,000, material from supplier 1 and no lubricationis $\left(13.80,19.09\right)$.

d.

To determine

Find the coefficient of multiple determination.

Give the conclusion stating the importance of lubrication regimen.

Answer to Problem 54E

The coefficient of multiple determination is 74.1%

The lubrication regimen is not important because there is not much of a difference in the coefficient of multiple determination even after removing the two variables corresponding to lubrication.

The two variables corresponding to lubrication shows no effect and need not be included in the model as long as the other predictors BHP and suppliers were retained.

Explanation of Solution

Given info:

The SSE after removing the variables corresponding to lubrication regimen is 48.426.

Calculation:

Coefficient of determination:

The coefficient of determination tells the total amount of variation in the dependent variable explained by the independent variable. It ranges from 0 to 1.

$R^2=1-\frac{\text{SSE}}{\text{SST}}$

Substitute 48.426 as SSE and 186.980 as SST.

$\begin{array}{c}{R}^2=1-\frac{\text{48}\text{.426}}{\text{186}\text{.980}}\\ =\frac{186.980-48.426}{186.980}\\ =\frac{138.554}{186.980}\\ =0.741\end{array}$

Thus, the coefficient of multiple determination is 74.1%

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_4={\beta }_5=0$

That is, the two dummy variables corresponding to lubrication are not significant to explain the variation in springback from wall opening angle.

Alternative hypothesis:

$H_a:\text{At least }\beta \text{'}\text{s}\ne 0$

That is, At least one of the two dummy variables corresponding to lubrication is significant to explain the variation in springback from wall opening angle.

Test statistic:

$f=\frac{\frac{\left(SS{E}_l-SS{E}_k\right)}{k-l}}{\frac{SS{E}_k}{\left[n-\left(k+1\right)\right]}}$

Where,

$SS{E}_k$ represents the sum of squares due to error obtained from the full model.

$SS{E}_l$ represents the sum of squares due to error obtained from the reduced model.

n represents the total number of observations,

k represents the number of predictors on the full model.

l represents the number of predictors on the reduced model.

Substitute 48.426for $SS{E}_k$, 42.065 for $SS{E}_l$, 36 for n, 5 for k and 3 for l .

$\begin{array}{c}f=\frac{\frac{\left(48.426-42.065\right)}{5-3}}{\frac{42.065}{\left[36-\left(5+1\right)\right]}}\\ =\frac{\frac{6.361}{2}}{\frac{42.065}{\left[36-6\right]}}\\ =\frac{3.1805}{\frac{42.065}{30}}\\ =\frac{95.415}{42.065}\end{array}$

$=2.268$

Critical value:

Software procedure:

• Click on Graph, select View Probability and click OK.
• Select F, enter 2 in numerator df and 30 in denominator df.
• Under Shaded Area Tab select Probability under Define Shaded Area By and select Right tail.
• Choose Probability value as 0.05.
• Click OK.

Output obtained from MINITAB is given below:

Conclusion:

Coefficient of determination:

The coefficient of determination for the whole model including the two variables corresponding to lubrication is 77.5% whereas the coefficient of multiple determination after removing the two variables corresponding to lubrication is 74.1%. Thus, there is a slight drop in coefficient of multiple determination.

Testing the hypothesis:

The test statistic value is 2.268 and the critical value is 3.316.

The test statistic is lesser than the critical value.

That is, $2.268\left(=\text{test statistic}\right)<3.316\left(=\text{critical value}\right)$.

Thus, the null hypothesis is not rejected,

Hence, there is sufficient evidence to conclude thatthe two dummy variables corresponding to lubrication are not significant to explain the variation in springback from wall opening angle.

e.

To determine

Identify whether the given model has improved than the model specified in part (d).

Answer to Problem 54E

Yes, the given model has improved than the model specified in part (d).

There is sufficient evidence to conclude thatthe addition of interaction terms is significant to explain the variation in springback from wall opening angle at 5% level of significance.

Explanation of Solution

Given info:

A regression model is built with the five predictors and the interactions between BHP and the four dummy variables.

The resulting SSE is 28.216 and the $R^2$ value is 0.849.

Calculation:

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_4={\beta }_5={\beta }_6={\beta }_7={\beta }_8=0$

That is, the addition of interaction terms is not significant to explain the variation in the dependent variable y.

Alternative hypothesis:

$H_a:\text{At least one of the }\beta \text{'}\text{s}\ne 0$

That is, at least one of theinteraction terms is significant to explain the variation in the dependent variable y.

The degrees of freedom for the regression would be 4.

The degrees of freedom for the error would be $30-4=26$.

Test statistic:

$\begin{array}{c}f=\frac{\frac{42.065-28.216}{4}}{\frac{28.216}{26}}\\ =\frac{\frac{13.849}{4}}{1.085}\\ =\frac{13.849}{4.34}\\ =3.191\end{array}$

Critical value:

Software procedure:

• Click on Graph, select View Probability and click OK.
• Select F, enter 4 in numerator df and 26 in denominator df.
• Under Shaded Area Tab select Probability under Define Shaded Area By and select Right tail.
• Choose Probability value as 0.05.
• Click OK.

Output obtained from MINITAB is given below:

Conclusion:

The test statistic value is 3.191 and the critical value is 2.743.

The test statistic is greater than the critical value.

That is, $3.191\left(=\text{test statistic}\right)>2.743\left(=\text{critical value}\right)$.

Thus, the null hypothesis is rejected,

Hence, there issufficient evidence to conclude thatthe addition of interaction terms is significant to explain the variation in springback from wall opening angle at 5% level of significance.