Chapter 14.3, Problem 46E

a.

To determine

Test whether the given model is useful or not at the 0.05 level of significance.

Answer to Problem 46E

There is convincing evidence that the given model is useful at the 0.05 level of significance.

Explanation of Solution

Calculation:

It is given that the variable y is absorption, $x_1$ is flour protein, and $x_2$ is starch damage. The sample size is 28 and the number of variables (k) is 2. The given regression equation is $y=19.4+1.44x_1+0.336x_2$.

1.

The model is $y=\alpha +{\beta }_1{x}_1+{\beta }_2{x}_2+e$.

2.

Null hypothesis:

$H_0:{\beta }_1={\beta }_2=0$

That is, there is no useful relationship between y and any of the predictors.

3.

Alternative hypothesis:

$H_a:$ At least one among ${\beta }_i^{\text{'}}\text{s}$ is not zero.

That is, there is a useful relationship between y and any of the predictors.

4.

Here, the significance level is $\alpha =0.05$.

5.

Test statistic:

$F=\frac{\frac{R^2}{k}}{\frac{\left(1-R^2\right)}{n-\left(k+1\right)}}$

Here, n is the sample size, and k is the number of variables in the model.

6.

Assumptions:

Since there is no availability of original data to check the assumptions, there is a need to assume that the variables are related to the model, and the random deviation is distributed normally with mean 0 and the fixed standard deviation.

7.

Calculation:

From the MINITAB output, the value of $R^2$ is 0.964.

The value of F-test statistic is calculated as follows:

$\begin{array}{c}F=\frac{\frac{R^2}{k}}{\frac{\left(1-R^2\right)}{n-\left(k+1\right)}}\\ =\frac{\frac{0.964}{2}}{\frac{\left(1-0.964\right)}{\left(28-\left(2+1\right)\right)}}\\ =\frac{0.482}{\frac{0.036}{25}}\\ =\frac{0.482\times 25}{0.036}\\ =\frac{12.05}{0.036}\\ =334.722\end{array}$

8.

P-value:

Software procedure:

Step-by-step procedure to find the P-value using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘F’ distribution.
• Enter the Numerator df as 2 and Denominator df as 25.
• Click the Shaded Area tab.
• Choose X Value and Right Tail for the region of the curve to shade.
• Enter the X value as 334.722.
• Click OK.

Output obtained using the MINITAB software is represented as follows:

From the MINITAB output, the P-value is 0.

9.

Conclusion:

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

Therefore, the P-value of 0 is less than the 0.05 level of significance.

Hence, reject the null hypothesis.

Thus, there is convincing evidence that the given model is useful at the 0.05 level of significance.

b.

To determine

Calculate a 95% confidence interval for ${\beta }_2$ and interpret it.

Answer to Problem 46E

The 95% confidence interval for ${\beta }_2$ is $\left(\underset{\_}{0.298,0.373}\right)$.

Explanation of Solution

Calculation:

Here, ${\beta }_2$ is the coefficient of variable $x_2$.

Since there is no availability of original data to check the assumptions, there is a need to assume that the variables are related to the model, and the random deviation is distributed normally with mean 0 and the fixed standard deviation.

The formula for confidence interval for ${\beta }_2$ is as follows:

$b_2\pm \left(t\text{ critical value}\right)s_{b_2}$.

Where, $b_2$ is the estimated value of ${\beta }_2$ and $s_{b_2}$ is the estimated standard deviation of $b_2$.

Degrees of freedom:

$\begin{array}{c}df=n-\left(k+1\right)\\ =28-\left(2+1\right)\\ =28-3\\ =25\end{array}$

Software procedure:

Step-by-step procedure to find the P-value using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• Enter the Degrees of freedom as 25.
• Click the Shaded Area tab.
• Choose Probability and Both for the region of the curve to shade.
• Enter the Probability value as 0.05.
• Click OK.

Output obtained using the MINITAB software is represented as follows:

From the MINITAB output, the critical value is 2.060.

From the given MINITAB output, the value of $b_2$ is 0.33563 and the value of $s_{b_2}$ is 0.01814.

The confidence interval for mean change in exam score is calculated as follows:

$\begin{array}{c}{b}_2\pm \left(t\text{ critical value}\right)s_{b_2}=0.33563\pm \left(2.060\right)0.01814\\ =0.33563\pm 0.03737\\ =\left(0.33563-0.03737,0.33563+0.03737\right)\\ =\left(0.29826,0.373\right)\\ \approx \left(0.298,0.373\right)\end{array}$

Thus, the 95% confidence interval for ${\beta }_2$ is $\left(\underset{\_}{0.298,0.373}\right)$.

There is 95% confident that the average increase in y is associated with 1-unit increase in starch damage that is between 0.298 and 0.373, when the other predictors are fixed.

c.

To determine

1. i. Test the hypothesis $H_0:{\beta }_1=0$ versus $H_a:{\beta }_1\ne 0$.
2. ii. Test the hypothesis $H_0:{\beta }_2=0$ versus $H_a:{\beta }_2\ne 0$

Answer to Problem 46E

It can be concluded that the quadratic term should not be eliminated from the model and simple linear model should not be sufficient.

Explanation of Solution

Calculation:

i.

1.

The predictor ${\beta }_1$ is the coefficient of $x_1$. Here, $x_1$ is the flour protein.

2.

Null hypothesis:

$H_0:{\beta }_1=0$.

3.

Alternative hypothesis:

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

4.

Here, the common significance levels are $\alpha =0.05,0.01,0.10$.

5.

Test statistic:

$t=\frac{b_1}{s_{b_1}}$

Here, $b_1$ is the estimated value of ${\beta }_1$, and $s_{b_1}$ is the estimated standard deviation of $b_1$.

6.

Assumptions:

The random deviations from the values by the population regression equation are distributed normally with mean 0 and fixed standard deviation.

7.

Calculation:

From the MINITAB output, the t test statistic value of $x_1$ is 6.95.

8.

P-value:

From the MINITAB output, the P-value for $x_1$ is 0.000.

9.

Conclusion:

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

Therefore, the P-value of 0 is less than any common levels of significance, such as 0.05, 0.01, and 0.10.

Hence, reject the null hypothesis.

Thus, there is convincing evidence that the flour protein variable is important and it is $\underset{\_}{{\beta }_1\ne 0}$.

ii.

1.

The predictor ${\beta }_2$ is the coefficient of $x_2$. Here, $x_2$ is the starch damage.

2.

Null hypothesis:

$H_0:{\beta }_2=0$.

3.

Alternative hypothesis:

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

4.

Here, the common significance levels are $\alpha =0.05,0.01,0.10$.

5.

Test statistic:

$t=\frac{b_2}{s_{b_2}}$

Here, $b_2$ is the estimated value of ${\beta }_2$, and $s_{b_2}$ is the estimated standard deviation of $b_2$.

6.

Assumptions:

The random deviations from the values by the population regression equation are distributed normally with mean 0 and fixed standard deviation.

7.

Calculation:

From the MINITAB output, the t test statistic value of $x_2$ is 18.51.

8.

P-value:

From the MINITAB output, the P-value for $x_1$ is 0.000.

9.

Conclusion:

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

Therefore, the P-value of 0 is less than any common levels of significance, such as 0.05, 0.01, and 0.10.

Hence, reject the null hypothesis.

Thus, there is convincing evidence that the starch damage variable is important and it is $\underset{\_}{{\beta }_2\ne 0}$.

d.

To determine

Explain whether both the independent variables are important or not.

Explanation of Solution

From the results of Part (c), there is evidence that the variables of flour protein and starch damage are important.

e.

To determine

Calculate a 90% confidence interval and interpret it.

Answer to Problem 46E

The 90% confidence interval is $\left(\underset{\_}{54.5084,56.2916}\right)$.

Explanation of Solution

Calculation:

It is given that the value of $x_1$ is 11.7 and the value of $x_2$ is 57. The standard deviation is 0.522.

Assume that the exam score is associated with the predictors according to the model. The random deviations from the values by the population regression equation are distributed normally with mean 0 and fixed standard deviation.

The formula for prediction interval for mean y value is as follows:

$\widehat{y}\pm \left(t\text{critical value}\right)s_{\widehat{y}}$.

The value of $\widehat{y}$ is calculated as follows:

$\begin{array}{c}y=19.4+1.44x_1+0.336x_2\\ =19.4+1.44\left(11.7\right)+0.336\left(57\right)\\ =19.4+16.848+19.152\\ =55.4\end{array}$

Degrees of freedom:

$\begin{array}{c}df=n-\left(k+1\right)\\ =28-\left(2+1\right)\\ =28-3\\ =25\end{array}$

Software procedure:

Step-by-step procedure to find the P-value using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• Enter the Degrees of freedom as 25.
• Click the Shaded Area tab.
• Choose Probability and Both for the region of the curve to shade.
• Enter the Probability value as 0.10.
• Click OK.

Output obtained using the MINITAB software is represented as follows:

From the MINITAB output, the critical value is 1.708.

The prediction interval for mean y value is calculated as follows:

$\begin{array}{c}\widehat{y}\pm \left(t\text{critical value}\right)s_{\widehat{y}}=55.4\pm \left(1.708\right)0.522\\ =55.4\pm 0.8916\\ =\left(55.4-0.8916,55.4+0.8916\right)\\ =\left(54.5084,56.2916\right)\end{array}$

Thus, the 90% confidence interval is $\left(\underset{\_}{54.5084,56.2916}\right)$.

There is 90% confident that the mean water absorption for wheat with 11.7 flour protein and 57 starch damage is between 54.5084 and 56.2916.

f.

To determine

Predict the water absorption for the shipment by 90% interval.

Answer to Problem 46E

The water absorption values for the particular shipment are between 53.3297 and 57.4703 by 90% prediction interval.

Explanation of Solution

Calculation:

It is given that the value of $x_1$ is 11.7 and the value of $x_2$ is 57. The standard deviation is 0.522.

Assume that the exam score is associated with the predictors according to the model. The random deviations from the values by the population regression equation are distributed normally with mean 0 and fixed standard deviation.

The formula for prediction interval for mean y value is as follows:

$\widehat{y}\pm \left(t\text{critical value}\right)\sqrt{s_e^2+s_{{}_{\widehat{y}}}^2}$.

From the given MINITAB output, the standard deviation is 1.094.

From Part e., the value of $\widehat{y}$ is 55.4.

From the MINITAB output in Part e., the critical value is 1.708.

The prediction interval for water absorption for shipment is calculated as follows:

$\begin{array}{c}\widehat{y}\pm \left(t\text{critical value}\right)\sqrt{s_e^2+s_{{}_{\widehat{y}}}^2}=55.4\pm \left(1.708\right)\sqrt{{1.094}^2+{0.522}^2}\\ =55.4\pm \left(1.708\right)\sqrt{1.1968+0.2725}\\ =55.4\pm \left(1.708\right)1.2121\\ =55.4\pm 2.0703\\ =\left(55.4-2.0703,55.4+2.0703\right)\\ =\left(53.3297,57.4703\right)\end{array}$

Thus, the 90% prediction interval is $\left(\underset{\_}{53.3297,57.4703}\right)$.

By prediction at 90% interval, the water absorption values for the particular shipment are between 53.3297 and 57.4703.