Chapter 8.1, Problem 2E

a.

To determine

Find the 95% confidence interval for the slope of Manganese.

Answer to Problem 2E

The 95% confidence interval for the slope of Manganese is $\underset{\_}{2.6196\le {\beta }_1\le 4.0206}$.

Explanation of Solution

Given info:

The data represents the MINITAB output of the regression model $\text{Tensile}\text{strength}={\beta }_0+{\beta }_1\text{Manganese+}{\beta }_2\text{Thickness}$ for a sample of 20 plates in order to determine the effect of thickness and manganese content in tensile strength of steel plates.

Calculation:

Multiple linear regression model:

A multiple linear regression model is given as $\widehat{y}=b_0+b_1{x}_1+\mathrm{...}+b_k{x}_k$ where $\widehat{y}$ is the predicted value of response variable, and $x_1,x_2,\mathrm{...},x_k$ are the k predictor variables. The quantities $b_1,b_2,\mathrm{...},b_k$ are the estimated slopes corresponding to $x_1,x_2,\mathrm{...},x_k$ respectively.$b_0$ is the estimated intercept of the line, from the sample data.

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

Let $x_1,x_2$ be the Thickness (mm) and Manganese content (in parts per thousand).

From the accompanying MINITAB output, the slope coefficient of Manganese is $b_1\left(\text{Manganese}\right)={\widehat{\beta }}_1=3.3201$.

Confidence interval:

The general formula for the confidence interval for the slope of the regression line is,

$CI={\widehat{\beta }}_i\pm t_{\frac{\alpha }{2},n-\left(k+1\right)}{s}_{{\widehat{\beta }}_i}$

Where, ${\widehat{\beta }}_i$ be the slope coefficient of the sample regression line, $s_{{\widehat{\beta }}_i}$ be the estimate of error standard deviation of the estimated slope coefficient.

From the accompanying MINITAB output, the estimate of error standard deviation of slope coefficient of Manganese is $s_{{\widehat{\beta }}_1}=0.33198$.

Critical value:

For 95% confidence level,

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\\ \frac{\alpha }{2}=\frac{0.05}{2}\\ =0.025\end{array}$

Degrees of freedom:

The number of plates that are sampled is $n=20$ and the number of predictor variables is $k=3$.

The degrees of freedom is,

$\begin{array}{c}d.f=n-\left(k+1\right)\\ =20-3\\ =17\end{array}$

From Table A.5 of the t- distribution in Appendix A, the critical value corresponding to the right tail area 0.025 and 17 degrees of freedom is 2.110.

Thus, the critical value is $\left(t_{{}_{\frac{\alpha }{2},n-\left(k+1\right)}}\right)=2.110$.

The 95% confidence interval is,

$\begin{array}{c}C.I={\widehat{\beta }}_i-\left(t_{{}_{\frac{\alpha }{2},n-\left(k+1\right)}}\times s_{{\widehat{\beta }}_i}\right)\le {\beta }_i\le {\widehat{\beta }}_i+\left(t_{{}_{\frac{\alpha }{2},n-\left(k+1\right)}}\times s_{{\widehat{\beta }}_i}\right)\\ ={\widehat{\beta }}_1-\left(t_{{}_{\frac{\alpha }{2},n-\left(k+1\right)}}\times s_{{\widehat{\beta }}_1}\right)\le {\beta }_1\le {\widehat{\beta }}_1+\left(t_{{}_{\frac{\alpha }{2},n-\left(k+1\right)}}\times s_{{\widehat{\beta }}_1}\right)\\ =\left(3.3201-\left(2.110\times 0.33198\right)\le {\beta }_1\le 3.3201+\left(2.110\times 0.33198\right)\right)\\ =\left(3.3201\pm 0.7005\right)\\ =\left(2.6196,4.0206\right)\end{array}$

Thus, the 95% confidence interval for the slope of Manganese is $\underset{\_}{2.6196\le {\beta }_1\le 4.0206}$.

Interpretation:

There is 95% confident that the average change in the tensile strength associated with 1 ppt increase in manganese lies between $2.6196\frac{\text{kg}}{{\text{mm}}^{\text{2}}}$ MPa and $4.0206\frac{\text{kg}}{{\text{mm}}^{\text{2}}}$ MPa given that feed thickness is fixed constant.

b.

To determine

Find the 99% confidence interval for the slope of Thickness.

Answer to Problem 2E

The 99% confidence interval for the slope of Thickness is $\underset{\_}{-0.7902\le {\beta }_2\le -0.0596}$.

Explanation of Solution

Calculation:

From the accompanying MINITAB output, the slope coefficient of Thickness is $b_2\left(\text{Thickness}\right)={\widehat{\beta }}_2=-0.4249$.

Confidence interval:

The general formula for the confidence interval for the slope of the regression line is,

$CI={\widehat{\beta }}_i\pm t_{\frac{\alpha }{2},n-\left(k+1\right)}{s}_{{\widehat{\beta }}_i}$

Where, ${\widehat{\beta }}_i$ be the slope coefficient of the sample regression line, $s_{{\widehat{\beta }}_i}$ be the estimate of error standard deviation of the estimated slope coefficient.

From the accompanying MINITAB output, the estimate of error standard deviation of slope coefficient of Thickness is $s_{{\widehat{\beta }}_2}=0.12606$.

Critical value:

For 99% confidence level,

$\begin{array}{c}1-\alpha =1-0.99\\ \alpha =0.01\\ \frac{\alpha }{2}=\frac{0.01}{2}\\ =0.005\end{array}$

Degrees of freedom:

The number of plates that are sampled is $n=20$ and the number of predictor variables is $k=3$.

The degrees of freedom is,

$\begin{array}{c}d.f=n-\left(k+1\right)\\ =20-3\\ =17\end{array}$

From Table A.5 of the t- distribution in Appendix A, the critical value corresponding to the right tail area 0.005 and 17 degrees of freedom is 2.898.

Thus, the critical value is $\left(t_{{}_{\frac{\alpha }{2},n-\left(k+1\right)}}\right)=2.898$.

The 95% confidence interval is,

$\begin{array}{c}C.I={\widehat{\beta }}_i-\left(t_{{}_{\frac{\alpha }{2},n-\left(k+1\right)}}\times s_{{\widehat{\beta }}_i}\right)\le {\beta }_i\le {\widehat{\beta }}_i+\left(t_{{}_{\frac{\alpha }{2},n-\left(k+1\right)}}\times s_{{\widehat{\beta }}_i}\right)\\ ={\widehat{\beta }}_2-\left(t_{{}_{\frac{\alpha }{2},n-\left(k+1\right)}}\times s_{{\widehat{\beta }}_2}\right)\le {\beta }_2\le {\widehat{\beta }}_2+\left(t_{{}_{\frac{\alpha }{2},n-\left(k+1\right)}}\times s_{{\widehat{\beta }}_2}\right)\\ =\left(-0.4249-\left(2.898\times 0.12606\right)\le {\beta }_1\le -0.4249+\left(2.898\times 0.12606\right)\right)\\ =\left(-0.4249\pm 0.3653\right)\\ =\left(-0.7902,-0.0596\right)\end{array}$

Thus, the 99% confidence interval for the slope of Thickness is $\underset{\_}{-0.7902\le {\beta }_2\le -0.0596}$.

Interpretation:

There is 99% confident that the average change in the tensile strength associated with 1 ppt increase in Thickness lies between $-0.7902\frac{\text{kg}}{{\text{mm}}^{\text{2}}}$ MPa and $-0.0596\frac{\text{kg}}{{\text{mm}}^{\text{2}}}$ MPa given that feed manganese is fixed constant.

c.

To determine

Test whether there is enough evidence to conclude that ${\beta }_1>3$.

Answer to Problem 2E

There is no sufficient evidence to conclude that ${\beta }_1>3$.

Explanation of Solution

Calculation:

From the accompanying MINITAB output, the slope coefficient of Manganese is $b_1\left(\text{Manganese}\right)={\widehat{\beta }}_1=3.3201$.

Here, the claim is that ${\beta }_1>3$.

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_1\le 3$

That is, 1 ppt increase in manganese tends to at most $3\frac{\text{kg}}{{\text{mm}}^{\text{2}}}$ increase in the true average of tensile strength, provided the effect of Thickness is accounted for.

Alternative hypothesis:

$H_1:{\beta }_1>3$

That is, increase in the true average of tensile strength due to 1 ppt increase in manganese will be greater than $3\frac{\text{kg}}{{\text{mm}}^{\text{2}}}$, provided the effect of Thickness is accounted for.

Test statistic:

The test statistic is,

$t=\frac{{\widehat{\beta }}_i-{\beta }_i}{s_{{\widehat{\beta }}_i}}\sim t_{\left(n-\left(k+1\right)\right)}$

Where, ${\widehat{\beta }}_i$ be the slope coefficient of the sample regression line, ${\beta }_i$ be the slope coefficient of the population regression line and $s_{{\widehat{\beta }}_i}$ be the estimate of error standard deviation of the estimated slope coefficient.

From the accompanying MINITAB output, the estimate of error standard deviation of slope coefficient of Manganese is $s_{{\widehat{\beta }}_1}=0.33198$.

Test statistic under null hypothesis:

Under the null hypothesis, the test statistic is obtained as follows:

$\begin{array}{c}t=\frac{{\widehat{\beta }}_i-{\beta }_i}{s_{{\widehat{\beta }}_i}}\sim t_{\left(n-\left(k+1\right)\right)}\\ =\frac{{\widehat{\beta }}_1-{\beta }_1}{s_{{\widehat{\beta }}_1}}\\ =\frac{3.3201-3}{0.33198}\\ =0.9642\end{array}$

Thus, the test statistic is 0.9642.

Degrees of freedom:

The number of plates that are sampled is $n=20$ and the number of predictor variables is $k=3$.

The degrees of freedom is,

$\begin{array}{c}d.f=n-\left(k+1\right)\\ =20-3\\ =17\end{array}$

Thus, the degree of freedom is 17.

Here, level of significance is not given.

So, the prior level of significance $\alpha =0.05$ can be used.

P-value:

Software procedure:

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

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution and enter 17 as degrees of freedom.
• Click the Shaded Area tab.
• Choose X-Value and Right Tail for the region of the curve to shade.
• Enter the X-value as 0.9642.
• Click OK.

Output obtained from MINITAB is given below:

From the output, the P- value is 0.1742.

Thus, the P- value is 0.1742.

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 0.1742 and $\alpha$ value is 0.05.

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

That is $0.1742\left(=P\right)>0.05\left(=\alpha \right)$.

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

Hence, 1 ppt increase in manganese tends to at most $3\frac{\text{kg}}{{\text{mm}}^{\text{2}}}$ increase in the true average of tensile strength, provided the effect of Thickness is accounted for.

Therefore, there is no sufficient evidence to conclude that ${\beta }_1>3$.

d.

To determine

Test whether there is enough evidence to conclude that ${\beta }_2<-0.1$.

Answer to Problem 2E

There is sufficient evidence to conclude that ${\beta }_2<-0.1$.

Explanation of Solution

Calculation:

From the accompanying MINITAB output, the slope coefficient of Thickness is $b_2\left(\text{Thickness}\right)=-0.4249$.

Here, the claim is that ${\beta }_2<-0.1$.

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_2\ge -0.1$

That is, 1 mm increase in thickness tends to at least $0.1\frac{\text{kg}}{{\text{mm}}^{\text{2}}}$ decrease in the true average of tensile strength, provided the effect of manganese is accounted for.

Alternative hypothesis:

$H_1:{\beta }_2<-0.1$

That is, decrease in the true average of tensile strength due to 1 mm increase in thickness will be less than $0.1\frac{\text{kg}}{{\text{mm}}^{\text{2}}}$, provided the effect of manganese is accounted for.

Test statistic:

The test statistic is,

$t=\frac{{\widehat{\beta }}_i-{\beta }_i}{s_{{\widehat{\beta }}_i}}\sim t_{\left(n-\left(k+1\right)\right)}$

Where, ${\widehat{\beta }}_i$ be the slope coefficient of the sample regression line, ${\beta }_i$ be the slope coefficient of the population regression line and $s_{{\widehat{\beta }}_i}$ be the estimate of error standard deviation of the estimated slope coefficient.

From the accompanying MINITAB output, the estimate of error standard deviation of slope coefficient of Thickness is $s_{{\widehat{\beta }}_2}=0.12606$.

Test statistic under null hypothesis:

Under the null hypothesis, the test statistic is obtained as follows:

$\begin{array}{c}t=\frac{{\widehat{\beta }}_i-{\beta }_i}{s_{{\widehat{\beta }}_i}}\sim t_{\left(n-\left(k+1\right)\right)}\\ =\frac{{\widehat{\beta }}_2-{\beta }_2}{s_{{\widehat{\beta }}_2}}\\ =\frac{-0.4249-\left(-0.1\right)}{0.12606}\\ =-2.577\end{array}$

Thus, the test statistic is -2.577.

Degrees of freedom:

The number of plates that are sampled is $n=20$ and the number of predictor variables is $k=3$.

The degrees of freedom is,

$\begin{array}{c}d.f=n-\left(k+1\right)\\ =20-3\\ =17\end{array}$

Thus, the degree of freedom is 17.

Here, level of significance is not given.

So, the prior level of significance $\alpha =0.05$ can be used.

P-value:

Software procedure:

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

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution and enter 17 as degrees of freedom.
• Click the Shaded Area tab.
• Choose X-Value and Left Tail for the region of the curve to shade.
• Enter the X-value as -2.577.
• Click OK.

Output obtained from MINITAB is given below:

From the output, the P- value is 0.009795.

Thus, the P- value is 0.009795.

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 0.009795 and $\alpha$ value is 0.05.

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

That is $0.009795\left(=P\right)<0.05\left(=\alpha \right)$.

By the rejection rule, reject the null hypothesis.

Hence, 1 mm increase in thickness tends to at least $0.1\frac{\text{kg}}{{\text{mm}}^{\text{2}}}$ decrease in the true average of tensile strength, provided the effect of manganese is accounted for.

Therefore, there is sufficient evidence to conclude that ${\beta }_2<-0.1$.