Chapter 13.4, Problem 40E

a.

To determine

Check whether there is a positive linear relationship between the minimum and maximum width of an object.

Answer to Problem 40E

There is convincing evidence that there is a positive linear relationship between the minimum and maximum width of an object.

Explanation of Solution

Calculation:

The given data provide the dimensions of 27 representative food products.

Here, $\beta$ indicates the slope of the population regression line relating the maximum width to the minimum width.

Null hypothesis:

$H_0:\beta =0$.

That is, there is no linear relationship between the minimum and maximum width of an object.

Alternative hypothesis:

$H_a:\beta >0$.

That is, there is a positive linear relationship between the minimum and maximum width of an object.

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

Test Statistic:

The formula for test statistic is as follows:

$t=\frac{b-\left(\text{hypothesized value}\right)}{s_b}$

In the formula, b denotes the estimated slope, $s_b$ denotes the standard deviation of b.

A standardized residual plot is shown below:

Standardized residual values and standardized residual plot:

Software procedure:

Step-by-step procedure to compute standardized residuals and its plot using MINITAB software:

• Select Stat > Regression > Regression > Fit Regression Model.
• In Response, enter the column of Maximum width.
• In Continuous Predictors, enter the columns of Minimum width.
• In Graphs, select Standardized under Residuals for Plots.
• In Results, select for all observations under Fits and diagnostics.
• In Residuals versus the variables, select Minimum width.
• Click OK.

Output obtained MINTAB software is given below:

From the standardized residual plot, it is observed that one point lies outside the horizontal band of 3 units from the central line of 0. The standardized residual for this outlier is 3.72, that is, for product 25.

Assumption:

Here, the assumption made is that, the simple linear regression model is appropriate for the data, even though there is one extreme standardized residual.

Test Statistic:

In the MINITAB output, the test statistic value is displayed in the column “T-value” corresponding to “Minimum width”, in the section “Coefficients”. The value is 13.53.

P-value:

From the above output, the corresponding P-value is 0.

Rejection rule:

If $P\text{-value}\le \alpha$ then reject the null hypothesis. Otherwise, do not reject the null hypothesis.

Conclusion:

The P-value is 0 and the level of significance is 0.05.

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

That is, $P\text{-value}\left(=0\right)<\alpha \left(=0.05\right)$.

Therefore, reject the null hypothesis.

Thus, there is convincing evidence that there is a positive linear relationship between the minimum and maximum width of an object.

b.

To determine

Compute and interpret $s_e$ .

Answer to Problem 40E

$s_e=0.67246$.

Explanation of Solution

Calculation:

From the MINITAB output in Part (a), it is clear that $s_e=0.67246$, as provided in the section “Model Summary”, under the column of “S”.

On an average, there is  67.246% deviation of the maximum width in the sample from the value predicted by least-squares regression.

c.

To determine

Find the 95% confidence interval for the mean maximum width of products for the minimum width of 6 cm.

Answer to Problem 40E

The 95% confidence interval for the mean maximum width of products for the minimum width of 6 cm is (5.708, 6.647).

Explanation of Solution

Calculation:

The confidence interval for $\alpha +\beta x^{*}$ is $(a+b{x}^{*})\pm (t\text{ critical value)}\cdot s_{a+b{x}^{*}}$, where $(a+b{x}^{*})$ is the point estimate value, the t-critical value is based on df = n – 2, and $s_{a+b{x}^{*}}$ is the estimated standard deviation.

From the MINITAB output in Part (a), the estimated linear regression line is $\widehat{y}=0.939+0.8731x$ and $s_e=0.672456$.

Point estimate:

The point estimate is calculated as follows:

$\begin{array}{c}\widehat{y}=0.939+0.8731x\\ =0.939+0.8731\left(6\right)\\ =0.939+5.2386\\ =6.1776\\ \approx 6.178\end{array}$

Estimated standard deviation:

For the given x values, the summation values are given in the following table:

Minimum width (X)	$X^2$
1.8	3.24
2.7	7.29
2	4
2.6	6.76
3.15	9.9225
1.8	3.24
1.5	2.25
3.8	14.44
5	25
4.75	22.5625
2.8	7.84
2.1	4.41
2.2	4.84
2.6	6.76
2.6	6.76
2.9	8.41
5.1	26.01
10.2	104.04
3.5	12.25
1.2	1.44
1.7	2.89
1.75	3.0625
1.7	2.89
1.2	1.44
1.2	1.44
7.5	56.25
4.25	18.0625
${\displaystyle \sum x}=83.6$	${{\displaystyle \sum x}}^2=367.5$

The value of $s_{xx}$ is calculated as follows:

$\begin{array}{c}{s}_{xx}={\displaystyle \sum x^2-\frac{{\left({\displaystyle \sum x}\right)}^2}{n}}\\ =367.5-\frac{{\left(83.6\right)}^2}{27}\\ =367.5-258.8504\\ =108.6496\\ \approx 108.65\end{array}$

Substitute $s_{xx}=108.65\text{ and }\overline{x}=3.096$. The estimated standard deviation of $a+b(30)$ is as follows:

$\begin{array}{c}{s}_{a+b{x}^{*}}=s_e\sqrt{\frac{1}{n}+\frac{{\left(x^{*}-\overline{x}\right)}^2}{s_{xx}}}\\ s_{a+b(6)}=0.67246\sqrt{\frac{1}{27}+\frac{{\left(6-3.096\right)}^2}{108.650}}\\ =0.67246\sqrt{0.0370+\frac{8.433216}{108.650}}\\ =0.67246\sqrt{0.0370+0.07762}\\ =0.67246\times 0.33855\\ =0.22766\\ \approx 0.228\end{array}$

Formula for degrees of freedom:

The formula for degrees of freedom is as follows:

$df=n-2$

The number of data value given is 27, that is $n=27$. Substitute $n=27$ in the degrees of freedom formula as follows:

$\begin{array}{c}df=27-2\\ =25\end{array}$

Critical value:

From the Appendix: Table of the t-critical values:

• Locate the value 25 in the degrees of freedom (df) column.
• Locate the 0.95 in the row of central area captured.
• The intersecting value that corresponds to df 25 with the confidence level 0.95 is 2.060.

Thus, the critical value for $df=25$ with two-tailed test is 2.060.

Substitute $a+b{x}^{*}=6.178$, $t-\text{critical value}=2.060$, and $s_{a+b{x}^{*}}=0.228$ in confidence interval as shown below:

$\begin{array}{c}6.178\pm (2.060\times 0.228)=\left(6.178-0.46968,6.178+0.46968\right)\\ =\left(5.708,6.647\right)\end{array}$

Therefore, one can be 95% confident that the mean maximum width of products with the minimum width of 6 cm will be between 5.708 cm and 6.647 cm.

d.

To determine

Find the 95% prediction interval for the mean maximum width of products with the minimum width of 6 cm.

Answer to Problem 40E

The 95% prediction interval for the mean maximum width of products with the minimum width of 6 cm is (4.716, 7.640).

Explanation of Solution

Calculation:

The confidence interval for $\alpha +\beta x^{*}$ is $(a+b{x}^{*})\pm (t\text{ critical value)}\cdot \sqrt{s_e^2+s_{a+b{x}^{*}}^2}$, where $(a+b{x}^{*})$ is the point estimate value, the t-critical value is based on df = n – 2, and $s_{a+b{x}^{*}}$ is the estimated standard deviation.

The estimated standard deviation of the amount by which a single y observation deviates from the value predicted by an estimated regression line is $\sqrt{s_e^2+s_{a+b{x}^{*}}^2}$.

Substitute $s_e^2={\left(0.67246\right)}^2,s_{a+b{x}^{*}}^2={\left(0.228\right)}^2$ in the estimated standard deviation formula.

$\begin{array}{c}\sqrt{s_e^2+s_{a+b{x}^{*}}^2}=\sqrt{{\left(0.67246\right)}^2+{\left(0.228\right)}^2}\\ =\sqrt{0.4522+0.051984}\\ =\sqrt{0.504184}\\ =0.710\end{array}$

From Part (c), the critical value for $df=25$ with two-tailed test is 2.060.

Substitute $a+b{x}^{*}=6.178$, $t-\text{critical value}=2.060$, and $\sqrt{s_e^2+s_{a+b{x}^{*}}^2}=0.710$ in confidence interval as shown below:

$\begin{array}{c}6.178\pm (2.060\times 0.710)=\left(6.178-1.4626,6.178+1.4626\right)\\ =\left(4.7154,7.640\right)\end{array}$

Therefore, the 95% prediction interval for the mean maximum width of products with the minimum width of 6 cm is (4.716, 7.640).