Chapter 13.4, Problem 44E

The article first introduced in Exercise 13.34 of Section 13.3 gave data on the dimensions of 27 representative food products.

1. a. Use the data set given there to test the hypothesis that there is a positive linear relationship between x = minimum width and y = maximum width of an object.
2. b. Calculate and interpret s_e.
3. c. Calculate a 95% confidence interval for the mean maximum width of products with a minimum width of 6 cm.
4. d. Calculate a 95% prediction interval for the maximum width of a food package with a minimum width of 6 cm.

13.34 The article “Vital Dimensions in Volume Perception: Can the Eye Fool the Stomach?” (Journal of Marketing Research [1999]: 313–326) gave the accompanying data on the dimensions of 27 representative food products (Gerber baby food, Cheez Whiz, Skippy Peanut Butter, and Ahmed’s tandoori paste, to name a few).

1. a. Fit the simple linear regression model that would allow prediction of the maximum width of a food container based on its minimum width.
2. b. Calculate the standardized residuals (or just the residuals if a computer program that doesn’t give standardized residuals is used) and make a residual plot to determine whether there are any outliers.
3. c. The data point with the largest residual is for a 1-liter Coke bottle. Delete this data point and determine the equation of the regression line. Did deletion of this point result in a large change in the equation of the estimated regression line?
4. d. For the regression line of Part (c), interpret the estimated slope and, if appropriate, the intercept.
5. e. For the data set with the Coke bottle deleted, are the assumptions of the simple linear regression model reasonable? Give statistical evidence.

To determine

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

Answer to Problem 44E

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

Explanation of Solution

Calculation:

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

1.

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

2.

Null hypothesis:

$H_0:\beta =0$.

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

3.

Alternative hypothesis:

$H_a:\beta >0$.

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

4.

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

5.

Test Statistic:

The formula for test statistic is,

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

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

6.

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 the 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, which is for the 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.

7.

Calculation:

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.

8.

P-value:

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

9.

Rejection rule:

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

Conclusion:

The P-value is 0.

The level of significance is 0.05.

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

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

Based on the rejection rule, reject the null hypothesis.

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

To determine

Compute and interpret $s_e$ .

Answer to Problem 44E

$s_e=0.67246$.

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 a 67.246% deviation of the maximum width in the sample from the value predicted by least-squares regression.

To determine

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

Answer to Problem 44E

The 95% interval for the mean maximum width of products, for a 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, 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,

$df=n-2$

The number of data values 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 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 the df 25 with 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 a minimum width of 6 cm will be between 5.708 cm and 6.647 cm.

To determine

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

Answer to Problem 44E

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