Chapter 14, Problem 58SE

Machine Maintenance. Jensen Tire & Auto is in the process of deciding whether to purchase a maintenance contract for its new computer wheel alignment and balancing machine. Managers feel that maintenance expense should be related to usage, and they collected the following information on weekly usage (hours) and annual maintenance expense (in hundreds of dollars).

1. a. Develop the estimated regression equation that relates annual maintenance expense to weekly usage.
2. b. Test the significance of the relationship in part (a) at a .05 level of significance.
3. c. Jensen expects to use the new machine 30 hours per week. Develop a 95% prediction interval for the company’s annual maintenance expense.
4. d. If the maintenance contract costs $3000 per year, would you recommend purchasing it? Why or why not?

To determine

Find the estimated regression equation that relates annual maintenance expense to weekly usage.

Answer to Problem 58SE

The estimated regression equation is $\widehat{\text{Expense}}=10.53+0.953\text{Usage}$.

Explanation of Solution

Calculation:

The data related to the weekly usage (hours) and Annual Maintenance Expense (in hundreds of dollars) to purchase a maintenance contract for new computer wheel alignment and balancing machine.

Regression:

Software procedure:

Step-by-step procedure to obtain the estimated regression equation using Excel:

• In Excel sheet, enter Size and Selling Price in different columns.
• In Data, select Data Analysis and choose Regression.
• In Input Y Range, select Expense.
• In Input X Range, select Usage.
• Select Labels.
• Click OK.

Output obtained using Excel is given below:

Thus, the estimated regression equation is $\widehat{\text{Expense}}=10.53+0.953\text{Usage}$.

To determine

Test for a significant relationship at $\alpha =0.05$.

Answer to Problem 58SE

There is a significant relationship between annual maintenance expense and weekly usage.

Explanation of Solution

Calculation:

State the test hypotheses.

Null hypothesis:

$H_0:{\beta }_1=0$

That is, there is no significant relationship between expense and usage.

Alternative hypothesis:

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

That is, there is a significant relationship between expense and usage.

From the output in Part (a) it is found that the F-test statistic is 47.62.

Level of significance:

The given level of significance is $\alpha =0.05$.

p-value:

From the output in pat (a) it is found that the p-value is 0.00.

Rejection rule:

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

Conclusion:

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.00\right)<\alpha \left(=0.05\right)$

Thus, the decision is “reject the null hypothesis”.

Therefore, the data provide sufficient evidence to conclude that there is a significant relationship between expense and usage.

Thus, annual maintenance expense and weekly usage are related.

To determine

Find a 95% prediction interval for the company’s annual maintenance expense.

Answer to Problem 58SE

The 95% prediction interval for the company’s annual maintenance expense is $\left(\$2,831,\text{\$4,993}\right)$.

Explanation of Solution

Calculation:

For a sample of size n, the degrees of freedom is given as $n-2$.

In this given problem, for sample of size 10, the degrees of freedom is as follows:

$\begin{array}{c}df=n-2\\ =10-2\\ =8\end{array}$

Thus, the degrees of freedom is 8.

Level of significance:

The given level of significance is $\alpha =0.05$.

For two-tails distribution, the value is as follows:

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

Form the table 2 of “t Distribution” in Appendix B, it is found that the value of t test statistic with level of significance 0.025 and degrees of freedom 8 is $t_{0.025,8}=2.306$

According to the regression equation $\widehat{\text{Expense}}=10.53+0.953\text{Usage}$, the value of $\widehat{y}*$ for $\text{Usage}=30$ is as follows:

$\begin{array}{c}\widehat{y}=10.53+0.953\left(30\right)\\ =10.53+28.59\\ =39.12\end{array}$

Thus, the possible value of dependent variable y when $x=30$, is 39.12.

The estimate of standard deviation corresponding to the prediction of the value of $\widehat{y}*$ is given as follows:

$s_{\text{Pred}}=s\sqrt{1+\frac{1}{n}+\frac{{\left(x*-\overline{x}\right)}^2}{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}}$

It is known for a sample size n that mean of a random variable x can be obtained as follows:

$\overline{x}=\frac{{\displaystyle \sum _1^n{x}_i}}{n}$

Thus, mean of the random variable x is given below:

$\begin{array}{c}\overline{x}=\frac{\left(13+10+20+28+32+17+24+31+40+38\right)}{10}\\ =\frac{253}{10}\\ =25.3\end{array}$

The value of ${\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}$ is calculated as follows:

$x_i$	${\left(x_i-\overline{x}\right)}^2$
13	151.29
10	234.09
20	28.09
28	7.29
32	44.89
17	68.89
24	1.69
31	32.49
40	216.09
38	161.29
Total	946.1

Here, it is found that, $\overline{x}=25.3$, ${\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}=946.1$.

From the output, the value of given MSE for sample of size 10 is $\text{MSE}=18.0593$.

The standard error of the estimate is obtained as follows:

$\begin{array}{c}s=\sqrt{\text{MSE}}\\ =\sqrt{18.0593}\\ =4.2496\end{array}$

Thus, the standard error of the estimate is 4.2496.

It is also found that for $x=30$,

$\begin{array}{c}{\left(x*-\overline{x}\right)}^2={\left(30-25.3\right)}^2\\ =22.09\end{array}$

For $x=30$, the standard deviation of an individual value of y is as follows:

$\begin{array}{c}{s}_{\text{Pred}}=s\sqrt{1+\frac{1}{n}+\frac{{\left(x*-\overline{x}\right)}^2}{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}}\\ =\left(4.2496\right)\sqrt{1+\frac{1}{10}+\frac{22.09}{946.1}}\\ =\left(4.2496\right)\sqrt{1+\frac{1}{10}+\frac{22.09}{190}}\\ =\left(4.2496\right)\sqrt{1.2163}\end{array}$

       $=4.6867$

Thus, the standard deviation of an individual value of y for $x=30$ is 4.6867.

The prediction interval for expected value of $y*$ is $\widehat{y}*\pm t_{\frac{\alpha }{2}}{s}_{\text{Pred}}$, where the confidence coefficient is $1-\alpha$ and $t_{\frac{\alpha }{2}}$ is the value of t distribution with degrees of freedom of $n-2$ for a sample of size n.

Therefore, the required prediction interval is given below:

$\begin{array}{c}\widehat{y}*\pm t_{\frac{\alpha }{2}}{s}_{\text{Pred}}=39.12\pm \left(2.306\right)\left(4.6867\right)\\ =39.12\pm 10.8151\\ \approx \left(28.31,49.93\right)\end{array}$

Thus, the 95% prediction interval for the company’s annual maintenance expense is $\left(\$2,831,\text{\$4,993}\right)$.

To determine

Explain whether it is recommended purchase a machine of the maintenance costs of $3,000 per year.

Explanation of Solution

Calculation:

The estimated regression equation is $\widehat{\text{Expense}}=10.53+0.953\text{Usage}$.

Substitute 30 for usage in estimated regression equation.

$\begin{array}{c}\widehat{\text{Expense}}=10.53+0.953\left(30\right)\\ =10.53+28.59\\ =39.12\end{array}$

In this case, the expected expense is 39.12 or $3,912. Therefore, it is recommended purchase a machine of the maintenance costs of $3,000 per year.