/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Let \(y=\) wear life of a bearin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 38

Let \(y=\) wear life of a bearing, \(x_{1}=\) oil viscosity, and \(x_{2}=\) load. Suppose that the multiple regression model relating life to viscosity and load is $$ Y=125.0+7.75 x_{1}+.0950 x_{2}-.0090 x_{1} x_{2}+\epsilon $$ a. What is the mean value of life when viscosity is 40 and load is 1100 ? b. When viscosity is 30 , what is the change in mean life associated with an increase of 1 in load? When viscosity is 40 , what is the change in mean life associated with an increase of 1 in load?

Short Answer

Expert verified
a. The mean life is 143.5. b. Change in mean life is -0.175 for viscosity 30, and -0.265 for viscosity 40.

Step by step solution

01

Substitute Values for Part (a)

For part (a), substitute \(x_1 = 40\) and \(x_2 = 1100\) into the regression model:\[ Y = 125.0 + 7.75(40) + 0.0950(1100) - 0.0090(40)(1100) \]
02

Calculate Each Term for Part (a)

Compute each term in the equation:- First term: \(125.0\)- Second term: \(7.75 \times 40 = 310.0\)- Third term: \(0.0950 \times 1100 = 104.5\)- Fourth term: \( -0.0090 \times 40 \times 1100 = -396.0\)
03

Solve for Mean Life in Part (a)

Add the computed values together to find the mean life:\[ Y = 125.0 + 310.0 + 104.5 - 396.0 \]\[ Y = 143.5 \]
04

Use Partial Derivative for Part (b)

To find the change in mean life with an increase of 1 in load, calculate:\[ \frac{\partial Y}{\partial x_2} = 0.0950 - 0.0090x_1 \]
05

Substitute \(x_1 = 30\) into Partial Derivative

For \(x_1 = 30\), substitute into the partial derivative:\[ \frac{\partial Y}{\partial x_2} = 0.0950 - 0.0090 \times 30 \]\[ = 0.0950 - 0.270 = -0.175 \]
06

Substitute \(x_1 = 40\) into Partial Derivative

For \(x_1 = 40\), substitute into the partial derivative:\[ \frac{\partial Y}{\partial x_2} = 0.0950 - 0.0090 \times 40 \]\[ = 0.0950 - 0.360 = -0.265 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Mean Value
The concept of mean value is crucial in multiple regression analysis. It helps us predict the average outcome of our dependent variable based on given independent variables. In this exercise, the mean value refers to the expected life (wear life) of a bearing. We use the regression equation to calculate it.

The regression model includes terms for oil viscosity \(x_1\) and load \(x_2\), which influence the life of the bearing. When they are plugged into the equation, each variable's contribution is accordingly adjusted. Hence, for part (a), with an oil viscosity of 40 and a load of 1100, the calculated mean life is 143.5 years.

Calculating the mean value helps us determine the average outcome, taking into account all the influencing factors. This step enables accurate predictions and decision-making in contexts where multiple variables affect the result.
Role of Partial Derivatives
Partial derivatives play a pivotal role in understanding how each independent variable impacts the dependent variable in a multiple regression model. They help us to calculate the rate of change of the mean life of a bearing with respect to the load or oil viscosity.

In this exercise, we specifically focus on the partial derivative with respect to the load \(x_2\). By taking the derivative \(\frac{\partial Y}{\partial x_2}\), we can isolate how changes in load affect the mean life, while keeping viscosity constant. This is especially useful when we want to assess only one variable's impact.

Here, when the partial derivative is computed, it reveals that the change in mean life of the bearing is slightly negative, indicating that an increase in load, holding viscosity constant, might decrease the life expectancy slightly.
Analyzing Change in Mean Life
Understanding the change in mean life involves analyzing how specific changes in the independent variables affect the dependent variable's expectation. Here, the load changes incrementally by 1, and its influence is observed with fixed viscosities of 30 and 40.

For a viscosity of 30, the partial derivative calculation shows \( -0.175\), suggesting a slight decrease in life expectancy per unit increase in load. When viscosity is increased to 40, the change becomes \(-0.265\), indicating a more pronounced decrease in mean life per additional unit of load.

These results tell us that as the viscosity increases, the load's impact on reducing wear life becomes more significant. This kind of analysis allows engineers and scientists to optimize factors for best performance and durability of components, like bearings, under specified conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The following data resulted from an experiment to assess the potential of unburnt colliery spoil as a medium for plant growth. The variables are \(x=\) acid extractable cations and \(y=\) exchangeable acidity/total cation exchange capacity ("Exchangeable Acidity in Unburnt Colliery Spoil," Nature, 1969: 161): \begin{tabular}{r|rrrrrrr} \(x\) & \(-23\) & \(-5\) & 16 & 26 & 30 & 38 & 52 \\ \hline\(y\) & \(1.50\) & \(1.46\) & \(1.32\) & \(1.17\) & \(.96\) & \(.78\) & \(.77\) \\ \(x\) & 58 & 67 & 81 & 96 & 100 & 113 & \\ \hline\(y\) & \(.91\) & \(.78\) & \(.69\) & \(.52\) & \(.48\) & \(.55\) & \end{tabular} Standardizing the independent variable \(x\) to obtain \(x^{\prime}=\) \((x-\bar{x}) / s_{x}\) and fitting the regression function \(y=\beta_{0}^{*}+\) \(\beta_{1}^{+} x^{\prime}+\beta_{2}^{*}\left(x^{\prime}\right)^{2}\) yielded the accompanying computer output. \begin{tabular}{ccc} Parameter & Estimate & Estimated SD \\ \hline\(\beta_{0}^{*}\) & \(.8733\) & \(.0421\) \\ \(\beta_{1}^{*}\) & \(-.3255\) & \(.0316\) \\ \(\beta_{2}^{*}\) & \(.0448\) & \(.0319\) \end{tabular}

Cardiorespiratory fitness is widely recognized as a major component of overall physical well-being. Direct measurement of maximal oxygen uptake \(\left(\mathrm{VO}_{2} \max \right)\) is the single best measure of such fitness, but direct measurement is timeconsuming and expensive. It is therefore desirable to have a prediction equation for \(\mathrm{VO}_{2} \max\) in terms of easily obtained quantities. Consider the variables $$ \begin{aligned} y &=\mathrm{VO}_{2} \max (\mathrm{L} / \mathrm{min}) \quad x_{1}=\text { weight }(\mathrm{kg}) \\ x_{2} &=\text { age }(\mathrm{yr}) \\ x_{3} &=\text { time necessary to walk } 1 \text { mile }(\mathrm{min}) \\ x_{4} &=\text { heart rate at the end of the walk (beats/min) } \end{aligned} $$ Here is one possible model, for male students, consistent with the information given in the article "Validation of the Rockport Fitness Walking Test in College Males and Females" (Research Quarterly for Exercise and Sport, 1994: 152-158): $$ \begin{aligned} &Y=5.0+.01 x_{1}-.05 x_{2}-.13 x_{3}-.01 x_{4}+\epsilon \\ &\sigma=.4 \end{aligned} $$ a. Interpret \(\beta_{1}\) and \(\beta_{3}\). b. What is the expected value of \(\mathrm{VO}_{2} \max\) when weight is \(76 \mathrm{~kg}\), age is \(20 \mathrm{yr}\), walk time is \(12 \mathrm{~min}\), and heart rate is \(140 \mathrm{~b} / \mathrm{m}\) ? c. What is the probability that \(\mathrm{VO}_{2} \max\) will be between \(1.00\) and \(2.60\) for a single observation made when the values of the predictors are as stated in part (b)?

An investigation of a die casting process resulted in the accompanying data on \(x_{1}=\) furnace temperature, \(x_{2}=\) die close time, and \(y=\) temperature difference on the die surface ("A Multiple-Objective Decision-Making Approach for Assessing Simultaneous Improvement in Die Life and Casting Quality in a Die Casting Process," Quality Engineering, 1994: 371-383). a. Carry out the model utility test. b. Calculate and interpret a \(95 \%\) confidence interval for \(\beta_{2}\), the population regression coefficient of \(x_{2}\). c. When \(x_{1}=1300\) and \(x_{2}=7\), the estimated standard deviation of \(\hat{y}\) is \(s_{\bar{Y}}=.353\). Calculate a \(95 \%\) confidence interval for true average temperature difference when furnace temperature is 1300 and die close time is \(7 .\) d. Calculate a \(95 \%\) prediction interval for the temperature difference resulting from a single experimental run with a furnace temperature of 1300 and a die close time of 7 .

a. Could a linear regression result in residuals \(23,-27,5\), \(17,-8,9\), and 15 ? Why or why not? b. Could a linear regression result in residuals \(23,-27,5\), \(17,-8,-12\), and 2 corresponding to \(x\) values \(3,-4,8\), \(12,-14,-20\), and 25 ? Why or why not?

Let \(y=\) sales at a fast-food outlet (1000s of \(\$$ ), \)x_{1}=\( number of competing outlets within a 1 -mile radius, \)x_{2}=\( population within a l-mile radius (1000s of people), and \)x_{3}\( be an indicator variable that equals 1 if the outlet has a drive-up window and 0 otherwise. Suppose that the true regression model is $$ Y=10.00-1.2 x_{1}+6.8 x_{2}+15.3 x_{3}+\epsilon $$ a. What is the mean value of sales when the number of competing outlets is 2 , there are 8000 people within a 1 -mile radius, and the outlet has a drive-up window? b. What is the mean value of sales for an outlet without a drive-up window that has three competing outlets and 5000 people within a I-mile radius? c. Interpret \)\beta_{3}$.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.