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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
Mean life is 143.5; change is -0.175 for viscosity 30, -0.265 for viscosity 40.

Step by step solution

01

Identify the Regression Equation

The given regression model is: \[ Y = 125.0 + 7.75x_1 + 0.0950x_2 - 0.0090x_1x_2 + \epsilon \] where \( Y \) is the predicted wear life, \( x_1 \) is the oil viscosity, \( x_2 \) is the load, and \( \epsilon \) is the error term.
02

Calculate Mean Life for Given Values

To find the mean life when viscosity is 40 and load is 1100, substitute \( x_1 = 40 \) and \( x_2 = 1100 \) into the regression equation: \[Y = 125.0 + 7.75(40) + 0.0950(1100) - 0.0090(40)(1100)\]. Simplifying, \[Y = 125.0 + 310 + 104.5 - 396 = 143.5.\] Therefore, the mean life is 143.5.
03

Derive Change in Life due to Load Increase

The change in mean life when load \( x_2 \) increases by 1 at a fixed viscosity \( x_1 \) can be obtained by taking the partial derivative of \( Y \) with respect to \( x_2 \): \[ \frac{\partial Y}{\partial x_2} = 0.0950 - 0.0090x_1. \]
04

Calculate Change in Life for Viscosity 30

Substitute \( x_1 = 30 \) into the derivative: \[ \frac{\partial Y}{\partial x_2} = 0.0950 - 0.0090(30) = 0.0950 - 0.27 = -0.175. \] Therefore, an increase of 1 in load decreases the mean life by 0.175 when viscosity is 30.
05

Calculate Change in Life for Viscosity 40

Substitute \( x_1 = 40 \) into the derivative: \[ \frac{\partial Y}{\partial x_2} = 0.0950 - 0.0090(40) = 0.0950 - 0.36 = -0.265. \] Therefore, an increase of 1 in load decreases the mean life by 0.265 when viscosity is 40.

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Key Concepts

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

Understanding the Regression Model
A multiple regression model is a statistical tool used to understand the relationship between one dependent variable and two or more independent variables. In the context of our exercise, we have the regression model: \[ Y = 125.0 + 7.75x_1 + 0.0950x_2 - 0.0090x_1x_2 + \epsilon \]. Here, \( Y \) represents the predicted life wear of a bearing, while \( x_1 \) and \( x_2 \) are the oil viscosity and load, respectively. This equation suggests that the wear life is influenced by oil viscosity, load, and an interaction term between viscosity and load. The interaction term \(-0.0090x_1x_2\) implies that the relationship between oil viscosity and life changes in response to different load levels. Each coefficient shows how much we expect \( Y \) to change for a one-unit change in one of the independent variables, holding the others constant.
  • Constant term (125.0): This is the expected mean wear life when both viscosity and load are zero.
  • Viscosity coefficient (7.75): Indicates how much the wear life changes as viscosity increases by one unit, assuming load remains constant.
  • Load coefficient (0.0950): Demonstrates the expected change in wear life with a one-unit increase in load, assuming viscosity remains constant.
Mean Value Calculation
Calculating the mean value of life involves substituting the given values of the independent variables into the regression model. For example, to compute the mean wear life when viscosity is 40 and load is 1100, substitute \( x_1 = 40 \) and \( x_2 = 1100 \) into the regression equation. Let's do that: \[Y = 125.0 + 7.75(40) + 0.0950(1100) - 0.0090(40)(1100)\]Simplifying this equation gives:
  • \( 125.0 + 310 \) for the constant term and viscosity contribution
  • \( + 104.5 \) for the load contribution
  • \( - 396 \) for the interaction adjustment
Therefore, the mean wear life is 143.5 units. This calculation helps in predicting the expected behavior of the dependent variable given specific values of the independent variables.
Introduction to Partial Derivatives
Partial derivatives are a powerful tool in calculus used to analyze how functions change with respect to one variable while holding others constant. In our regression model, we use the partial derivative to find how changes in load (\( x_2 \)) affect the predicted wear life (\( Y \)) for a fixed viscosity (\( x_1 \)). To find this, we take the derivative of \( Y \) with respect to \( x_2 \): \[\frac{\partial Y}{\partial x_2} = 0.0950 - 0.0090x_1\]This tells us how much \( Y \) is expected to change per one-unit increase in \( x_2 \), depending on \( x_1 \). The expression includes a term \(- 0.0090x_1\), which shows that the effect of load on wear life depends on the viscosity level. By substituting different values of \( x_1 \) (e.g., viscosity of 30 or 40), we can see varying impacts on life. For instance:
  • When viscosity is 30, an increase of 1 in load decreases the mean life by 0.175.
  • When viscosity is 40, the decrease in mean life is 0.265 for the same load increase.
This highlights the sensitivity of wear life to load changes at different viscosity levels, providing deeper insights into how the dependent and independent variables interact.

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Most popular questions from this chapter

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