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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. 147.5; b. Changes are -0.175 at viscosity 30 and -0.265 at viscosity 40.

Step by step solution

01

Substitute for Part (a)

We need to find the mean value of life when the oil viscosity is 40 and the load is 1100. Substitute these values into the regression equation. \[Y = 125.0 + 7.75 \times 40 + 0.0950 \times 1100 - 0.0090 \times 40 \times 1100\]
02

Compute Each Term for Part (a)

Calculate each term in the substituted regression equation:1. Compute \(7.75 \times 40 = 310\).2. Compute \(0.0950 \times 1100 = 104.5\).3. Compute \(0.0090 \times 40 \times 1100 = 396\).
03

Calculate Mean for Part (a)

Substitute the computed values into the equation and solve:\[Y = 125.0 + 310 + 104.5 - 396\]\[Y = 543.5 - 396\]\[Y = 147.5\]Thus, the mean value of life is 147.5 when viscosity is 40 and load is 1100.
04

Understand Part (b)

We need to understand how the mean life changes with respect to a unit increase in load, with a focus on the term involving \(x_2\), which is the load. Here, the direct effect on load is given by the coefficient of \(x_2\), which is 0.095. However, we must also consider the interaction term \(-0.0090 x_{1}\) when viscosity \(x_1\) varies.
05

Calculate Change in Mean Life for Part (b) with Viscosity 30

Substitute \(x_1 = 30\) in the interaction term and calculate the change in mean life:- Calculate the change from \(0.095 x_2 - 0.0090 \times 30 = 0.095 - 0.27 = -0.175\).- Therefore, the change in mean life for a unit increase in load \(x_2\) when viscosity \(x_1\) is 30 is \(-0.175\).
06

Calculate Change in Mean Life for Part (b) with Viscosity 40

Substitute \(x_1 = 40\) in the interaction term and calculate the change in mean life:- Calculate the change from \(0.095 x_2 - 0.0090 \times 40 = 0.095 - 0.36 = -0.265\).- Therefore, the change in mean life for a unit increase in load \(x_2\) when viscosity \(x_1\) is 40 is \(-0.265\).

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Key Concepts

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

Regression Coefficients
In multiple regression analysis, understanding regression coefficients is key to interpreting how each predictor variable affects the dependent variable. - **Intercept**: The regression equation starts with an intercept, which in this case is 125.0. This represents the mean value of the dependent variable, wear life, when all predictor variables (oil viscosity and load) are equal to zero.- **Coefficient for Oil Viscosity \, \(x_1\)**: The coefficient here is 7.75. This tells us that, holding the load constant, for each unit increase in viscosity, the mean wear life increases by 7.75 units. - **Coefficient for Load \, \(x_2\)**: The coefficient 0.095 indicates that, with the viscosity held constant, each additional unit of load increases the mean wear life by 0.095 units.- **Interaction Term \, \(-0.0090\)**: The presence of the interaction term \(-0.0090 \, x_1 \, x_2\) means that the effect of the load on wear life varies depending on the level of oil viscosity. This term subtracts from the mean life, meaning that for a certain viscosity, the effect of additional load on life decreases as viscosity increases.
Understanding these coefficients helps predict the wear life accurately given any oil viscosity and load combination.
Interaction Terms
Interaction terms are used to explore whether the effect of one explanatory variable depends on the level of another explanatory variable. In our multiple regression model:- The interaction term \(-0.0090 \, x_1 \, x_2\) is crucial in understanding how oil viscosity and load jointly affect the wear life of a bearing.- This term implies that the impact of load on the wear life isn't just additive but also depends on the value of oil viscosity. - For example, at a viscosity of 30, the direct impact of load on wear life (before accounting for the interaction) is given by the coefficient 0.095. However, the interaction term modulates this effect, making the change in mean life with increasing load negative.- With higher viscosity, such as 40, the interaction effect intensifies the reduction in life, resulting in a change from a positive increase to a net negative proportion.Thus, understanding interaction terms enables us to grasp complex relationships where variables do not operate independently, but rather, their effects are intertwined.
Mean Value Calculation
Mean value calculation in regression is integral in determining the expected outcome of the dependent variable given specific values of independent variables. This is how we calculate using our model:- **Substitute Given Values**: First, plug the given levels of oil viscosity and load into the equation.- **Simplify the Equation**: Start calculating each term separately. - For example, when viscosity \(x_1=40\) and load \(x_2=1100\), compute each variable term. The results you need to compute are: \(7.75 \times 40\), \(0.0950 \times 1100\), and \(-0.0090 \times 40 \times 1100\). - **Combine Terms**: Add the constant and calculated results and subtract the interaction results to find the mean life. The solution to the above calculations reveals the expected wear life given certain oil viscosity and load parameters. For instance, the exercise concluded with a mean life of 147.5 units for specific conditions. This precise calculation helps in understanding how input variations affect predictions in multiple regression contexts.

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