/*! 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 5 You obtain the multiple regressi... [FREE SOLUTION] | 91Ó°ÊÓ

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You obtain the multiple regression equation \(\hat{y}=5+3 x_{1}-4 x_{2}\) from a set of sample data. (a) Interpret the slope coefficients for \(x_{1}\) and \(x_{2}\). (b) Determine the regression equation with \(x_{1}=10\). Graph the regression equation with \(x_{1}=10\). (c) Determine the regression equation with \(x_{1}=15 .\) Graph the regression equation with \(x_{1}=15\) (d) Determine the regression equation with \(x_{1}=20 .\) Graph the regression equation with \(x_{1}=20 .\) (e) What is the effect of changing the value \(x_{1}\) on the graph of the regression equation?

Short Answer

Expert verified
For \(x_{1}=10\) the equation is \hat{y}=35-4x_{2}\, for \(x_{1}=15\) it is \hat{y}=50-4x_{2}\, and for \(x_{1}=20\) it is \hat{y}=65-4x_{2}\. Changing \(x_{1}\) moves the regression line vertically.

Step by step solution

01

Interpret Slope Coefficients

The slope coefficient for \(x_{1}\) is 3. This indicates that for every one unit increase in \(x_{1}\), \(y\) increases by 3 units, holding \(x_{2}\) constant. The slope coefficient for \(x_{2}\) is -4. This indicates that for every one unit increase in \(x_{2}\), \(y\) decreases by 4 units, holding \(x_{1}\) constant.
02

Regression Equation with \(x_{1}=10\)

Substitute \(x_{1}=10\) into the original regression equation: \hat{y} = 5 + 3(10) - 4x_{2} = 35 - 4x_{2}\. This is the new regression equation.
03

Graph the Regression Equation with \(x_{1}=10\)

Graph the equation \hat{y} = 35 - 4x_{2}\. It is a straight line with a y-intercept of 35 and a slope of -4. For every one unit increase in \(x_{2}\), \(y\) decreases by 4 units.
04

Regression Equation with \(x_{1}=15\)

Substitute \(x_{1}=15\) into the original regression equation: \hat{y} = 5 + 3(15) - 4x_{2} = 50 - 4x_{2}\. This is the new regression equation.
05

Graph the Regression Equation with \(x_{1}=15\)

Graph the equation \hat{y} = 50 - 4x_{2}\. This line has a y-intercept of 50 and a slope of -4. For every one unit increase in \(x_{2}\), \(y\) decreases by 4 units.
06

Regression Equation with \(x_{1}=20\)

Substitute \(x_{1}=20\) into the original regression equation: \hat{y} = 5 + 3(20) - 4x_{2} = 65 - 4x_{2}\. This is the new regression equation.
07

Graph the Regression Equation with \(x_{1}=20\)

Graph the equation \hat{y} = 65 - 4x_{2}\. This line has a y-intercept of 65 and a slope of -4. For every one unit increase in \(x_{2}\), \(y\) decreases by 4 units.
08

Effect of Changing \(x_{1}\)

Changing the value of \(x_{1}\) shifts the regression equation vertically without changing the slope. As \(x_{1}\) increases, the y-intercept of the regression line increases, causing the line to shift upwards. Conversely, as \(x_{1}\) decreases, the y-intercept decreases, shifting the line downwards.

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Key Concepts

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

slope coefficient interpretation
When dealing with multiple regression analysis, slope coefficients are crucial as they show the relationship between independent variables ( x_{1}, x_{2}, etc.) and the dependent variable ( y ). In the given regression equation , the interpretation of the slope coefficients provides insights into how changes in each predictor influence the outcome.
The slope coefficient for x_{1} is 3 . This means that if increases by 1 unit, increases by 3 units, assuming x_{2} remains constant.
Conversely, the slope coefficient for x_{2} is -4 . This implies that for every 1-unit increase in x_{2} , decreases by 4 units, holding x_{1} constant.
These interpretations help in understanding the individual impact of each independent variable on the dependent variable within the model.
regression equation transformation
Altering the values of independent variables transforms the regression equation, providing varied perspectives of the model. Considering the original regression equation an example transformation occurs by setting x_{1} to specific values.
Let's start with x_{1} set to 10 :
Substituting x_{1}=10 into the equation results in:
= 35 - 4 x_{2}
This new equation shows the relationship between y and x_{2} with x_{1} fixed at 10 .
Transforming the regression equation with x_{1}=15 :
Substituting this value, we get:
= 50 - 4 x_{2}
Again, this portrays a different relationship between y and x_{2} with x_{1} held constant at 15 .
Finally, for x_{1}=20 :
Substitution results in:
= 65 - 4 x_{2}
This showcases another variation of the equation where x_{1} is set to 20 .
Each transformed equation retains the original slopes of 3 for x_{1} and -4 for x_{2} , demonstrating how fixed values of one variable affect the outcome.
graphing regression equations
Graphing regression equations is an effective way to visualize the relationships predicted by the regression model. Let's look at the equations derived by setting x_{1} to different values.
For x_{1}=10 , the regression equation is:
= 35 - 4 x_{2}
This is a straight line with a y-intercept at 35 and a slope of -4 . Each unit increase in x_{2} results in a 4-unit decrease in y .
For x_{1}=15 :
= 50 - 4 x_{2}
The line now has a y-intercept of 50 , maintaining the same slope of -4 .
Finally, for x_{1}=20 :
= 65 - 4 x_{2}
Here, the y-intercept is 65 , with the slope staying -4 .
Each graph visually depicts how fixing x_{1} at increasing levels shifts the line upwards on the graph, affecting the predictions. The slope remains unchanged, indicating that the rate at which y changes relative to x_{2} is consistent regardless of the fixed value of x_{1} . This visual interpretation can be very useful in understanding the dynamics of the regression model in practical applications.

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Most popular questions from this chapter

Concrete As concrete cures, it gains strength. The following data represent the 7 -day and 28 -day strength (in pounds per square inch) of a certain type of concrete: $$ \begin{array}{cc|cc} \begin{array}{l} \text { 7-Day } \\ \text { Strength, } x \end{array} & \begin{array}{l} \text { 28-Day } \\ \text { Strength, } y \end{array} & \begin{array}{l} \text { 7-Day } \\ \text { Strength, } x \end{array} & \begin{array}{l} \text { 28-Day } \\ \text { Strength, } \boldsymbol{y} \end{array} \\ \hline 2300 & 4070 & 2480 & 4120 \\ \hline 3390 & 5220 & 3380 & 5020 \\ \hline 2430 & 4640 & 2660 & 4890 \\ \hline 2890 & 4620 & 2620 & 4190 \\ \hline 3330 & 4850 & 3340 & 4630 \\ \hline \end{array} $$ (a) Treating the 7 -day strength as the explanatory variable, \(x\), determine the estimates of \(\beta_{0}\) and \(\beta_{1}\). (b) Compute the standard error of the estimate. (c) Determine \(s_{b_{1}}\). (d) Assuming the residuals are normally distributed, test whether a linear relation exists between 7 -day strength and 28 -day strength at the \(\alpha=0.05\) level of significance. (e) Assuming the residuals are normally distributed, construct a \(95 \%\) confidence interval for the slope of the true leastsquares regression line. (f) What is the estimated mean 28 -day strength of this concrete if the 7 -day strength is 3000 psi?

Tires The following data represent the cost of tires (in dollars) along with a variety of potential explanatory variables. Slalom time is the amount of time it took for a 3-series BMW to get through a slalom track, lap time is the amount of time it took the same car to complete a \(1 / 3\) -mile lap, and stopping distance is the distance it took the BMW to stop on wet pavement traveling 60 miles per hour. Find the best regression model using each of the three techniques presented in the section. What do you notice? $$ \begin{array}{lcccccc} \text { TIRE } & \begin{array}{c} \text { Cost } \\ \text { (dollars) } \end{array} & \text { MPG } & \begin{array}{c} \text { Slalom Time } \\ \text { (seconds) } \end{array} & \begin{array}{c} \text { Lap Time } \\ \text { (seconds) } \end{array} & \begin{array}{c} \text { Stopping } \\ \text { Distance } \\ \text { (feet) } \end{array} & \begin{array}{c} \text { Cornering } \\ \text { g-Force } \end{array} \\ \hline \text { BFGoodrich g-Force Sport COMP-2 } & 114 & 30.5 & 5.13 & 30.24 & 80.0 & 0.90 \\ \hline \text { Bridgestone Potenza RE760 Sport } & 126 & 30.2 & 5.08 & 30.14 & 79.4 & 0.91 \\ \hline \text { Firestone Firehawk Wide Oval Indy } 500 & 111 & 30.4 & 5.16 & 30.58 & 83.3 & 0.88 \\ \hline \text { Yokohama S.drive } & 119 & 31.0 & 5.20 & 30.61 & 82.2 & 0.90 \\\ \hline \text { Bridgestone Turanza Serenity Plus } & 154 & 32.2 & 5.10 & 31.13 & 90.4 & 0.84 \\ \hline \text { Continental PureContact } & 134 & 32.7 & 5.15 & 31.18 & 91.2 & 0.85 \\ \hline \text { Michelin Primacy MXV4 } & 135 & 32.3 & 5.15 & 31.18 & 90.2 & 0.85 \\ \hline \text { Yokohama AVID Ascend } & 134 & 32.3 & 5.17 & 31.11 & 91.4 & 0.86 \\ \hline \end{array} $$

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Why is it important to perform graphical as well as analytical analyses when analyzing relations between two quantitative variables?

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