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8.4 Multiple Regression8.4 Multiple Regression

Lecture Unit 8

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8.4 Introduction

• In this section we extend simple linear regression where we had one explanatory variable, and allow for any number of explanatory variables.

• We expect to build a model that fits the data better than the simple linear regression model.

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• We shall use computer printout to – Assess the model

• How well it fits the data• Is it useful• Are any required conditions violated?

– Employ the model• Interpreting the coefficients• Predictions using the prediction equation• Estimating the expected value of the dependent variable

Introduction

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Coefficients

Dependent variable Independent variables

Random error variable

Multiple Regression Model

• We allow for k explanatory variables to potentially be related to the response variable

y = 0 + 1x1+ 2x2 + …+ kxk +

The Multiple Regression ModelIdea: Examine the linear relationship between 1 response variable (y) & 2 or more explanatory variables (xi)

εxβxβxββy kk22110

kk22110 xbxbxbby

Population model:

Y-intercept Population slopes Random Error

Estimated (or predicted)

value of yEstimated slope coefficients

Estimated multiple regression model:

Estimatedintercept

Simple Linear Regression

Random Error for this x value

y

x

Observed Value of y for xi

Predicted Value of y for xi

0 1y b b x

xi

Slope = b1

Intercept = b0

εi

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Multiple Regression, 2 explanatory variables

•X

•X

•*

•*•*

•*

•*•*

•*•*

•*

•*

•*

•*

•*

•Y

•*

•*

•*

•*

•2

•1

•Least Squares

Plane (instead of

line)

•Scatter of points around plane are random error.

Multiple Regression ModelTwo variable model

y

x1

x2

22110 xbxbby yi

yi

<

e = (yi – yi)<

x2i

x1i The best fit equation, y , is found by minimizing the sum of squared errors, e2

<

Sample observation

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• The error is normally distributed.• The mean is equal to zero and the standard

deviation is constant ( for all values of y. • The errors are independent.

Required conditions for the error variable

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– If the model assessment indicates good fit to the data, use it to interpret the coefficients and generate predictions.

– Assess the model fit using statistics obtained from the sample.

– Diagnose violations of required conditions. Try to remedy problems when identified.

8.4 Estimating the Coefficients and Assessing the Model

• The procedure used to perform regression analysis:– Obtain the model coefficients and statistics using statistical

software.

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• Predicting final exam scores in BUS/ST 350– We would like to predict final exam scores in 350.– Use information generated during the semester.– Predictors of the final exam score:

• Exam 1• Exam 2• Exam 3• Homework total

Estimating the Coefficients and Assessing the Model, Example

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• Data were collected from 203 randomly selected students from previous semesters• The following model is proposed

final exam = exam1 exam2exam3hwtot

Estimating the Coefficients and Assessing the Model, Example

exam 1 exam2 exam3 hwtot finalexm80 60 80 159 7280 70 75 359 7695 70 90 330 8490 100 100 359 9270 60 80 272 6490 70 70 344 8490 85 90 351 8885 35 90 200 7685 55 70 251 6040 80 95 293 64

Regression StatisticsMultiple R 0.618439R Square 0.38246679Adjusted R Square 0.36999137Standard Error 11.5122313Observations 203

ANOVAdf SS MS F Significance F

Regression 4 16252.40443 4063 30.66 7.32692E-20Residual 198 26241.23104 132.5Total 202 42493.63547

Coefficients Standard Error t Stat P-value Lower 95% Upper 95%Intercept 0.04978935 8.17368799 0.006 0.995 -16.06886586 16.16844455exam 1 0.10021107 0.075633398 1.325 0.187 -0.048939306 0.249361453exam2 0.15413733 0.072271404 2.133 0.034 0.011616858 0.296657794exam3 0.29600913 0.066724619 4.436 2E-05 0.16442702 0.427591244hwtot 0.10771069 0.022685084 4.748 4E-06 0.062975308 0.15244607213

This is the sample regression equation (sometimes called the prediction equation)This is the sample regression equation (sometimes called the prediction equation)

Regression Analysis, Excel Output

Final exam score = 0.0498 + 0.1002exam1 + 0.1541exam2 + 0.2960exam3 +0.1077hwtot

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• b0 = 0.0498. This is the intercept, the value of y when all

the variables take the value zero. Since the data range of all the independent variables do not cover the value zero, do not interpret the intercept.

• b1 = 0.1002. In this model, for each additional point on

exam 1, the final exam score increases on average by

0.1002 (assuming the other variables are held

constant).

Interpreting the Coefficients

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• b2 = 0.1541. In this model, for each additional point on exam 2, the final exam score increases on average by 0.1541 (assuming the other variables are held constant).

• b3 = 0.2960. For each additional point on exam 3, the final exam score increases on average by 0.2960 (assuming the other variables are held constant).

• b4 = 0.1077. For each additional point on the homework, the final exam score increases on average by 0.1077 (assuming the other variables are held constant).

Interpreting the Coefficients

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• Predict the average final exam score of a student with the following exam scores and homework score:– Exam 1 score 75,– Exam 2 score 79,– Exam 3 score 85,– Homework score 310

– Use trend function in ExcelFinal exam score =0.0498 + 0.1002(75) +0.1541(79) + 0.2960(85) + 0.1077(310) = 78.2857

Final Exam Scores, Predictions

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Model Assessment

• The model is assessed using three tools:– The standard error of the residuals – The coefficient of determination– The F-test of the analysis of variance

• The standard error of the residuals participates in building the other tools.

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• The standard deviation of the residuals is estimated by the Standard Error of the Residuals:

• The magnitude of s is judged by comparing it to

1knSSE

s

Standard Error of Residuals

.y

Regression StatisticsMultiple R 0.618439R Square 0.38246679Adjusted R Square 0.36999137Standard Error 11.5122313Observations 203

ANOVAdf SS MS F Significance F

Regression 4 16252.40443 4063 30.66 7.32692E-20Residual 198 26241.23104 132.5Total 202 42493.63547

Coefficients Standard Error t Stat P-value Lower 95% Upper 95%Intercept 0.04978935 8.17368799 0.006 0.995 -16.06886586 16.16844455exam 1 0.10021107 0.075633398 1.325 0.187 -0.048939306 0.249361453exam2 0.15413733 0.072271404 2.133 0.034 0.011616858 0.296657794exam3 0.29600913 0.066724619 4.436 2E-05 0.16442702 0.427591244hwtot 0.10771069 0.022685084 4.748 4E-06 0.062975308 0.15244607219

Regression Analysis, Excel OutputStandard error of the residuals; sqrt(MSE) (standard error of the residuals)2: MSE=SSE/198

Sum of squares of residuals SSE

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• From the printout, s = 11.5122….• Calculating the mean value of y we have• It seems s is not particularly small. • Question:

Can we conclude the model does not fit the data well?

78.84y

Standard Error of Residuals

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• The proportion of the variation in y that is explained by differences in the explanatory variables x1, x2, …, xk

• R = 1 – (SSE/SSTotal)• From the printout, R2 = 0.382466…• 38.25% of the variation in final exam score is explained by

differences in the exam1, exam2, exam3, and hwtot explanatory variables. 61.75% remains unexplained.

• When adjusted for degrees of freedom, Adjusted R2 = 36.99%

Coefficient of Determination R2

(like r2 in simple linear regression

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• We pose the question:Is there at least one explanatory variable linearly related to the response variable?

• To answer the question we test the hypothesis

H0: 1 = 2 = … = k=0

H1: At least one i is not equal to zero.

• If at least one i is not equal to zero, the model has some validity.

Testing the Validity of the Model

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• The hypotheses are tested by what is called an F test shown in the Excel output below

Testing the Validity of the Final Exam Scores Regression Model

k =n–k–1 = n-1 =

ANOVAdf SS MS F Significance F

Regression 4 16252.404 4063 30.66 7.32692E-20Residual 198 26241.231 132.5Total 202 42493.635

P-value

SSR

SSE MSE=SSE/(n-k-1)

MSR=SSR/k

MSR/MSE

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[Variation in y] = SSR + SSE. Large F results from a large SSR. Then, much of the variation in y is explained by the regression model; the model is useful, and thus, the null hypothesis H0 should be rejected. Reject H0 when P-value < 0.05

Testing the Validity of the Final Exam Scores Regression Model

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The P-value (Significance F) < 0.05Reject the null hypothesis.

Testing the Validity of the Final Exam Scores Regression Model

ANOVAdf SS MS F Significance F

Regression 4 16252.404 4063 30.66 7.32692E-20Residual 198 26241.231 132.5Total 202 42493.635

Conclusion: There is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. At least one of the i is not equal to zero. Thus, at least one explanatory variable is linearly related to y. This linear regression model is valid

Conclusion: There is sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. At least one of the i is not equal to zero. Thus, at least one explanatory variable is linearly related to y. This linear regression model is valid

Coefficients Standard Error t Stat P-valueIntercept 0.04978935 8.17368799 0.006 0.995145915exam 1 0.10021107 0.075633398 1.325 0.186712117exam2 0.15413733 0.072271404 2.133 0.034176157exam3 0.29600913 0.066724619 4.436 1.51714E-05hwtot 0.10771069 0.022685084 4.748 3.93288E-06

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• The hypothesis for each i is

• Excel printout

H0: i 0H1: i 0 d.f. = n - k -1

Test statistic0

i

i

b

bt

s

Testing the Coefficients