1. Lecture Unit 8
8.4 Multiple Regression

2. 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.

3. Introduction 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

4. Multiple Regression Model
We allow for k explanatory variables to potentially be related to the response variable
y = b0 + b1x1+ b2x2 + …+ bkxk + e
Coefficients
Random error variable
Dependent variable
Independent variables

5. The Multiple Regression Model
Idea: Examine the linear relationship between 1 response variable (y) & 2 or more explanatory variables (xi)
Population model:
Y-intercept
Population slopes
Random Error
Estimated multiple regression model:
Estimated
(or predicted)
value of y
Estimated
intercept
Estimated slope coefficients

6. Simple Linear Regressiony
Observed Value of y for xi εi
Slope = b1
Predicted Value of y for xi
Random Error for this x value
Intercept = b0 xi x

7. Multiple Regression, 2 explanatory variables* Y 2 1
Least Squares Plane (instead of line)
Scatter of points around plane are random error.

8. Multiple Regression Model
Two variable model y
Sample observation yi yi
<
e = (yi – yi)
<
x2i x2
x1i
<
The best fit equation, y ,
is found by minimizing the
sum of squared errors, e2 x1

9. Required conditions for the error variable
The error e is normally distributed.
The mean is equal to zero and the standard deviation is constant (se) for all values of y.
The errors are independent.

10. 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.
Diagnose violations of required conditions. Try to remedy problems when identified.
Assess the model fit using statistics obtained from the sample.
If the model assessment indicates good fit to the data, use it to interpret the coefficients and generate predictions.

11. Estimating the Coefficients and Assessing the Model, Example
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

12. Estimating the Coefficients and Assessing the Model, Example
Data were collected from 203 randomly selected students from previous semesters
The following model is proposed final exam = b0 + b1exam1 + b2exam2 + b3exam3 + b4hwtot
exam 1
exam2
exam3
hwtot
finalexm 80 60
159 72 70 75
359 76 95 90
330 84
100 92
272 64
344 85
351 88 35
200 55
251 40
293

13. Regression Analysis, Excel Output
This is the sample regression equation
(sometimes called the prediction equation)
Final exam score =
exam exam exam hwtot

14. Interpreting the Coefficients
b0 = 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 = In this model, for each additional point on exam 1, the final exam score increases on average by (assuming the other variables are held constant).

15. Interpreting the Coefficients
b2 = In this model, for each additional point on exam 2, the final exam score increases on average by (assuming the other variables are held constant).
b3 = For each additional point on exam 3, the final exam score increases on average by (assuming the other variables are held constant).
b4 = For each additional point on the homework, the final exam score increases on average by (assuming the other variables are held constant).

16. Final Exam Scores, Predictions
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 Excel
Final exam score =
(75) (79) (85) (310) =

17. 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.

18. Standard Error of Residuals
The standard deviation of the residuals is estimated by the Standard Error of the Residuals:
The magnitude of se is judged by comparing it to

19. Regression Analysis, Excel Output
Standard error of the residuals; sqrt(MSE)
(standard error of the residuals)2: MSE=SSE/198
Sum of squares of residuals SSE

20. Standard Error of Residuals
From the printout, se = ….
Calculating the mean value of y we have
It seems se is not particularly small.
Question: Can we conclude the model does not fit the data well?

21. Coefficient of Determination R2 (like r2 in simple linear regression
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 = …
38.25% of the variation in final exam score is explained by differences in the exam1, exam2, exam3, and hwtot explanatory variables % remains unexplained.
When adjusted for degrees of freedom, Adjusted R2 = 36.99%

22. Testing the Validity of the Model
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: b1 = b2 = … = bk=0
H1: At least one bi is not equal to zero.
If at least one bi is not equal to zero, the model has some validity.

23. Testing the Validity of the Final Exam Scores Regression Model
The hypotheses are tested by what is called an F test shown in the Excel output below
MSR/MSE
P-value
k =
n–k–1 =
n-1 =
SSR
MSR=SSR/k
MSE=SSE/(n-k-1)
SSE

24. Testing the Validity of the Final Exam Scores Regression Model
[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

25. Testing the Validity of the Final Exam Scores Regression Model
Conclusion: There is sufficient evidence to reject
the null hypothesis in favor of the alternative hypothesis.
At least one of the bi is not equal to zero. Thus, at least one explanatory variable is linearly related to y.
This linear regression model is valid
The P-value (Significance F) < 0.05
Reject the null hypothesis.

26. Testing the Coefficients
The hypothesis for each bi is
Excel printout
H0: bi = 0
H1: bi ¹ 0
Test statistic
d.f. = n - k -1