1. Simple Linear Regression

2. Introduction
In Chapter we examine the relationship between interval variables via a mathematical equation.
The motivation for using the technique:
Forecast the value of a dependent variable (y) from the value of independent variables (x1, x2,…xk.).
Analyze the specific relationships between the independent variables and the dependent variable.

3. The Model The model has a deterministic and a probabilistic components
House
Cost
Building a house costs about
$75 per square foot.
House cost = (Size)
Most lots sell
for $25,000
House size

4. The Model However, house cost vary even among same size houses!
Since cost behave unpredictably, we add a random component.
House
Cost
Most lots sell
for $25,000
House cost = (Size)
+ e
House size

5. The Model The first order linear model y = dependent variable
x = independent variable
b0 = y-intercept
b1 = slope of the line
e = error variable
b0 and b1 are unknown population parameters, therefore are estimated
from the data. y
Rise
b1 = Rise/Run
Run b0 x

6. Estimating the Coefficients
The estimates are determined by
drawing a sample from the population of interest,
calculating sample statistics.
producing a straight line that cuts into the data. y w
Question: What should be considered a good line? w w w w
w w w w w
w w
w w w x

7. The Least Squares (Regression) Line
A good line is one that minimizes the sum of squared differences between the
points and the line.

8. The Least Squares (Regression) Line
Sum of squared differences =
(2 - 1)2 +
(4 - 2)2 +
( )2 +
( )2 = 6.89
Sum of squared differences =
(2 -2.5)2 +
( )2 +
( )2 +
( )2 = 3.99
Let us compare two lines 4
(2,4)
The second line is horizontal w w
(4,3.2) 3
2.5 2 w
(1,2)
(3,1.5) w
The smaller the sum of
squared differences
the better the fit of the
line to the data. 1 2 3 4

9. The Estimated Coefficients
To calculate the estimates of the line coefficients, that minimize the differences between the data points and the line, use the formulas:
The regression equation that estimates
the equation of the first order linear model
is:

10. The Simple Linear Regression Line
Example 17.2 (Xm17-02)
A car dealer wants to find the relationship between the odometer reading and the selling price of used cars.
A random sample of 100 cars is selected, and the data recorded.
Find the regression line.
Independent variable x
Dependent variable y

11. The Simple Linear Regression Line
Solution
Solving by hand: Calculate a number of statistics
where n = 100.

12. The Simple Linear Regression Line
Solution – continued
Using the computer (Xm17-02)
Tools > Data Analysis > Regression > [Shade the y range and the x range] > OK

13. The Simple Linear Regression Line
Xm17-02

14. Interpreting the Linear Regression -Equation
17067
No data
The intercept is b0 = $17067.
This is the slope of the line.
For each additional mile on the odometer,
the price decreases by an average of $0.0623
Do not interpret the intercept as the
“Price of cars that have not been driven”

15. Error Variable: Required Conditions
The error e is a critical part of the regression model.
Four requirements involving the distribution of e must be satisfied.
The probability distribution of e is normal.
The mean of e is zero: E(e) = 0.
The standard deviation of e is se for all values of x.
The set of errors associated with different values of y are all independent.

16. but the mean value changes with x
The Normality of e
E(y|x3)
The standard deviation remains constant, m3
b0 + b1x3
E(y|x2)
b0 + b1x2 m2
E(y|x1)
but the mean value changes with x m1
b0 + b1x1
From the first three assumptions we have:
y is normally distributed with mean
E(y) = b0 + b1x, and a constant standard deviation se x1 x2 x3

17. Assessing the Model
The least squares method will produces a regression line whether or not there are linear relationship between x and y.
Consequently, it is important to assess how well the linear model fits the data.
Several methods are used to assess the model. All are based on the sum of squares for errors, SSE.

18. Sum of Squares for Errors
This is the sum of differences between the points and the regression line.
It can serve as a measure of how well the line fits the data. SSE is defined by
A shortcut formula

19. Standard Error of Estimate
The mean error is equal to zero.
If se is small the errors tend to be close to zero (close to the mean error). Then, the model fits the data well.
Therefore, we can, use se as a measure of the suitability of using a linear model.
An estimator of se is given by se

20. Standard Error of Estimate, Example
Calculate the standard error of estimate for Example 17.2, and describe what does it tell you about the model fit?
Solution
Calculated before
It is hard to assess the model based
on se even when compared with the
mean value of y.

21. The slope is not equal to zero
Testing the slope
When no linear relationship exists between two variables, the regression line should be horizontal. q q q q q q q q q q q q
Linear relationship.
Linear relationship.
Linear relationship.
Linear relationship.
No linear relationship.
Different inputs (x) yield
the same output (y).
Different inputs (x) yield
different outputs (y).
The slope is not equal to zero
The slope is equal to zero

22. Testing the Slope We can draw inference about b1 from b1 by testing
H0: b1 = 0
H1: b1 = 0 (or < 0,or > 0)
The test statistic is
If the error variable is normally distributed, the statistic is Student t distribution with d.f. = n-2.
where
The standard error of b1.

23. Testing the Slope, Example
Test to determine whether there is enough evidence to infer that there is a linear relationship between the car auction price and the odometer reading for all three-year-old Tauruses, in Example Use a = 5%.

24. Testing the Slope, Example
Solving by hand
To compute “t” we need the values of b1 and sb1.
The rejection region is t > t.025 or t < -t.025 with n = n-2 = 98. Approximately, t.025 = 1.984

25. Testing the Slope, Example
Xm17-02
Using the computer
There is overwhelming evidence to infer
that the odometer reading affects the
auction selling price.

26. Coefficient of determination
To measure the strength of the linear relationship we use the coefficient of determination.

27. Coefficient of determination
To understand the significance of this coefficient note:
The regression model
Explained in part by
Overall variability in y
Remains, in part, unexplained
The error

28. Coefficient of determinationy2
Two data points (x1,y1) and (x2,y2)
of a certain sample are shown. y
Variation in y = SSR + SSE y1 x1 x2
Total variation in y =
Variation explained by the
regression line
+ Unexplained variation (error)

29. Coefficient of determination
R2 measures the proportion of the variation in y that is explained by the variation in x.
R2 takes on any value between zero and one.
R2 = 1: Perfect match between the line and the data points.
R2 = 0: There are no linear relationship between x and y.

30. Coefficient of determination, Example
Find the coefficient of determination for Example 17.2; what does this statistic tell you about the model?
Solution
Solving by hand;

31. Coefficient of determination
Using the computer From the regression output we have
65% of the variation in the auction
selling price is explained by the
variation in odometer reading. The
rest (35%) remains unexplained by
this model.

32. 17.6 Finance Application: Market Model
One of the most important applications of linear regression is the market model.
It is assumed that rate of return on a stock (R) is linearly related to the rate of return on the overall market.
R = b0 + b1Rm +e
Rate of return on a particular stock
Rate of return on some major stock index
The beta coefficient measures how sensitive the stock’s rate
of return is to changes in the level of the overall market.

33. The Market Model, Example
Example 17.6 (Xm17-06)
Estimate the market model for Nortel, a stock traded in the Toronto Stock Exchange (TSE).
Data consisted of monthly percentage return for Nortel and monthly percentage return for all the stocks.
This is a measure of the stock’s
market related risk. In this sample,
for each 1% increase in the TSE return, the average increase in Nortel’s return is .8877%.
This is a measure of the total market-related risk embedded in the Nortel stock.
Specifically, 31.37% of the variation in Nortel’s
return are explained by the variation in the
TSE’s returns.

34. Using the Regression Equation
Before using the regression model, we need to assess how well it fits the data.
If we are satisfied with how well the model fits the data, we can use it to predict the values of y.
To make a prediction we use
Point prediction, and
Interval prediction

35. Point Prediction Example 17.7
Predict the selling price of a three-year-old Taurus with 40,000 miles on the odometer (Example 17.2).
A point prediction
It is predicted that a 40,000 miles car would sell for $14,575.
How close is this prediction to the real price?

36. Interval Estimates
Two intervals can be used to discover how closely the predicted value will match the true value of y.
Prediction interval – predicts y for a given value of x,
Confidence interval – estimates the average y for a given x.
The prediction interval
The confidence interval

37. Interval Estimates, Example
Example continued
Provide an interval estimate for the bidding price on a Ford Taurus with 40,000 miles on the odometer.
Two types of predictions are required:
A prediction for a specific car
An estimate for the average price per car

38. Interval Estimates, Example
Solution
A prediction interval provides the price estimate for a single car:
t.025,98
Approximately

39. Interval Estimates, Example
Solution – continued
A confidence interval provides the estimate of the mean price per car for a Ford Taurus with 40,000 miles reading on the odometer.
The confidence interval (95%) =

40. The effect of the given xg on the length of the interval
As xg moves away from x the interval becomes longer. That is, the shortest interval is found at x.

41. The effect of the given xg on the length of the interval
As xg moves away from x the interval becomes longer. That is, the shortest interval is found at x.

42. The effect of the given xg on the length of the interval
As xg moves away from x the interval becomes longer. That is, the shortest interval is found at x.

43. Coefficient of Correlation
The coefficient of correlation is used to measure the strength of association between two variables.
The coefficient values range between -1 and 1.
If r = -1 (negative association) or r = +1 (positive association) every point falls on the regression line.
If r = 0 there is no linear pattern.
The coefficient can be used to test for linear relationship between two variables.

44. Testing the coefficient of correlation
To test the coefficient of correlation for linear relationship between X and Y
X and Y must be observational
X and Y are bivariate normally distributed X Y

45. Testing the coefficient of correlation
When no linear relationship exist between the two variables, r = 0.
The hypotheses are:
H0: r = 0 H1: r ¹ 0
The test statistic is:
The statistic is Student t distributed with d.f. = n - 2, provided the variables
are bivariate normally distributed.

46. Testing the Coefficient of correlation
Foreign Index Funds (Index)
A certain investor prefers the investment in an index mutual funds constructed by buying a wide assortment of stocks.
The investor decides to avoid the investment in a Japanese index fund if it is strongly correlated with an American index fund that he owns.
From the data shown in Index.xls should he avoid the investment in the Japanese index fund?

47. Testing the Coefficient of correlation
Foreign Index Funds
A certain investor prefers the investment in an index mutual funds constructed by buying a wide assortment of stocks.
The investor decides to avoid the investment in a Japanese index fund if it is strongly correlated with an American index fund that he owns.
From the data shown in Index.xls should he avoid the investment in the Japanese index fund?

48. Testing the Coefficient of Correlation, Example
Solution
Problem objective: Analyze relationship between two interval variables.
The two variables are observational (the return for each fund was not controlled).
We are interested in whether there is a linear relationship between the two variables, thus, we need to test the coefficient of correlation

49. Testing the Coefficient of Correlation, Example
Solution – continued
The hypotheses H0: r = 0 H1: r ¹ 0.
Solving by hand:
The rejection region: |t| > ta/2,n-2 = t.025,59-2 »
The sample coefficient of correlation: Cov(x,y) = ; sx = .0509; sy = r = cov(x,y)/sxsy=.491
The value of the t statistic is
Conclusion: There is sufficient
evidence at a = 5% to infer that
there are linear relationship
between the two variables.

50. Testing the Coefficient of Correlation, Example
Excel solution (Index)