1 Simple Linear Regression Chapter 17

Introduction In Chapters 17 to 19 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 (x 1, x 2,…x k.). –Analyze the specific relationships between the independent variables and the dependent variable.

3 House size House Cost Most lots sell for $25,000 Building a house costs about $75 per square foot. House cost = (Size) 17.2 The Model The model has a deterministic and a probabilistic components

4 House cost = (Size) House size House Cost Most lots sell for $25,000   However, house cost vary even among same size houses! 17.2 The Model Since cost behave unpredictably, we add a random component.

The Model The first order linear model y = dependent variable x = independent variable  0 = y-intercept  1 = slope of the line  = error variable x y 00 Run Rise   = Rise/Run  0 and  1 are unknown population parameters, therefore are estimated from the data.

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.           Question: What should be considered a good line? x y

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

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 Example 17.2 (Xm17-02)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 The Simple Linear Regression Line

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

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

13 The Simple Linear Regression Line Xm17-02

14 This is the slope of the line. For each additional mile on the odometer, the price decreases by an average of $ Interpreting the Linear Regression - Equation The intercept is b 0 = $ No data Do not interpret the intercept as the “Price of cars that have not been driven” 17067

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

16 The Normality of  From the first three assumptions we have: y is normally distributed with mean E(y) =  0 +  1 x, and a constant standard deviation   From the first three assumptions we have: y is normally distributed with mean E(y) =  0 +  1 x, and a constant standard deviation     0 +  1 x 1  0 +  1 x 2  0 +  1 x 3 E(y|x 2 ) E(y|x 3 ) x1x1 x2x2 x3x3  E(y|x 1 )  The standard deviation remains constant, but the mean value changes with x

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 –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 Sum of Squares for Errors –A shortcut formula

19 –The mean error is equal to zero. –If   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   as a measure of the suitability of using a linear model. –An estimator of   is given by s  Standard Error of Estimate

20 Example 17.3 –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 s  even when compared with the mean value of y. Standard Error of Estimate, Example

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

22 We can draw inference about  1 from b 1 by testing H 0 :  1 = 0 H 1 :  1 = 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. The standard error of b 1. where Testing the Slope

23 Example 17.4 –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  = 5%. Testing the Slope, Example

24 Solving by hand –To compute “t” we need the values of b 1 and s b1. –The rejection region is t > t.025 or t < -t.025 with = n-2 = 98. Approximately, t.025 = Testing the Slope, Example

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

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

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

28 Coefficient of determination x1x1 x2x2 y1y1 y2y2 y Two data points (x 1,y 1 ) and (x 2,y 2 ) of a certain sample are shown. Total variation in y = Variation explained by the regression line + Unexplained variation (error) Variation in y = SSR + SSE

29 Coefficient of determination R 2 measures the proportion of the variation in y that is explained by the variation in x. R 2 takes on any value between zero and one. R 2 = 1: Perfect match between the line and the data points. R 2 = 0: There are no linear relationship between x and y.

30 Example 17.5 –Find the coefficient of determination for Example 17.2; what does this statistic tell you about the model? Solution –Solving by hand; Coefficient of determination, Example

31 – 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. Coefficient of determination

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 =  0 +  1 R m +  Rate of return on a particular stockRate 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 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. Example 17.6 (Xm17-06)Xm17-06

34 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 17.7 Using the Regression Equation Before using the regression model, we need to assess how well it fits the data.

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). –It is predicted that a 40,000 miles car would sell for $14,575. – How close is this prediction to the real price? A point prediction

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 confidence interval –The prediction 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 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%) = Interval Estimates, Example

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

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

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

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 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 Testing the coefficient of correlation X Y

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

46 Testing the Coefficient of correlation Foreign Index Funds (Index)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 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 Testing the Coefficient of Correlation, Example

49 Solution – continued –The hypotheses H 0 :   0 H 1 :   0. –Solving by hand: The rejection region: |t| > t  /2,n-2 = t.025,59-2  The sample coefficient of correlation: Cov(x,y) = ; s x =.0509; s y = 0512 r = cov(x,y)/s x s y =.491 The value of the t statistic is Conclusion: There is sufficient evidence at  = 5% to infer that there are linear relationship between the two variables. Testing the Coefficient of Correlation, Example

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

51 Spearman Rank Correlation Coefficient The Spearman rank test is a nonparametric procedure. The procedure is used to test linear relationships between two variables when the bivariate distribution is nonnormal. Bivariate nonnormal distribution may occur when –at least one variable is ordinal, or –both variables are interval but at least one variable is not normal.

52 –The hypotheses are: H 0 :  s  0 H 1 :  s  0 –The test statistic is where ‘a’ and ‘b’ are the ranks of x and y respectively. –For a large sample (n > 30) r s is approximately normally distributed Spearman Rank Correlation Coefficient

53 Spearman Rank Correlation Coefficient, Example Example 17.8 (Xm17-08)Xm17-08 –A production manager wants to examine the relationship between: Aptitude test score given prior to hiring, and Performance rating three months after starting work. –A random sample of 20 production workers was selected. The test scores as well as performance rating was recorded.

54 Spearman Rank Correlation Coefficient, Example Scores range from 0 to 100Scores range from 1 to 5

55 Spearman Rank Correlation Coefficient, Example Solution –The problem objective is to analyze the relationship between two variables. (Note: Performance rating is ordinal.) –The hypotheses are: H 0 :  s = 0 H 1 :  s = 0 –The test statistic is r s, and the rejection region is |r s | > r critical (taken from the Spearman rank correlation table).

56 Spearman Rank Correlation Coefficient, Example Ties are broken by averaging the ranks. –Solving by hand Rank each variable separately. Calculate s a = 5.92; s b =5.50; cov(a,b) = Thus r s = cov(a,b)/[s a s b ] =.379. The critical value for  =.05 and n = 20 is.450.

57 Conclusion: Do not reject the null hypothesis. At 5% significance level there is insufficient evidence to infer that the two variables are related to one another. Conclusion: Do not reject the null hypothesis. At 5% significance level there is insufficient evidence to infer that the two variables are related to one another. Spearman Rank Correlation Coefficient, Example

58 Excel Solution (Data Analysis Plus; Xm17-08)Xm17-08 Spearman Rank Correlation Coefficient, Example > 0.05

Regression Diagnostics - I The three conditions required for the validity of the regression analysis are: –the error variable is normally distributed. –the error variance is constant for all values of x. –The errors are independent of each other. How can we diagnose violations of these conditions?

60 Residual Analysis Examining the residuals (or standardized residuals), help detect violations of the required conditions. Example 17.2 – continued: –Nonnormality. Use Excel to obtain the standardized residual histogram. Examine the histogram and look for a bell shaped. diagram with a mean close to zero.

61 For each residual we calculate the standard deviation as follows: A Partial list of Standard residuals Standardized residual ‘i’ = Residual ‘i’ Standard deviation Residual Analysis

62 It seems the residual are normally distributed with mean zero Residual Analysis

63 Heteroscedasticity When the requirement of a constant variance is violated we have a condition of heteroscedasticity. Diagnose heteroscedasticity by plotting the residual against the predicted y The spread increases with y ^ y ^ Residual ^ y

64 Homoscedasticity When the requirement of a constant variance is not violated we have a condition of homoscedasticity. Example continued

65 Non Independence of Error Variables – A time series is constituted if data were collected over time. –Examining the residuals over time, no pattern should be observed if the errors are independent. –When a pattern is detected, the errors are said to be autocorrelated. –Autocorrelation can be detected by graphing the residuals against time.

66 Patterns in the appearance of the residuals over time indicates that autocorrelation exists Time Residual Time Note the runs of positive residuals, replaced by runs of negative residuals Note the oscillating behavior of the residuals around zero. 00 Non Independence of Error Variables

67 Outliers An outlier is an observation that is unusually small or large. Several possibilities need to be investigated when an outlier is observed: –There was an error in recording the value. –The point does not belong in the sample. –The observation is valid. Identify outliers from the scatter diagram. It is customary to suspect an observation is an outlier if its | standard residual | > 2

The outlier causes a shift in the regression line … but, some outliers may be very influential An outlier An influential observation

69 Procedure for Regression Diagnostics Develop a model that has a theoretical basis. Gather data for the two variables in the model. Draw the scatter diagram to determine whether a linear model appears to be appropriate. Determine the regression equation. Check the required conditions for the errors. Check the existence of outliers and influential observations Assess the model fit. If the model fits the data, use the regression equation.