Chapter 17 Simple Linear Regression 1 17

Chapter 17 Simple Linear Regression 1

17. 1 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, …xk. ). – Analyze the specific relationships between the independent variables and the 2 dependent variable.

17. 2 The Model The model has a deterministic and a probabilistic components House Cost Most lots sell for $25, 000 out ab s ost c ize) se t. (S ou a h re foo 0 + 75 g ldin r squa 2500 Bui pe t= 75 e cos $ ous H House size 3

17. 2 The Model However, house cost vary even among same size houses! Since cost behave unpredictably, House Cost Most lots sell for $25, 000 we add a random component. +e House cost = 25000 + 75(Size) House size 4

17. 2 The Model • The first order linear model y = dependent variable x = independent variable y b 0 = y-intercept b 1 = slope of the line e = error variable b 0 and b 1 are unknown population parameters, therefore are estimat from the data. Rise b 1 = Rise/Run x 5

17. 3 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 y w Question: What should b data. w w w considered a good line? w w w x 6

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

The Least Squares (Regression) Line Sum of squared differences- = 2 + - 2)2 (1. 5 - 3)2 + - 4)2 = 6. 89 (2 1) (4 + (3. 2 Sum of squared differences-2. 5)2 + - 2. 5)2 (1. 5 - 2. 5)2 (3. 2 - 2. 5)2 = 3. 99 (2 = (4 + + 3 2. 5 2 Let us compare two lines The second line is horizonta (2, 4) w 4 w (4, 3. 2) w (1, 2) w (3, 1. 5) 1 1 2 3 4 The smaller the sum of squared differences the better the fit of the line to the data. 8

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 estimate the equation of the first order linear m is: 9

The Simple Linear Regression Line • Example 17. 2 (Xm 17 -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, Independent Dependent variable x variable y 10

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

The Simple Linear Regression Line • Solution – continued – Using the computer (Xm 17 -02) Tools > Data Analysis > Regression > [Shade the y range and the x range] > OK 12

The Simple Linear Regression Line Xm 17 -02 13

Interpreting the Linear Regression -Equation 17067 0 No data The intercept is b 0 = $17067. This is the slope of the line. For each additional mile on the odometer, the price decreases by an average of $0. 06 Do not interpret the intercept as the “Price of cars that have not been driven” 14

17. 4 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 15

The Normality of e E(y|x 3) The standard deviation remains constant, m 3 b 0 + b 1 x 3 E(y|x 2) b 0 + b 1 x 2 m 2 but the mean value changes with E(y|x 1) x b 0 + b 1 x 1 m 1 From the first three assumptions we x 1 have: y is normally distributed with mean E(y) = b 0 + b 1 x, and a constant standard deviation se x 2 x 3 16

17. 5 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. 17

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 18

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 19

Standard Error of Estimate, Example • 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 ba on se even when compared with mean value of y. 20

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

Testing the Slope • We can draw inference about b 1 from b 1 by testing H 0: b 1 = 0 H 1: b 1 = 0 (or < 0, or > 0) – The test statistic is where The standard error of b 1. – If the error variable is normally distributed, 22 the statistic is Student t distribution with d. f. = n-2.

Testing the Slope, Example • Example 17. 4 – Test to determine whethere 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 17. 2. Use a = 5%. 23

Testing the Slope, Example • Solving by hand – To compute “t” we need the values of b 1 and sb 1. – The rejection region is t > t. 025 or t < -t. 025 with n = n -2 = 98. Approximately, t. 025 = 1. 984 24

Testing the Slope, Example. Xm 17 -02 • Using the computer There is overwhelming evidence to infer that the odometer reading affects the auction selling price. 25

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

Coefficient of determination • To understand the significance of this coefficient note: d laine xp Ey Overall variability inem R ains y The art b in p , in p art, u n expl regression model The d aine error 27

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

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 29 data points.

Coefficient of determination, Example • 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; 30

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. 31

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 = b 0 + b 1 Rm +e Rate of return on a particular stock Rate of return on some major stock ind The beta coefficient measures how sensitive the stock’s rate of return is to changes in the level of the overall market. 32

The Market Model, Example 17. 6 (Xm 17 -06) • Estimate the market model for Nortel, a stock traded in the Toronto Stock Exchange (TSE). • Data consisted of monthly percentage This is a measure of the return for Nortel and monthly percentage stock’s return. This alla measure of the total market-related ri for is the stocks. market related risk. In this sample, for each 1% increase in the TSE return, the average increase in Nortel’s return is. 8877%. 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. 33

17. 7 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 34

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? 35

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 – The prediction – The confidence given x. interval 36

Interval Estimates, Example • Example 17. 7 - 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 37

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

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%) = 39

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. 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

17. 8 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. 43

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 Y X 44

Testing the coefficient of correlation • When no linear relationship exist between the two variables, r = 0. • The hypotheses are: H 0: r = 0 H 1: 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. 45

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 46 fund?

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 47 fund?

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 whethere is a linear relationship between the two variables, thus, we need to test the coefficient of correlation 48

Testing the Coefficient of Correlation, Example • Solution – continued – The hypotheses H 0: r = 0 H 1: r ¹ 0. – Solving by hand: The value of the t statistic is Conclusion: There is sufficient evidence at a = 5% to infer tha there are linear relationship • The rejection region: |t| > ta/2, n-2 = t. 025, 59 -2 » 2. 000. between the two variables. • The sample coefficient of correlation: Cov(x, y) =. 001279; sx =. 0509; sy = 0512 r = cov(x, y)/sxsy=. 491 49

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

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. 51

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

Spearman Rank Correlation Coefficient, Example • Example 17. 8 (Xm 17 -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. 53

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

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: r s = 0 • H 1: r s = 0 – The test statistic is rs, and the rejection region is |rs| > rcritical (taken from the Spearman rank 5 correlation table). 5

Spearman Rank Correlation Coefficient, Example Ties are broken by averaging the ranks. – Solving by hand • Rank each variable separately. • Calculate sa = 5. 92; sb =5. 50; cov(a, b) = 12. 34 • Thus rs = cov(a, b)/[sasb] =. 379. 56

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

Spearman Rank Correlation Coefficient, Example • Excel Solution (Data Analysis Plus; Xm 17 -08) > 0. 05 58

17. 9 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? 59

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. 60

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

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

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. + ^ y ++ Residual + + + ++ + + + ^ y ^ The spread increases with y + ++ + + 63

Homoscedasticity • When the requirement of a constant variance is not violated we have a condition of homoscedasticity. • Example 18. 2 - continued 64

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. 65

Non Independence of Error Variables Patterns in the appearance of the residuals over time indicates that autocorrelation exists. Residual + ++ + 0 + + ++ + 0 + Time + + Note the runs of positive residuals, Note the oscillating behavior of the replaced by runs of negative residuals around zero. 66

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 67

An outlier An influential observation + + + + +++++ … but, some outliers may be very influential + + + + The outlier causes a shift in the regression line 68

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. 69