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Simple Linear Regression 1. review of least squares procedure 2. inference for least squares lines 1 Introduction • We will examine the relationship between quantitative variables x and y 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. 2 The Model The model has a deterministic and a probabilistic components House Cost Most lots sell for $25,000 House size 3 The Model However, house cost vary even among same size houses! Since cost behave unpredictably, House Cost we add a random component. Most lots sell for $25,000 House size 4 The Model • The first order linear model y b0 b1x e y = dependent variable x = independent variable b0 = y-intercept b1 = slope of the line e = error variable y b0 and b1 are unknown population parameters, therefore are estimated from the data. Rise b0 b1 = Rise/Run Run x 5 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 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 - 1)2 + (4 - 2)2 +(1.5 - 3)2 + (3.2 - 4)2 = 6.89 Sum of squared differences = (2 -2.5)2 + (4 - 2.5)2 + (1.5 - 2.5)2 + (3.2 - 2.5)2 = 3.99 4 3 2.5 2 Let us compare two lines The second line is horizontal (2,4) w w (4,3.2) (1,2) w 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 Alternate formula for the slope b1 To calculate the estimates of the slope and intercept of the least squares line , use the formulas: b1 SS xy SS xx b0 y b1 x SS xy x y xy i i SS xx x 2 i i i n The regression equation that estimates the equation of the first order linear model is: i n x sy b1 r sx 2 (n 1) sx2 ŷ b0 b1 x 9 The Simple Linear Regression Line • Example: – 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. Car Odometer Price 1 37388 14636 2 44758 14122 3 45833 14016 4 30862 15590 5 31705 15568 6 34010 14718 . . . Independent Dependent . . . variable x variable y . . . 10 The Simple Linear Regression Line • Solution – Solving by hand: Calculate a number of statistics 2 x 36,009.45; y 14,822.823; SS xx xi SS xy 2 x i n 43,528, 690 xy (x y ) i i i i n 2, 712,511 where n = 100. b1 SS xy (n 1) sx2 2, 712,511 .06232 43,528, 690 b0 y b1 x 14,822.82 (.06232)(36, 009.45) 17, 067 ŷ b0 b1x 17,067 .0623 x 11 The Simple Linear Regression Line • Solution – continued – Using the computer 1. Scatterplot 2. Trend function 3. Tools > Data Analysis > Regression 12 The Simple Linear Regression Line SUMMARY OUTPUT Regression Statistics Multiple R 0.8063 R Square 0.6501 Adjusted R Square 0.6466 Standard Error 303.1 Observations 100 yˆ 17,067 .0623 x ANOVA df Regression Residual Total SS MS 1 16734111 16734111 98 9005450 91892 99 25739561 CoefficientsStandard Error t Stat Intercept 17067 169 100.97 Odometer -0.0623 0.0046 -13.49 F Significance F 182.11 0.0000 P-value 0.0000 0.0000 13 Odometer Line Fit Plot 16000 Price 17067 Interpreting the Linear Regression Equation 0 15000 14000 No data 13000 Odometer yˆ 17,067 .0623 x The intercept is b0 = $17067. Do not interpret the intercept as the “Price of cars that have not been driven” This is the slope of the line. For each additional mile on the odometer, the price decreases by an average of $0.0623 14 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. 15 The Normality of e E(y|x3) The standard deviation remains constant, m3 b0 + b1x3 E(y|x2) b0 + b1x2 m2 but the mean value changes with x b0 + b1x1 E(y|x1) m1 From the first three assumptions we have: x1 y is normally distributed with mean E(y) = b0 + b1x, and a constant standard deviation se x2 x3 16 Assessing the Model • The least squares method will produces a regression line whether or not there is a 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 n SSE ( y i ŷ i ) 2 . i 1 – A shortcut formula SSE yi2 b0 yi b1 xi yi 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 S tan dard Error of Estimate se SSE n2 19 Standard Error of Estimate, Example • Example: – Calculate the standard error of estimate for the previous example and describe what it tells you about the model fit. • Solution SSE 9, 005, 450 SSE 9, 005, 450 se 303.13 n2 98It is hard to assess the model based on se even when compared with the mean value of y. s e 303.1 y 14,823 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 q 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 b1 from b1 by testing H0: b1 = 0 H1: b1 = 0 (or < 0,or > 0) – The test statistic is b1 b1 t s b1 The standard error of b1. where sb1 se SS xx – If the error variable is normally distributed, the statistic is Student t distribution with d.f. = n-2. 22 Testing the Slope, Example • 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 the previous example . Use a = 5%. 23 Testing the Slope, Example • Solving by hand – To compute “t” we need the values of b1 and sb1. b1 .0623 sb1 t se (n 1) s x2 303 .1 .00462 (99 )( 43,528 ,690 ) b1 b1 .0623 0 13 .49 .00462 sb1 – 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 • Using the computer Price Odometer SUMMARY OUTPUT 14636 37388 14122 44758 Regression Statistics 14016 45833 Multiple R 0.8063 15590 30862 R Square 0.6501 There is overwhelming evidence to infer 15568 31705 Adjusted R Square 0.6466 14718 34010 Standard Error 303.1 that the odometer reading affects the 14470 45854 Observations 100 auction selling price. 15690 19057 15072 40149 ANOVA 14802 40237 df SS MS F Significance F 15190 32359 Regression 1 16734111 16734111 182.11 0.0000 14660 43533 Residual 98 9005450 91892 15612 32744 Total 99 25739561 15610 34470 14634 37720 Coefficients Standard Error t Stat P-value 14632 41350 Intercept 17067 169 100.97 0.0000 15740 24469 Odometer -0.0623 0.0046 -13.49 0.0000 25 Coefficient of determination – To measure the strength of the linear relationship we use the coefficient of determination. ( x x )( y y i i 2 R 2 2 sx s y 2 SSE or R 1 2 ( yi y ) 2 Note that the coefficient of determination is r2 26 Coefficient of determination • To understand the significance of this coefficient note: The regression model Overall variability in y The error 27 Coefficient of determination y2 Two data points (x1,y1) and (x2,y2) of a certain sample are shown. y y1 x1 Total variation in y = ( y1 y ) 2 ( y 2 y ) 2 Variation in y = SSR + SSE x2 Variation explained by the + Unexplained variation (error) regression line (ŷ1 y) 2 (ŷ 2 y) 2 (y1 ŷ1 ) 2 (y 2 ŷ 2 ) 2 28 Coefficient of determination • R2 measures the proportion of the variation in y that is explained by the variation in x. 2 R 1 SSE (y i y) (y y) ( y i y ) 2 SSE 2 i 2 SSR (y i y) 2 • 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. 29 Coefficient of determination, Example • Example – Find the coefficient of determination for the used car price –odometer example.what does this statistic tell you about the model? • Solution – Solving by hand; ( xi x )( yi y R 2 2 sx s y 2 2 [ 2,712,511]2 (43,528,688)(259,996) .6501 30 Coefficient of determination – Using the computer From the regression output we have SUMMARY OUTPUT Regression Statistics Multiple R 0.8063 R Square 0.6501 Adjusted R Square 0.6466 Standard Error 303.1 Observations 100 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. ANOVA df Regression Residual Total Intercept Odometer SS 16734111 9005450 25739561 MS 16734111 91892 CoefficientsStandard Error 17067 169 -0.0623 0.0046 t Stat 100.97 -13.49 1 98 99 F Significance F 182.11 0.0000 P-value 0.0000 0.0000 31 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 32 Point Prediction • Example – Predict the selling price of a three-year-old Taurus with 40,000 miles on the odometer. A point prediction ŷ 17067 .0623 x 17067 .0623(40,000) 14,575 – It is predicted that a 40,000 miles car would sell for $14,575. – How close is this prediction to the real price? 33 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 yˆ ta 2 se 1 ( xg x ) 2 1 n ( xi x )2 – The confidence interval yˆ ta 2 se ( xg x ) 2 1 n ( xi x ) 2 34 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 35 Interval Estimates, Example • Solution – A prediction interval provides the price estimate for a single car: yˆ ta 2 se t.025,98 ( xg x ) 2 1 1 n ( xi x )2 Approximately 1 (40,000 36,009) 2 [17,067 .0623(40000)] 1.984(303.1) 1 14,575 605 100 4,309,340,310 36 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%) = ŷ t a 2 s e 1 n ( x g x)2 ( x i x)2 1 (40, 000 36, 009) 2 [17, 067 .0623(40000)] 1.984(303.1) 14,575 70 100 4,309,340,310 37 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. ŷ b0 b1x g ŷ t a 2 s e 2 1 ( x g x) n (n 1)s 2x x 38 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. ŷ b0 b1x g ŷ( x g x 1) ŷ( x g x 1) ŷ t a 2 s e ŷ t a 2 s e 2 1 ( x g x) n (n 1)s 2x 1 12 n (n 1)s 2x x 1 x 1 x ( x 1) x 1 ( x 1) x 1 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. ŷ b0 b1x g x 2 x x2 ( x 2) x 2 ( x 2) x 2 ŷ t a 2 s e 2 1 ( x g x) n (n 1)s 2x ŷ t a 2 s e 1 12 n (n 1)s 2x ŷ t a 2 s e 1 22 n (n 1)s 2x 40 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? 41 Residual Analysis • Examining the residuals (or standardized residuals), help detect violations of the required conditions. • Example – 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. 42 Residual Analysis ObservationPredicted Price Residuals Standard Residuals 1 14736.91 -100.91 -0.33 2 14277.65 -155.65 -0.52 3 14210.66 -194.66 -0.65 4 15143.59 446.41 1.48 5 15091.05 476.95 1.58 For each residual we calculate the standard deviation as follows: A Partial list of Standard residuals s ri s e 1 hi where Standardized residual ‘i’ = Residual ‘i’ 1 ( x i x)2 hi Standard deviation 2 n (n 1)s x 43 Residual Analysis Standardized residuals 40 30 20 10 0 -2 -1 0 1 2 More It seems the residual are normally distributed with mean zero 44 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 + + + + + + + + + + + + ++ + + + + + + + + + + + + + + + The spread increases with ^y y^ ++ + ++ ++ ++ + + ++ + + 45 Homoscedasticity • When the requirement of a constant variance is not violated we have a condition of homoscedasticity. • Example - continued Residuals 1000 500 0 13500 -500 14000 14500 15000 15500 16000 -1000 Predicted Price 46 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. 47 Non Independence of Error Variables Patterns in the appearance of the residuals over time indicates that autocorrelation exists. Residual Residual + ++ + 0 + + + + + + + + + + ++ + + + Time Note the runs of positive residuals, replaced by runs of negative residuals + + + 0 + + + + Time + + Note the oscillating behavior of the residuals around zero. 48 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 49 An outlier + + + + + + + + + An influential observation +++++++++++ … but, some outliers may be very influential + + + + + + + The outlier causes a shift in the regression line 50 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. 51