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Simple Linear Regression Least squares procedure 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 (Example) 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 . . x variable . y variable . . . 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 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 12 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 13 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. 14 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 15 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. 16 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 17 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 SSE se n2 18 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 98 It 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 19 “p-values” and Significance Levels Each independent variable has “p-value” or significance level. The p-value tells how likely it is that the coefficient for that independent variable emerged by chance and does not describe a real relationship. A p-value of .05 means that there is a 5% chance that the relationship emerged randomly and a 95% chance that the relationship is real. It is generally accepted practice to consider variables with a p-value of less than .1 as significant, though the only basis for this cutoff is convention 20 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 21 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? 22 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 t 2 se 1 ( xg x ) 2 1 n ( xi x )2 – The confidence interval yˆ t 2 se ( xg x ) 2 1 n ( xi x ) 2 23 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 24 Interval Estimates, Example • Solution – A prediction interval provides the price estimate for a single car: yˆ t 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 25 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 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 26 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. 27 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 28 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^ ++ + ++ ++ ++ + + ++ + + 29 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 30 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. 31 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. 32 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 33 An outlier + + + + + + + + + An influential observation +++++++++++ … but, some outliers may be very influential + + + + + + + The outlier causes a shift in the regression line 34 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. 35