2 IntroductionIn 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 HouseCostBuilding a house costs about$75 per square foot.House cost = (Size)Most lots sellfor $25,000House size
4 The Model However, house cost vary even among same size houses! Since cost behave unpredictably, we add a random component.HouseCostMost lots sellfor $25,000House cost = (Size)+ eHouse size
5 The Model The first order linear model y = dependent variable x = independent variableb0 = y-interceptb1 = slope of the linee = error variableb0 and b1 are unknown population parameters, therefore are estimatedfrom the data.yRiseb1 = Rise/RunRunb0x
6 Estimating the Coefficients The estimates are determined bydrawing a sample from the population of interest,calculating sample statistics.producing a straight line that cuts into the data.ywQuestion: What should be considered a good line?wwwww w w www ww wwx
7 The Least Squares (Regression) Line A good line is one that minimizes the sum of squared differences between thepoints and the line.
8 The Least Squares (Regression) Line Sum of squared differences =(2 - 1)2 +(4 - 2)2 +( )2 +( )2 = 6.89Sum of squared differences =(2 -2.5)2 +( )2 +( )2 +( )2 = 3.99Let us compare two lines4(2,4)The second line is horizontalww(4,3.2)32.52w(1,2)(3,1.5)wThe smaller the sum ofsquared differencesthe better the fit of theline to the data.1234
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 estimatesthe equation of the first order linear modelis:
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 xDependent variable y
11 The Simple Linear Regression Line SolutionSolving by hand: Calculate a number of statisticswhere n = 100.
12 The Simple Linear Regression Line Solution – continuedUsing the computer (Xm17-02)Tools > Data Analysis > Regression > [Shade the y range and the x range] > OK
14 Interpreting the Linear Regression -Equation 17067No dataThe 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 $0.0623Do 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 eE(y|x3)The standard deviation remains constant,m3b0 + b1x3E(y|x2)b0 + b1x2m2E(y|x1)but the mean value changes with xm1b0 + b1x1From the first three assumptions we have:y is normally distributed with meanE(y) = b0 + b1x, and a constant standard deviation sex1x2x3
17 Assessing the ModelThe 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 byA 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?SolutionCalculated beforeIt is hard to assess the model basedon se even when compared with themean value of y.
21 The slope is not equal to zero Testing the slopeWhen no linear relationship exists between two variables, the regression line should be horizontal.qqqqqqqqqqqqLinear relationship.Linear relationship.Linear relationship.Linear relationship.No linear relationship.Different inputs (x) yieldthe same output (y).Different inputs (x) yielddifferent outputs (y).The slope is not equal to zeroThe slope is equal to zero
22 Testing the Slope We can draw inference about b1 from b1 by testing H0: b1 = 0H1: b1 = 0 (or < 0,or > 0)The test statistic isIf the error variable is normally distributed, the statistic is Student t distribution with d.f. = n-2.whereThe 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 handTo 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 Xm17-02Using the computerThere is overwhelming evidence to inferthat the odometer reading affects theauction 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 modelExplained in part byOverall variability in yRemains, in part, unexplainedThe error
28 Coefficient of determination y2Two data points (x1,y1) and (x2,y2)of a certain sample are shown.yVariation in y = SSR + SSEy1x1x2Total variation in y =Variation explained by theregression 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?SolutionSolving by hand;
31 Coefficient of determination Using the computer From the regression output we have65% of the variation in the auctionselling price is explained by thevariation in odometer reading. Therest (35%) remains unexplained bythis 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 +eRate of return on a particular stockRate of return on some major stock indexThe beta coefficient measures how sensitive the stock’s rateof 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’smarket 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’sreturn are explained by the variation in theTSE’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 usePoint prediction, andInterval 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 predictionIt is predicted that a 40,000 miles car would sell for $14,575.How close is this prediction to the real price?
36 Interval EstimatesTwo 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 intervalThe confidence interval
37 Interval Estimates, Example Example continuedProvide 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 carAn estimate for the average price per car
38 Interval Estimates, Example SolutionA prediction interval provides the price estimate for a single car:t.025,98Approximately
39 Interval Estimates, Example Solution – continuedA 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 YX and Y must be observationalX and Y are bivariate normally distributedXY
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 0The test statistic is:The statistic is Student t distributed with d.f. = n - 2, provided the variablesare 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 FundsA 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 SolutionProblem 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 – continuedThe 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=.491The value of the t statistic isConclusion: There is sufficientevidence at a = 5% to infer thatthere are linear relationshipbetween the two variables.
50 Testing the Coefficient of Correlation, Example Excel solution (Index)