Question

Question 1 A developer who specialises in summer cottage properties is considering purchasing a large tract of land adjoining a lake. The current owner of the tract has already subdivided the land into separate building lots and has prepared the lots by removing some of the trees. The developer wants to forecast the value of each lot. From previous experience, she knows that the most important factors affecting the price of the lot are size, number of mature trees, and distance to the lake. From a nearby area, she gathers the relevant data for some recently sold lots. These data are stored in a worksheet file. (Column 1 = price in thousands of dollars; column 2 = lot size in thousands of square feet; column 3 = number of mature trees; column 4 = distance to the lake in feet.) A multiple regression analysis was performed. The Excel printout is shown below. SUMMARY OUTPUT Regression Statistics Multiple R a R Square 0.2425 Adjusted R Square 0.2019 Standard Error 40.24 Observations ANOVA F SS 29030 90694 MS 9677 Significance F 0.0013 Regression Residual Total 5.97 Coefficients Stat P-value Standard Error Intercept Lot size Trees Distance 51.391 0.700 0.679 -0.378 i j 0.195 2.19 1.25 2.96 k 0.0331 0.2156 0.0045 0.0577 (a) Find the numbers a - k. (14 marks) (b) What is the value of the standard error of estimate? Interpret its value. (8 marks) (c) What is the value of the coefficient of multiple determination? What does this statistic tell you? (8 marks) (d) Test the overall validity of the model at 0.01 level of significance. (8 marks) (e) Test to determine whether each of the independent variables is linearly related to the price of the lot at 0.05 level of significance. (12 marks)

Answer 1

SSReg = mSReg - 29.30 – 9677. af 9677 e - 3 • we know for degree of freedom of no of observation - no of regressor 1 d = 56.1 erem Besidual = Sb Residual - 90694, 56. df e = 11619.536. o odegree of freedom for Total sun of square is no of ab. number of observation-1 - 6o- If = 59 756. = ssRegression & SsResidual J- 29030 + 90694 18 = 119724 o trotstat = Confliu'ent". standard error corffiu ent. t- Stat Standarderror . th= 51.901 – 23.4.04 05 2

2. = 0.2294 2.96 = -0.378.-1.938 0.195 la) (6) a = 0.4924. =-1.938 9- K = 0.4924+ 1.938 = 2:43.04 Standerror linterriept) = 23. 46621 Standard error (let size) = 0.56. Standard error (trees) = 0.2294 Deietania taszere Standard error (distance) = 0.195 her Nalue of standard error oltre avoragodio tance that the agevored Que lopsegree of conefficient measures how preosely the model estimates the comefficient's unknown value. (e) rame of coresfiyent of determination is Roy R Square = 0.2425 ine. 24025% of the model variability loco variability in the response. is explained by model!