Need help
Suppose the Sherwin-williams Company is interested in developing regression model with paint sales (Y) as the dependent variable and selling price (P) as the independent variable. Complete the following worksheet and then use it to determine the estimated regression line. Selling Price Sales Sales Region ($/Gallon) (x 1000 Gal) TIM 15 160 2,400 225 25.600 13.5 220 2,970 182.25 48/400 16.5 140 2,310 272.25 19.600 14.5 190 2,755 210.25 36.100 17 130 2,210 289 16,900 16 160 2,560 256 25,600 13 190 2,470 169 36.100 18 150 2,700 324 22.500 12 210 2,520 144 44,100 10 15.5 190 2,945 240.25 36,100 Total 151 1,740 25,840 2,312 Regression Parameters Estimations Slope (8) Intercept (a) In words, for a dollar increase in the selling price, the expected sales will by gallons in a given sales region.What is the standard error of the estimate (s.)? O 14.309 O 15.674 O 17.086 What is the estimate of the standard deviation of the estimated slope (s;)? O 3.025 O 2.533 O 2.775 Can you reject the hypothesis (at the 0.05 level of significance) that there is no relationship (i.e., 8 = 0) between the variables? (Hir (0.025,8 = 2.306) Yes O No Complete the following worksheet and then use it to calculate the coefficient of determination.(u - 7) 15 160 175.361 1.452 196.000 13.5 220 195.768 473.846 2,116.000 16.5 140 154.953 362.704 1,156.000 14.5 190 142.163 66.635 256,00 0 17 130 146.150 668.222 1,936.000 6 16 160 161.755 149.940 196.000 11 190 202.571 816.302 256.000 18 150 134.545 1,556.697 576.000 9 12 210 216.176 1,778.815 1,296.000 10 15.5 190 164.558 29.615 256,000 Total 5,904.712 7 8,240.000 The coefficient of determination (r?) is According to the regression model, which of the following is the best estimate together with the 95 percent prediction interval of paint sales [in thousands of gallons) in a sales region where the selling price is $12.507 O 209.315 + 2(3.025) Q 389.410 + 2[3.025) 389.440 4 2(17.086] O 209.315 4 2(17.086) What is the price elasticity of demand at a selling price of $12.507 O -0.06 O -0.81 0 -0.54 0 -0.05