9) A set of data is available on 392 vehicles. The goal is to find a model to predict the variable logmpg = the log of miles per gallon. The possible independent variables are: cylinders = Number of cylinders between 4 and 8, displacement = Engine displacement (cu. inches), horsepower = Engine horsepower, weight = Vehicle weight (Ibs.), acceleration = Time to accelerate from 0 to 60 mph (sec.), and year = Model year (modulo 100). An analyst fits two models. The output for these two models is provided below, use that output for this problem. Model 1: logmpg = Bo + Bicylinders + 82displacement + 83horsepower + BAweight + Byaccele: The regression equation is logmpg = 1.83 - 0.0233 cylinders + 0.000287 displacement - 0.00100 horsepower - 0.000264 weight - 0.00121 acceleration + 0.0296 year Predictor Coef SE Coef T P Constant 1.8275 0.1685 10.85 0.000 Cylinders -0.02328 0.01175 -1.98 0.048 Displacement 0.0002874 0.0002603 1.10 0.270 Horsepower -0.0010020 0.0004894 -2.05 0.041 Weight -0.00026427 0.00002370 -11.15 0.000 Acceleration -0.001214 0.003609 -0.34 0.737 Year 0.029649 0.001861 15.93 0.000 S = 0.121512 R-Sq = 87.4% R-Sq(adj) = 87.2% Analysis of Variance Source DF SS MS F P Regression 6 39.5254 6.5876 446.16 0.000 Residual Error 385 5.6845 0.0148 Total 391 45.2100 Model 2: logmpg = Bo + Bicylinders + Byhorsepower + 83weight + Bayear The regression equation is logmpg = 1.77 - 0.0139 cylinders - 0.000724 horsepower - 0.000258 weight + 0.0296 year Predictor Coef SE Coef P Constant 1.7709 0.1486 11.92 0.000 Cylinders -0.013908 0.008589 -1.62 0.106 Horsepower -0.0007236 0.0003469 -2.09 0.038 Weight -0.00025751 0.00001858 -13.86 0.000 Year 0.029559 0.001850 15.98 0.000 S = 0.121422 R-Sq = 87.4% R-Sq(adj) = 87.2% Analysis of Variance Source DF SS MS F P Regression 4 39.5043 9.8761 669.86 0.000 Residual Error 387 5.7057 0.0147 Total 391 45.2100 a) For model 1, test the null hypothesis that all the coefficients for all the variables are all equal to 0 in other words that B1 = . . . = 86 = 0