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Choose the one alternative that best completes the statement or answers the question.
-In a comprehensive road test on all new car models, one variable measured is the time it takes a car to accelerate from 0 to 60 miles per hour. To model acceleration time, a regression analysis is conducted on a random sample of 129 new cars.
TIME60: y=\quad y = Elapsed time (in seconds) from 0mph0 \mathrm { mph } to 60mph60 \mathrm { mph }
MAX: x1=\quad \mathrm { x } _ { 1 } = Maximum speed attained (miles per hour)
Initially, the simple linear model E(y) =β0+β1x1\mathrm { E } ( \mathrm { y } ) = \beta _ { 0 } + \beta _ { 1 \mathrm { x } 1 } was fit to the data. Computer printouts for the analysis are given below:
UNWEIGHTED LEAST SQUARES LINEAR REGRESSION OF TIME60
 PREDICTOR  VARIABLES  COEFFICIENT  STD ERROR  STUDENT’S T  P  CONSTANT 18.71710.6370829.380.0000 MAX 0.083650.0049117.050.0000\begin{array}{l|c|c|c|c}\text { PREDICTOR } & & & & \\\text { VARIABLES } & \text { COEFFICIENT } & \text { STD ERROR } & \text { STUDENT'S T } & \text { P } \\\hline \text { CONSTANT } & 18.7171 & 0.63708 & 29.38 & 0.0000 \\\text { MAX } & -0.08365 & 0.00491 & -17.05 & 0.0000\end{array}

 R-SQUARED 0.6960 RESID. MEAN SQUARE (MSE)  1.28695 ADJUSTED R-SQUARED 0.6937 STANDARD DEVIATION 1.13444\begin{array} { l l l l } \text { R-SQUARED } & 0.6960 & \text { RESID. MEAN SQUARE (MSE) } & 1.28695 \\ \text { ADJUSTED R-SQUARED } & 0.6937 & \text { STANDARD DEVIATION } & 1.13444 \end{array}


 SOURCE  DF  SS  MS  F  P  REGRESSION 1374.285374.285290.830.0000 RESIDUAL 127163.4431.28695 TOTAL 128537.728\begin{array}{l|r|c|c|c|c}\text { SOURCE } & \text { DF } & \text { SS } & \text { MS } & \text { F } & \text { P } \\\hline \text { REGRESSION } & 1 & 374.285 & 374.285 & 290.83 & 0.0000 \\\text { RESIDUAL } & 127 & 163.443 & 1.28695 & & \\\text { TOTAL } & 128 & 537.728 & & &\end{array}

CASES INCLUDED 129 MISSING CASES 0
Approximately what percentage of the sample variation in acceleration time can be explained by the simple linear model?


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