mid1practice2soln - Midterm 1 Practice Problems 2 With...

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Midterm 1 Practice Problems 2 With Solutions (1) Regress a Wreck A statistician is trying to learn what factors affect the price of a used car. Her Y variable is the price of the car. She is considering several possible predictor variables. They are X 1 , the original value of the car, X 2 , the mileage on the car, X 3 , the number of repairs that have been done on the car, and X 4 , the number of seat belts in the car. (a) For each of the four possible predictor variables the statistician has obtained the correlation of Y and X, and the covariance of Y and X. Cor ( Y, X 1 ) = . 795 Cov ( Y, X 1 ) = 3 , 688 , 147 Cor ( Y, X 2 ) = . 789 Cov ( Y, X 2 ) = 149 . 155 Cor ( Y, X 3 ) = . 539 Cov ( Y, X 3 ) = 1186 . 4 Cor ( Y, X 4 ) = . 004 Cov ( Y, X 4 ) = 7 . 6 Say what a plot of Y vs X should look like in each case. Solution: Variable X 1 (original value) has a strong positive correlation with Y so the plot should show a clear upward trend. Variable X 4 (number of seat-belts) has a correlation with Y that is close to 0 and so the plot should be nearly flat– i.e. not showing a clear relationship between X and Y. Variable X 2 (mileage) has the stronger of the two negative correlations (closer to -1) so the plot should show the stronger of the two downward trends. The points would less spread out about the line than in the plot for X 3 , number of repairs. (b) Rank the variables X 1 , X 2 , X 3 , X 4 in terms of how good a job you expect them to do of pre- dicting Y based on the values given in part (a) (NOT on your common sense opinion!) Order them from best predictor to worst predictor and briefly explain your reasoning. Solution: The strength of the relationship is determined by the correlation. (Note: The covariance is not good for comparing strengths of relationships because different units can affect what is a “big” covariance!) The sign is irrelevant to the strength of the relationship–it only determines the direction of the relationship. Here original price, X 1 , has the highest correlation in absolute value at .795, followed by mileage, X 2 , at -.789, repairs, X 3 , at -.539, and seat-belts, X 4 , at -.004. The stronger the relationship, the better a predictor the variable will be. Therefore original value will be the best predictor followed by mileage, number of repairs, and number of seat-belts. (c) To simplify matters the statistician has fit two regressions, one of price (Y) on mileage ( X 2 ) and one of price (Y) on the number of repairs the car has had ( X 3 ). Printouts for these regressions are given on the following page. Give three numbers from the printouts that tell you which predictor, X 2 or X 3 is doing a better job and briefly explain why that number tells you it is doing a better job. Do the numbers confirm your prediction from part (b)? (Note: Comparing R-squared from print- out 1 to R-squared from printout 2 counts as one number. You should give three pairs/comparisons.) Solution: Mileage, X 2 is the better predictor. We can see this using almost any number from the printout that follows. For instance, R 2 = 63 . 7% for mileage and only 29.1% for repairs–and 1
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a higher R 2 means we have done a better job explaining variability. We have Root MSE = 831.7
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