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Note: I just need part 6a&b

The case involves predicting the value score of a car based on the price of the car, five-year cost/mile, road-test score, an

The case involves predicting the value score of a car based on the price of the car, five-year cost/mile, road-test score, and predicted reliability. A car with a value score of 1.0 is considered to be "average- value." A car with a value score of 2.0 is considered to be twice as good a value as a car with a value score of 1.0; a car with a value score of 0.5 is considered half as good as average; and so on. The data for 20 family sedans, contained in the textbook data file FamilySedans.xlsx, is shown below TestPredicted Value Price () Cost/Mile Score Reliability Score 0.59 Nissan Altima 2.5 S (4-cyl.) Kia Optima LX (2.4) Subaru Legacy 2.5i Premium 23,970 21,885 23,830 32,360 23,730 22,035 21,800 23,625 0.59 Honda Accord LX-P (4-cyl.) 0.56 0.58 0.56 0.57 0.57 Hyundai Sonata GLS (2.4) Ford Fusion SE (4-cyl.) Chevrolet Malibu LT (4-cyl.) Kia Optima SX (2.0T) Ford Fusion SEL (V6) Nissan Altima 3.5 SR (V6) Hyundai Sonata Limited (2.0T) Honda Accord EX-L (V6) Mazda6 s Grand Touring (V6) Ford Fusion SEL (V6, AWD) Subaru Legacy 3.6R Limited Chevrolet Malibu LTZ (V6) Chrysler 200 Limited (V6) Chevrolet Impala LT (3.6) 1.55 29,050 28,400 30,335 28,090 28,695 30,790 30,055 30,094 28,045 27,825 28,995 1.36
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Answer #1

Input the data in SPSS by defining value score as y, price as x1, cost/mile as x2, road test score as x3 and predicted reliability as x4 in the variable view.

6.a. We perform regression of y on each x1,x2,x3 and x4 individually.

Analyze-Regression-Linear-Dependent(y)- Independent (x1)--ok

Similarly repeat it for x2, x3 and x4.

The results show that adjusted R^2 of y on x1 is 0.290, that on x2, x3 and x4 is 0.486, 0.123 and 0.074 respectively.

Hence the dependent variable y is best explained by independent variable x2 since the adjusted R^2 is maximum for it.

b. The regression line is thus given as, y=2.942-2.312(x2)

Hence for Ford Fusion Hybrid, value score, y=2.942-2.312(0.63)=1.48544.

Standard error of the estimate is 0.14154

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