Question

Car   Weight (pounds), x   Miles per Gallon, y
1   3,765   19
2   3,984   18
3   3,530   20
4   3,175   22
5   2,580   26
6   3,730   18
7   2,605   25
8   3,772   18
9   3,310   20
10   2,991   24
11   2,752   25

The accompanying data represent the weights of various domestic cars and their gas mileages in the city. The linear correlation coefficient between the weight of a car and its miles per gallon in the city is r= -0.97 The least-squares regression line treating weight as the explanatory variable and miles per gallon as the response variable is y = -0.0061x +41.3297. Complete parts (a) through (c) below. B! Click the icon to view the data table. O B. A. A Residual OC. A Residual OD. A Residual A Residual o 27 2 287 07-.-6 07•.-.- o+---- -- 227__'o 13 16- -24 2500 3250 4000 Weight (pounds) 2500 3250 4000 Weight (pounds) 2500 3250 4000 Weight (pounds) 2500 3250 4000 Weight (pounds) (c) Interpret the coefficient of determination and comment on the adequacy of the linear model. by the linear model. The least-squares regression model appears to be based on the residual plot. % of the variance in (Round to one decimal place as needed.)

Answer 1

- 19 Given the regression equation is y=-0.oogiat 41.3297. The residual is obtained by: (yirŷ from data! weight (2) MPG(Y) yio 9 3765 -0.0061* 3765+41.3297= 18.3632 0.6368 3984 18 17.0273 0.9727 350 20 19.7967 0-2033 3175 22 21.9622 0.0378 2580 25.5917 0.4083 3730 18.5767 -0.57679-4233 2 605 25.4392 -0.4392 3772 18.3205 -0.3205 3310 21.1387 -1.1387 2991 24 23.0846 0.9154 2752 24.5425 0.4575 20. ... Correct graph of residual is in B. c) The correlation coefficient is r=-0.97 => 82= 0.940g #coefficient of Peter ination It indicates that i 94.09% of the variance in miles per Gallon is explain by the linear model. The least square regression model is appears to be normally distributed based on the residual plot. since the residual points are random, they does not exibits any fined pattern.