Question

Problem 8 (18 points) An individual wanted to determine the relation that might exist between speed and miles per gallon of an automobile. Let Xbe the average speed of a car on the highway measured in miles per hour andlet Y represent the miles per gallon of the automobile The following data is collected 50 28 60 62 20 65 26 25 20 17 d Predict the miles per gallon of a car traveling 63 miles per hour e Predict the average speed of a car whose fuel mileage is 23 miles per gallon f Find r squared. What does this statistic tell us about between average speed and miles per gallon?

Answer 1

We now draw up the following table for using method of least squares

Sr. No.	x	x2	y	xy
1	50	2500	28	1400
2	55	3025	26	1430
3	55	3025	25	1375
4	60	3600	22	1320
5	60	3600	20	1200
6	62	3844	20	1240
7	65	4225	17	1105
8	65	4225	15	975
N = 8	∑x=472	∑x2=28044	∑y=173	∑xy=10045

∑y = aN+b∑x 173 = 8a + 472b                ……(i)

∑xy = a∑x + b ∑x2 10045 =472a + 28044b   …(ii)

Solving (i) and (ii) we get,

a = 70.39                              b = -0.826

now the line of regression of x on y is given as,

y = a + bx             i.e. y = 70.39 – 0.826x

a. To find y at x = 63

Y = 70.39 – 0.826(63) = 18.352 mils/gal

b. To find x at y = 23

23 = 70.39 – 0.826x

x = 57.372 miles /hr​​​​​​

c. To find R²

x	y	(y-y*)2	y'	(y-y’)2
50	28	40.64	29.09	55.73
55	26	19.14	24.96	11.12
55	25	11.39	24.96	11.12
60	22	0.14	20.83	0.63
60	20	2.64	20.83	0.63
62	20	2.64	19.18	5.99
65	17	21.39	16.70	24.26
65	15	43.89	16.70	24.26
Mean (y*)	21.625	∑ = 141.88		∑ = 133.73

R2133.73 = 141.88 = 0.943

The value of R²indicates that the regression line matches very closely the actual observed values. i.e it is 94.3 % accurate