Question

A Ministry of Transportation investigation on driving speed and gasoline consumption for midsize vehicles (in miles per gallon), resulted in the following data:

Speed (Miles per Hour) 55       50      25      60      `25    30     55      40      50      30

Miles per Gallon             25      26      35      21       32    30     23      25      25      28

1. Compute the covariance. ________________
2. Compute the coefficient of correlation. __________________
3. Based on (a) and (b), what conclusion can you reach about the relationship between driving speed and fuel consumption?

Answer 1

a) Covariance { {55,50,25,60,25,30,55,40,50,30} , {25,26,35,21,32,30,23,25,25,28} }

Let,

Speed (Miles per Hour) = x

Miles per Gallon = y

x	y	( x- x̄ ) = ( x – 42)	(y - ȳ )= (y – 27)	( x- x̄ ) (y - ȳ )	( x- x̄ )²	(y - ȳ )²
55	25	13	-2	-26	169	4
50	26	8	-1	-8	64	1
25	35	-17	8	-136	289	64
60	21	18	-6	-108	324	36
25	32	-17	5	-85	289	25
30	30	-12	3	-36	144	9
55	23	13	-4	-52	169	16
40	25	-2	-2	4	4	4
50	25	8	-2	-16	64	4
30	28	-12	1	-12	144	1
∑ x = 420	∑y = 270	∑(x- x̄ ) = 0	∑( y - ȳ) = 0	∑( x- x̄ ) (y - ȳ ) = - 475	∑ (x- x̄ )² = 1660	∑( y - ȳ)² = 164

Mean x̄ = ∑ x / n

               = 420 / 10

         x̄ = 42

Mean ȳ = ∑y / n

               = 270 / 10

         ȳ = 27

Population Cov (x, y) = ∑( x- x̄ ) (y - ȳ ) / n

                                      = - 475 / 10

                                      = -47.5

Population Cov (x, y) = -47.5

Sample Cov (x, y) = ∑( x- x̄ ) (y - ȳ ) / ( n – 1)

                                 = - 475 / 9

                                 = - 52.7778

Sample Cov (x, y) = - 52.7778

b. Correlation Coefficient r:

r = ∑( x- x̄ ) (y - ȳ ) / ( √ ∑ (x- x̄ )²) ( √∑( y - ȳ)² )

r = ( - 475 ) / (√1660)(√164)

r = - 0.9104

c. Since the value of r = - 0.9104, there is A strong negative linear relationship between Speed (Miles per Hour) and Miles per Gallon