Question

Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of pounds), and let y be the miles per gallon (mpg).

x	30	43	31	47	23	40	34	52
y	30	21	22	13	29	17	21	14

Complete parts (a) through (e), given Σx = 300, Σy = 167, Σx² = 11,908, Σy² = 3761, Σxy = 5885, and

r ≈ −0.888.

(a) Draw a scatter diagram displaying the data.

(b) Verify the given sums Σx, Σy, Σx², Σy², Σxy, and the value of the sample correlation coefficient r. (Round your value for r to three decimal places.)

Σx =
Σy =
Σx2 =
Σy2 =
Σxy =
r =

(c) Find x, and y. Then find the equation of the least-squares line = a + bx. (Round your answers for x and y to two decimal places. Round your answers for a and b to three decimal places.)

x	=
y	=
	=	+  x

(d) Graph the least-squares line. Be sure to plot the point (x, y) as a point on the line.

--

(e) Find the value of the coefficient of determination r². What percentage of the variation in y can be explained by the corresponding variation in x and the least-squares line? What percentage is unexplained? (Round your answer for r² to three decimal places. Round your answers for the percentages to one decimal place.)

r2 =
explained	%
unexplained	%

(f) Suppose a car weighs x = 42 (hundred pounds). What does the least-squares line forecast for y = miles per gallon? (Round your answer to two decimal places.)
mpg

Answer 1

Solution given that __x 30 +3 3। ५+ २3 40 34 5२ Y 300 २३ 13 २१ ।। २) 14 Let x be the weight of the car y be the miles per gallon (mpg) 6 x x x + y (K- X) (y ) १00 १०3 ६२ 43 31 47 23 ५० 34 १00 - १०० 18५१ ५५। १61 ५५ २२0११ ७१५ ।600 २६१ डक ५५। १ ०.२ a - 3 २ 13 २१ । । -25 5.5 -6. 5 १.5 - 14.5 2.5 - 3.5 662 680 ५ 2,7०५१ १२६५.८ -6.६ Ex-300 y:67 21-॥१०8 (x-7001-4) - 3761 Exy : 5885 - 0.6 Ey - 300 By - 167 871 - 11,908 १५- 3,761 exy : 5,885

Note O Sum of x = 300 meam value of x = { = 300 8 = 37.5 Sum of y = 167 mean value of y = 167 = 20.87

(-) (-1) ५-१)- 84.64 56 25 -69 30.25 ५२.२४ ।.44 -7.8 6084 -24. १०.25 61.24 - 08.1 14.44 -१.5 210.25 6:२९ 12.25 20 25 0.04 ५6.24 (५-१): - 0.2 -66.9 -377.5 १-) - 658 2 74.92- Calculate the value of the sample correlation coefficient r ४. ६ (x-)-1) V(x-7)- CY-50- -377.5 VC6se) (07५.१३) - 322.5 V180897-4

-377-5 4283203 - 0.8876 obeys to Rigy corresponunsy = -0.888 im 8 -0.888 Sum of x = 300 ; cmean) - 300% = 37.5 Sum of y = 167 ; Ý (mean) = 1677€ -20.67 Sum of squales (x-x)² = 658 Cy-90+= 27-4a Press the buttons to have light and MUSIC.FI Sum of ploducts. CX-5) [y-ġ) = -377.5 regression Equation ŷ = a +bx Where b= sum of products Sum of squales - 377.5 = -0.57 658

as mean of y-cbcmain of $)] 20.97 - [(-0.53)(37.5)] = 20.87- [(-21-375)] - 20.87 +21.375 to play correspurruary = 42.245 ŷ = 42.245 + (-0.57)X = 42-245-0.57% Dracs the buttons to have light and music.. 10 20 30 40 50

© Y-.0.888 82=40:888) = 0.7885 Hence 78.851. of the variation iny can be explained by the corresponding variation in 80 81.15 is unexplained. 82= 0.1885 Explained = 78-85./. unexplained. 21.15.1 t) 3 9. 945 + (-01-24 = 42.245 -0.57(%) where x=42 – 42.245 - (0:57)(42) -42.245 - 23.94 - 17.305