Question

You are the foreman of the Bar-S cattle ranch in Colorado. A neighboring ranch has calves for sale, and you are going to buy some calves to add to the Bar-S herd. How much should a healthy calf weigh? Let x be the age of the calf (in weeks), and let y be the weight of the calf (in kilograms).

x	2	5	10	16	26	36
y	39	46	79	100	150	200

Complete parts (a) through (e), given Σx = 95, Σy = 614, Σx² = 2357, Σy² = 82,378, Σxy = 13,798, and r ≈ 0.998.

(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

(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) The calves you want to buy are 19 weeks old. What does the least-squares line predict for a healthy weight? (Round your answer to two decimal places.)

kg

Answer 1

Part a)

Part b)

	X	Y	X * Y	X2	Y2
	2	39	78	4	1521
	5	46	230	25	2116
	10	79	790	100	6241
	16	100	1600	256	10000
	26	150	3900	676	22500
	36	200	7200	1296	40000
Total	95	614	13798	2357	82378

r = 0.998

Part c)

X̅ = Σ( Xi / n ) = 95/6 = 15.83
Y̅ = Σ( Yi / n ) = 614/6 = 102.33

Equation of regression line is Ŷ = a + bX


b = 4.780
a =( Σ Y - ( b * Σ X) ) / n
a =( 614 - ( 4.7798 * 95 ) ) / 6
a = 26.654
Equation of regression line becomes Ŷ = 26.654 + 4.780 X

Part e)

Coefficient of Determination
R2 = r2 = 0.997
Explained variation = 0.997* 100 = 99.7%
Unexplained variation = 1 - 0.997* 100 = 0.3%

Part f)

When X = 19
Ŷ = 26.654 + 4.78 X
Ŷ = 26.654 + ( 4.78 * 19 )
Ŷ = 117.47