Question

Consider the following data collected from a sample of 12 American black bears:

Length (cm)	139.0	138.0	139.0	120.5	149.0	141.0	141.0	150.0	166.0	151.5	129.5	150.0
Weight (kg)	110	60	90	60	85	100	95	85	155	140	105	110

(a) Sketch a scatterplot of the data. Treat length as the explanatory variable. Describe the association.

(b) Construct the equation for the line of best fit.

(c) Estimate the weight of a bear which measures 142.5 cm in length.

(d) What percent of the variation in the bears’ weights can be described by the differences in their lengths?

Answer 1

	Length (X)	Weight (Y)	X * Y	X2	Y2
	139	110	15290	19321	12100
	138	60	8280	19044	3600
	139	90	12510	19321	8100
	120.5	60	7230	14520.25	3600
	149	85	12665	22201	7225
	141	100	14100	19881	10000
	141	95	13395	19881	9025
	150	85	12750	22500	7225
	166	155	25730	27556	24025
	151.5	140	21210	22952.25	19600
	129.5	105	13597.5	16770.25	11025
	150	110	16500	22500	12100
Total	1714.5	1195	173257.5	246447.8	127625

Part a)

There is strong and positive relationship between length and weight.

Part b)

Equation of regression line is Ŷ = a + bX

b = ( 12 * 173257.5 - 1714.5 * 1195 ) / ( 12 * 246447.75 - ( 1714.5 )²)
b = 1.694
a =( Σ Y - ( b * Σ X) ) / n
a =( 1195 - ( 1.6942 * 1714.5 ) ) / 12
a = -142.471
Equation of regression line becomes Ŷ = -142.4709 + 1.6942 X

Part c)

When X = 142.5
Ŷ = -142.471 + 1.694 X
Ŷ = -142.471 + ( 1.694 * 142.5 )
Ŷ = 98.92

Part d)

r = 0.704

Coefficient of Determination
R² = r² = 0.495
Explained variation = 0.495* 100 = 49.5%
Unexplained variation = 1 - 0.495* 100 = 50.5%

49.5% of the variation in the bears’ weights can be described by the differences in their lengths.