Question

Do heavier cars really use more gasoline? Suppose 5 cars were chosen at random. Let x be the weight of the car (in hundreds of pounds) and y be gas mileage (in miles per gallon).

X- 34,46,33,47,23

y- 17,29,23,28,21

- What is the value of the correlation coefficient? Based on the correlation coefficient, as the weight of the car increases, does the gas mileage tend to increase or decrease?

-Find the equation of the least-squares regression line.

-What percentage of the variation in y can be explained by the corresponding variation in x and the least-squares line?

- Predict the gas mileage for a car that weighs x = 30 (hundred pounds). Is this interpolation or extrapolation?

Answer 1

X Values
∑ = 183
Mean = 36.6
∑(X - M_x)² = SS_x = 401.2

Y Values
∑ = 118
Mean = 23.6
∑(Y - M_y)² = SS_y = 99.2

X and Y Combined
N = 5
∑(X - M_x)(Y - M_y) = 151.2

R Calculation
r = ∑((X - M_y)(Y - M_x)) / √((SS_x)(SS_y))

r = 151.2 / √((401.2)(99.2)) = 0.7579

As r is positive, so as x increases y also increases

Sum of X = 183
Sum of Y = 118
Mean X = 36.6
Mean Y = 23.6
Sum of squares (SS_X) = 401.2
Sum of products (SP) = 151.2

Regression Equation = ŷ = bX + a

b = SP/SS_X = 151.2/401.2 = 0.3769

a = M_Y - bM_X = 23.6 - (0.38*36.6) = 9.8066

ŷ = 0.3769X + 9.8066

Here r=0.7579

So r^2=0.7579^2=0.5744

So 57.44% of the variation in y can be explained by the corresponding variation in x and the least-squares line

For x=30,

ŷ = (0.3769*30) + 9.8066=21.1136