Question

You are given the following regression equation for a scatter plot which The displays data for X= weight of car (in pounds) and y= Miles per gallon in City:

Equation: y=-0.006x + 42.825 and r2= 0.7496.

a. find the value of r2 based on the information given.

b. Based on your value of r, what conclusion can you make about the correlation of this data?

c. What does the value of r2 tell you about the regression? d. Use the regression equation to estimate miles per gallon for a car that weighs 3000 pounds e. Use the regression equation to estimate the weight of a car that gets 30 miles to the gallon.

Answer 1

The given regression equation is,

y=-0.006x + 42.825

where, X= weight of car (in pounds) and y= Miles per gallon in City.

Also r² = 0.7496

(a). The value of r² is 0.7496.

(b). The correlation can be calculated as,

$r = \sqrt {{r^2}} = \sqrt {0.7496} = 0.8658$

Since the sign of the slope coefficient 0.006 is negative , so the correlation value is -0.8658 which can be interpreted as strong negative association between the variables x= weight of car (in pounds) and y= Miles per gallon in City.

(c). The value of r² tells about the regression is that 74.96% of the variation in miles per gallon in city can be explained by the weight of car (in pounds).

(d). Using the regression equation, the estimated miles per gallon for a car that weighs 3000 pounds is given as,

$\begin{array}{l} \hat y = - 0.006x + 42.825\\ \hat y = - 0.006\left( {3000} \right) + 42.825\\ \hat y = - 18 + 42.825\\ \hat y = 24.825 \end{array}$

Therefore, the estimated miles per gallon for a car that weighs 3000 pounds is 24.825.

(e). Using the regression equation, the estimated the weight of a car that gets 30 miles to the gallon is given as,

y = -0.006.x + 42.825 30 = -0.006.1 + 42.825 30 – 42.825 = -0.006.0 -12.825 = -0.006.0 12.825 = 0.006.0 12.825 0.006 I = 2137.5 12138 pounds 2

Therefore, the estimated weight of a car that gets 30 miles to the gallon is 2138 pounds.