Question

We are interested in exploring the relationship between the weight of a vehicle and its fuel efficiency (gasoline mileage). The data in the table show the weights, in pounds, and fuel efficiency, measured in miles per gallon, for a sample of 12 vehicles.

Weight	Fuel Efficiency
2710	24
2570	27
2620	29
2750	38
3000	23
3410	24
3640	21
3700	27
3880	22
3900	19
4060	18
4710	15

e.) What percent of the variation in fuel efficiency is explained by the variation in the weight of the vehicles, using the regression line? (Round your answer to the nearest whole number.)

g.) For the vehicle that weighs 3000 pounds, find the residual (y − ŷ). (Round your answer to two decimal places.)

i.) Remove the outlier from the sample data. Find the new correlation coefficient and coefficient of determination. (Round your answers to two decimal places.)

correlation coefficient
coefficient of determination

Find the new best fit line. (Round your answers to four decimal places.)
ŷ =

Answer 1

Answer to the question)

Part e)

Enter data in excel

click on data tab, click on data analysis, select regression, click ok

We get the following window on screen:

.

Input Y range as fuel efficiency

Input X range as weight

tick mark the check box next to label

click ok

Following outcome is obtained:

.

Hence the percent of variation explained is provided by the value of R square = 0.57379*100 = 57.379%

Hence 57.379% of the variation in fuel efficiency is explained by this linear regression model

.

Part g)

The regression equation is : Fuel efficiency = 46.5882 -0.0066* Weight

Given : weight = 3000

On plugging this value we get :

Predicted fuel efficieny ( y ^) = 46.5882 - 0.0066 *3000

y^ = 26.7882

Actual value y = 23

Residual = y - y^

Residual = (23 - 26.7882)

Residual = -3.7882 ~-3.79

.

Part i)

Plot manually the scatter plot for this data: with weights on x axis and fuel efficiency on y axis

.

From this we get to know that it outlier value is : (2750 , 38)

On removing this row , and working on the rest of the values, repeating the same steps as above we get:

Correlation coefficient = 0.830025

Coefficient of determination = R square = 0.6889

New equation of line:

Y = 40.4879 -0.00514*x

where Y = fuel efficiency , and x = weight