Question

Does the weight of a vehicle affect the gas mileage? The following random sample was collected where weight of a vehicle (hundreds of pounds) and y = miles per gallon * (lbs) y (mpg) 26 22.0 35 16. 1 29 1 8.8 39 15.7 20 23.4 1. Based on a scatter diagram, would you estimate the linear correlation coefficient to be (a) close to -1 (c) close to 1 (b) close to 0 and negative (d) close to 0 and positive (e) Cannot determine 2. What is the equation for the least squares line? (a) y = -32.55x +0.448 (a) y = 32.55x +0.448 (b) y = -32.55x -0.448 (d) y = -0.448x + 32.55 (e) y = 0.448x - 32.55 3. Compute the coefficient of determination. (a) -0.941 (C) -0.970 (b) 0.941 (d) 0.970 (e) 0.965 4. Compute the standard error of estimate S. (a) 0.965 (c) 0.941 (b) -0.970 (d) 1.975 (e) 0.065 5. If a vehicle weighs 2200 pounds, what does the least squares line predict for the miles per gallon? (a) 22.7 (c) 42.4 (b) 66.0 (d) 22.0 (e) Cannot determine

Show your work

Answer 1

The scatter plot is as shown below

X (lbs) Y(mpg) 22 16.1 Y(mpg) vs X(lbs) 35 29 18.8 15.7 23.4 20 0 5 10 15 20 25 30 35 40 45

Question (1)

Cleary the regression line or least squares is in a downward direction which implies that as value of x increses the value of y decreases

So the linear correlation coefficient will be close to -1 since the points are very close to the regression line or least squares line

So Answer is option A

Question (2)

The scatter plot with the equation is as follows

The equation is y = -0.448x + 32.55

X (lbs) Y(mpg) 22 Y(mpg) vs X(lbs) 16.1 18.8 y=-0.4479x + 32.548 15.7 23.4 5 10 15 20 25 30 35 40 45

So Answer is Option D

Question (C)

The scatter plot with equation and R-square or coefficient of determination is as follows

The R2 value or coefficient of determination is 0.9411

Y(mpg) vs X(lbs) X (lbs) 26 35 29 39 20 Y(mpg) 22 16.1 18.8 15.7 23.4 y = -0.4479x + 32.548 R=0.9411 0 5 10 15 20 25 30 35 40 45

So Answer is Option B

Question (D)

The regression output is as shown below

SUMMARY OUTPUT Y(mpg) 22 16.1 18.8 15.7 23.4 Regression Statistics Multiple R 0.97012153 R Square 0.941135784 Adjusted R Square 0.921514378 Standard Error 0.965410149 Observations 5 ANOVA F 47.96474894 MS 44.7039497 0.93201676 Significance F 0.00617185 Regression Residual Total SS 44.70394973 2.796050269 47.5 Intercept X (lbs) Coefficients 32.54847397 -0.447935368 Standard Errort Stat 1.975158593 16.4789167 0.064677655 -6.92565874 P-value 0.000486359 0.00617185 Lower 95% 26.2626378 -0.653768533 Upper 95% 38.83431013 -0.242102203

THe Standard error of the estimate from the Regression statistics section in the above image is 0.96541

So Answer is Optin A

Question (5)

The equation from Question (2) is y = - 0.448x + 32.55

Given vehicle weight is 2200 pounds

So x = 22 (since it is in hundreds of punds)

y = 0.448 * 22 + 32.55

y = -9.856 + 32.55

y = 22.694

y = 22.7

So Answer is Option A