Question

Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of pounds), and let y be the miles per gallon (mpg). x 29 44 33 47 23 40 34 52 y 32 20 26 13 29 17 21 14 Complete parts (a) through (e), given Σx = 302, Σy = 172, Σx2 = 12,064, Σy2 = 4036, Σxy = 6066, and r ≈ −0.902. (a) Draw a scatter diagram displaying the data. Flash Player version 10 or higher is required for this question. You can get Flash Player free from Adobe's website. (b) Verify the given sums Σx, Σy, Σx2, Σy2, Σxy, and the value of the sample correlation coefficient r. (Round your value for r to three decimal places.) Σx = Σy = Σx2 = Σy2 = Σxy = r = (c) Find x, and y. Then find the equation of the least-squares line = a + bx. (Round your answers for x and y to two decimal places. Round your answers for a and b to three decimal places.) x = y = = + x (d) Graph the least-squares line. Be sure to plot the point (x, y) as a point on the line. (e) Find the value of the coefficient of determination r2. What percentage of the variation in y can be explained by the corresponding variation in x and the least-squares line? What percentage is unexplained? (Round your answer for r2 to three decimal places. Round your answers for the percentages to one decimal place.) r2 = explained % unexplained % (f) Suppose a car weighs x = 45 (hundred pounds). What does the least-squares line forecast for y = miles per gallon? (Round your answer to two decimal places.) mpg

Answer 1

Answer a)

Scatter Plot between X and Y 35 30 25 20 15 10 10 20 30 40 50 60 x- Weight of the car (in hundreds of pounds)

Answer b)

Step 1: Find X- Y, X and Y2 as it was done in the table below. 841 1936 1089 29 32 928 1024 676 169 841 289 441 196 26 13 29 17 21 611 667 680 714 1600 1156 34 52 Step 2: Find the sum of every column to get ΣΧ-302 , ΣΥ=172 , Σ , ΣΧ-12064 , ΣΥ-4036 X-Y 6066 Step 3: Use the following formula to work out the correlation coefficient. 8.6066-302-172 -0.9017 8-1 2064-3022-ו8 . 4036-1722

Based on above calculation values are as follows:

Σx = 302

Σy = 172

Σx² = 12064

Σy2 = 4036

Σxy = 6066

r = −0.902

Answer c)

Following is the calculation of for y-intercept (a) and slope (b):

Step 1: Find X.Y and X2 as it was done in the table belov. 841 1936 1089 29 32 880 858 611 667 680 714 26 13 29 1600 40 1156 14 52 Step 2: Find the sum of every column ΣΧ-302 , ΣΥ-172 , Σ , ΣΧ2-12064 X2 12064 X 302 X-Y 6066 Step 3: Use the following equations to find a and b ΣΥΣΧ2-ΣΧΣΧΥ 172.12064-302-6066 8-12064-3022 45.794 n.ΣΧΥ-Σχ.ΣΥ 8-6066-302 . 172 -0.644 2 n.ΣΧ2-(ΣΧ)2 8.12064 _ (302) Step 4: Substitute a and b in regression equation formula y 45.794 0.644-z

Answer d)

Following is graph showing regression line:

y = 45.794-0.644 x 36 5A 32 30 26 24 УЗ. 20 18 16 14 12 10 60 50 45 35 30 25 15

Answer e)

Coefficient of Determination r² = -0.902*-0.902 = 0.813

Interpretation

Since coefficient of determination is 0.813, so it can be said that 81.3% percentage of the variation in y can be explained by the corresponding variation in x.

The percentage of variation that is unexplained = 100 - 81.3% = 18.7%

Answer f)

At x = 45, we can predict y using regression equation:

y = 45.794 - 0.644*45

y = 16.814

y = 16.81 miles per gallon