The following data were collected during a study of consumer buying patterns: Ob
ID: 463631 • Letter: T
Question
The following data were collected during a study of consumer buying patterns: Obtain a linear regression line for the data. (Round your intermediate calculations and final answers to 3 decimal places.) What percentage of the variation is explained by the regression line? (Do not round intermediate calculations. Round your answer to the nearest whole percent. Omit the "%" sign in your response.) Use the equation determined in part b to predict the expected value of y for x = 41. (Round your intermediate calculations and final answers to 3 decimal places.)Explanation / Answer
b.
b = [(n*sum of x*y) - (sum of x*sum of y)/(n*sum of x^2)-(sum of x)^2)] = [(13*29864 - 352*1068)/(13*11388 - 352^2)] = 388232-375936/148044-123904 = 12296/24140 = 0.51.
Thus b = 0.51.
a = [(sum of y*sum of x^2)-(sum of x*sum of x*y)/(n*sum of x^2)-(sum of x)^2]
= (1068*11388)-(352*29864)/(13*11388)-(352)^2
= 1650256/24140 = 68.362
Thus a = 68.362 and b = 0.509
Thus the linear regression line is in the form y = a+bx will be y = 68.362+0.509x
c. For this we will have to find r^2 i.e. coefficient of determination.
r = {(n*sum of xy)-(sum of x*sum of y)/[(n*sum of x^2 - (sum of x)^2)*(n*sum of y^2 - (sum of y)^2)]^0.5}
= {(13*29864 - 352*1068)/[(13*11388 - 352^2)*(13*88596-1068^2)]^0.5}
= 12296/16386.99
= 0.7504
Thus r^2 = 0.7504^2 = 0.5630
Thus 56.30% or 56% (rounded off to the nearest whole %) of the variation in the dependent variable is explained by the independent variable.
d. y = 68.362+0.509x
when x = 41, y will be = 68.362+(0.509*41)
= 68.362+20.884
= 89.246
Observation x y xy x^2 1 12 78 936 144 2 21 76 1,596 441 3 44 83 3,652 1,936 4 29 81 2,349 841 5 49 93 4,557 2,401 6 46 99 4,554 2,116 7 29 85 2,465 841 8 13 77 1,001 169 9 18 70 1,260 324 10 19 69 1,311 361 11 17 84 1,428 289 12 25 87 2,175 625 13 30 86 2,580 900 Total 352.00 1,068.00 29,864.00 11,388.00