The following data were collected during a study of consumer buying patterns: Ob

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

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