Question

Use the given data to find a.) Equation of the regression line (Round to two decimal places as needed) b.) correlation coefficient (Round to two decimal places as needed) x 5 8 10 11 15 22 2436 38 40 y 50 28 60 66 62 70 88 112 124 140 TTTT Paragraph ▼ Arial 3(12pt) ▼ ·ー·ー·T·ノー4 ▼ HTML ESS Pathi p

Answer 1

Mean (x) = X^bar = 1/n sigma x_i = 209/10 = 20.9 2) Mean(y) = y^bar = 1/n sigma y_i = 800/10 = 80 3) variance (x) = 1/n sigma x^2_i = x^bar^2 = 5935/10 - (20.9)^2 = = 593.5 - 436.81 = 156.69 4) S middot D (x) = squareroot var(x) = 12.5175 5) variance(y) = 1/n sigma y^2_i - y^bar^2 = 75248/10 - 80^2 = 7524.8 - 6400 = 1124.8 6) S middot d = squareroot var(y) = 33.53880 7) cov(xy) = 1/n sigma x_iy_i - x^bary^bar = 18825/10 - (20.9)(80) = 1882.5 -1 675 cov(xy) = 210.5 \r\n Correlation coefficient between X & y r_xy = cov(xy)/sigm a_x middot sigm a_y = 210.5/12.5175 times 33.5380 = 210.5/419.811915 r = 0.50141502 = 0.50 regression expectation coefficient by x = r middot sigma_y/sigma_x = (0.50141502) (33.5380/12.5175) b_yx = 1 .3434 b_xy = r middot sigma_x/sigma_y = (0.50141502)(12.5175/33.5380) b_xy = 0.1871. Regression line equation of x on y (x