Question

Question 1 1 pts question 1- based on the sample data summarized by the following information: n = 5, L r = 25, x2 = 165 Sy=24, y = 134, ry=144 1<x<9 1. Compute the linear correlation coefficient r- 0.875 0.635 0.576 O 0.738 0.925

Answer 1

n=5 Ex=25, Ix² = 165, Ey=24, 5y2=134 Sny = 144 12229 7 linear conelation coefficient r = Iny ñ 144 (5x4.8) - 0.875189949 n S(529.7?)($99.9°) | 165. 5²) ( 134 - 4.8²) r=0.875 120 =0.6 27 y=mxtb men S*4 - Sx Sy nJX² (50) 5+144 - (25*24) 5x165 (25) 200 b= Eymer 24-0.6+25 9 = 1.8 5 n ŷ = 0.6x + 1.8 37 gol confidance interval for slope Bi MS Res f MSRes = {yi? - B. Saiy; B. I ta, (n=ck+11) nas Res = {4;² - B. sniti 134 - 0.6x144 476 = 11.9 n-1 Enir n-1 5-1 4 0.6 + tol, (5-61+)) 11.9 165 0.6 + tool, 3 +0.2685539278 {to., 3 = 2.353 ::0.6 + 2.353 * 0.268553 927). ..0.6 +0.551 0.6 + 0.6319073921 47 Coefficient of determination 2 R² = 82= (0.875)2 = 0.765625 = 0.766 نی where rois linear correlation coefficient