Question

3 JU The grades of a class of 9 students on a midterm report (x) and on the final examination () are as follows 67 99 96 94 81 72 71 50 77 68 99 99 85 47 34 78 66 82 x=85. Compute a 95% confidence interval for the mean response Hy/x when

Answer 1

solution, given that, x=85, n=g. XY ♡ W 67 68 4489 4624 4559 gg 9g 9801 9801 9801 96 gg 9216 9801 9504 94 85 8856 7225 79go 81 47 6561 220g 3807 72 34 5184 1156 2448 71 78 5041 6084 5538 2500 4356 oo 77 82 5929 €724 6314 Simple linear gegoession with t predictor ei= Bot Bixit Ei. whese, Bo and Bi segression coeffn Therefore, estimated reg oessio eqn will be 50 ?i= Bot Bixi Ểo = 7 - Bi T. $i = Sexy

Sxx = x; 2 (Zxi)? Sxx = 57557 - (707) Sxx = 57557 - 55538 078 Sxx = 2018-22 sex = eiz- (2 Syp = 51980- (658)2 Syy: 51980 - 48104-12 See = 3872.88 Sxp = ZXp; - (Exi) (EPI) Sxy = 53281 - (707)(658) g

Sxy= 53281- 51689.56 Sxy = 159 1.44 . sxy = 54 1591-44 2018-22 SXX Bi = 0.7885 Ipi - 8 zxi 458 - (0.7885) 6707) - B. - 100.5305 = 11.1700 Therefore, estimated degression equation will be = (11-1700) + (0.7885) xi Here, Xn= 85 "Therefore 95% confidence interval for elyly will be

SXX . MSE = SSE n-2 SSE: Spy- sxe Sxx = 3872.88 - (1591.44) $ 2018-22 SSE = 3872-88- 1254.9084 SSE = 2617.9716 MSE = SSE n-2 2617-9716 MSE = 373.9959.

coherei X = 85 en - poedicted value of gesponse whes predictor is xn Thereforei n = 11-1700 + (0-7885)985) P = 78:1925 These fese i confidence interxcel will be ovo K = 0.05 → 2/2 = 0.025 orn to-025; 7 = 2.3646 l SX ån-tx/210-2 [ MSE (1 + (xh-xi Ent tx 210-2 [ Mst ( ħr (X-X) ) } .... (xb-7)2 = (85-78.55)2 (Xn-x)2 = 41.6025 c= IMSE ( ħt CXnA

c= 49.2603 C = 7.0187 Therefore confidence interval for elyix will be I en talin-2 *c, eht tx/21n-2&c } 78.41925 - (2.3646) (710187), 78.1925+ (2.3646) (7.0187) } ltyoff ou5961, Therefore, lpix € 61.5961/ 94.78897