Question

Current Attempt in Progress Use the normal distribution to find a confidence interval for a difference in proportions p 1 -p 2 given the relevant sample results. Assume the results come from random samples. A 90% confidence interval for p 1 -p 2 given that p^1 = 0.25 with n 1 = 60 and p^2 = 0.40 with n 2 = 80 Give the best estimate for p 1 -p 2, the margin of error, and the confidence interval. Round your answer for the best estimate to two decimal places and round your answers for the margin of error and the confidence interval to three decimal places. Best estimate: i Margin of error: i Confidence interval : i to i

Answer 1

Solution in Ocke the normal dilansition to find a Confickence intervalli- Rescrired information is given by ho - P2 0.40 Pi= 0.05 ny Bo na Go Elcolate :- @ Find the poirst Estimate: - the point edinate by casing given data is lale have to obtained Elere, > point Eftimate ics, point Estimate = h PR to.95-0:40] -0.15 Therefore, -0.15 Point Estimate is Morgin Error We have to obtained the margin of Error bucking given cok is Gere, - orgin of Error its

fit-pij pą Ci -62] + Kal-E= zcritical Ne nu cohere, RelC.I -1-906 zeriti al = Zeala] szto. 05) Sto.ao 67099.1 & co.10 ale- Co.103 la Tal2-0.05 from een @, we get 0.0511-0.05) El.bulla +0.4001-0.410] 80 60 Abb9:11 = (6.95*0.75] 60 + .40*0.60 od 0-24 0.1875 + -ב-9רומט X 29 BD its = [l.604qx 3.1054037@4103] Etli buleg * V6.195*60-3] trouua * 0.07866) = - 0.1287 ง OilQq 19. Eslo. 198+ Therefore, 0.109 reorgin of Error is

3 @ Confidence inkilal : ale have to obtained thi Conifidence in local is Ekre --> Confidence interlal is, CIE X £19.E =t-0.151 0.1290 et-0.15-0.109 -0.15+0:19] = [-0.272) 0.021] Therefore, 0.00) to 0.279 95.8. f citu (sample recullen ☺ 06 the tu distribution and the gilen Rexrired information is given by > nav bypothesis asis, Ito: Oly chec Attiradila bypothesis (Cicet : ltt < licLeft-laie] X2 = 11.2 Xi = 66 3.4 Si= 41 nos 05 n25

Calculate :- a Gilem the host Stastic and the probleem @ Test Statistic :- hk hale to obtained the test Glaticlic by casing gian data is, Gere, - tecl Clatistic is t-test- (xi-i2 Gpaxtfitto ] 622 cohere, Op GHG1²+62) Esitno-oj t951) ece. 1)+605+63.432 to5+05-0) (p²= tau* 16.01] + [aux 11.86 ] UB 14] 403.44 + oHf.414. 416

p_Kalce ti-test, df=O1+60-03] 680.66 116 Gp' - lll. 185 Fin En @ Cue get 6:6-11.0] 161.165 * a - 35 2.6 14.185.6 [0.08] -26 - I 16616 -2.6 1.065 It= -0.014 Therefore, trest 2.614 palu - he have to obtained the plalu, by cising given dake is Ebre, - Valu. ils, -[-2.4I, df [80+50-0)) -t-9.414, 418 ] 0.0092

(6 Therefore, 0.0099 p-Valee is