Question

2. In a random sample of 432 employees at non-profit organizations, 387 said they were "highly satisfied" with their job. At for-profit companies, 391 of 528 respondents said the same. Using a 99% confidence level, build a confidence interval to test the claim that employees at non-profit orga- nizations are more likely to be "very satisfied" with their work. (a) X, = (b) n = (c) 2 =— (d) x2 = - (e) n2 = (1) P2 = SE, = VB:42,782) = 0.0147 SE, = 13,147,78) = 0.0191 SE= = V SE? + SE? (8) Ô– P2=— (h) SE, = — (i) z* = — (1) MoE = — (k) Cl = ( —

Answer 1

EXCEL (Mega Stat) output: Hypothesis test for two independent proportions - p1 0.8958 387/432 387. 432 P2 Po 0.7405 0.8104 p (as decimal) 391/528 778/960 p (as fraction) 391 778. X 528 960 n 0.1553 difference 0. hypothesized difference 0.0254 std. error 6.11 z 1.01E-09 p-value (two-tailed) 0.0933 confidence interval 99.% lower 0.2173 confidence interval 99. % upper 0.062 margin of error (a). x4 = 387 (6). n = 432

(c). = 0.8958 (d). xz = 391 (e). n = 528 (1). 2 =0.7405 SE, = SE + SE = V(0.0147)* +(0.0191) = 0.0005809 = 0.0254 (g). Ô1 - 2 = 0.1553 (h). SE, = 0.0254 1). 2* = 2.58 ). MoE = 0.062 (k). CI=(0.0933.0.2173).