COMM 291 Application of Statistics in Business (JanApr 2010) ASSIGNMENT 4 cod 1:00 pm, Friday, March 10, 2010 uncertainty 1. Practice with the Normal tables a) Pr(z ? ___0.53__ ) = 0.70 b) Pr(z = 1) = ___0__ c) Pr(0.5 ? z ?1.5) = __0.9332- 0.6915 =_0.24___ d) Pr(-2.3 ? z ? 2.3) = __1- 2* ( 0.0107) = 0.98____ e) Pr(|z| ? 1.26) = __2* 0.1056 = 0.21____ f) Pr( ___-0.91___ ? z ?1.5) = 0.75 Pr (z 30) = Pr ( z > (30 25) / sigma ) = 0.231 (30-25)/ (sigma) = 0.73 S = 6.85 Var = s^2 = 46.91 b) Pr ( m < 18) = Pr ( z < (18-25)/ 7 ) = Pr (z < -1) = 0.1587 c) Pr ( m < 18) = Pr ( z< ( 18 25) / 3.5) = Pr ( z < -2) = 0.0228 Pr (m > 30) = Pr (z > (30-25) / 3.5) = Pr (z > 1.43) = 0.9236 d) Sigma / n-square understructure = 7 / 5 = 1.4 Pr ( m-bar > 28) = Pr ( z > (m-bar mu) / 1.4 ) = Pr ( z > (28 25) / 1.4) = Pr (z > 2.14) = 0.0166 misgiving 3 a) enquiry 1: let p have-to doe with the property of garmentors who currently call for at least $10,000 investe d in the billet market p-head = 0.65 SE (p head) = satisfying root (0.65 * 0.35/ 692) = 0.018 ME = z* X 0.018 = 1.96 X 0.018 = 0.03528 95% agency breakup is (0.61472, 0.68528) We are 95% sure that (0.61472, 0.
68528) will be back the population residuum of investors who currently have at least $10,000 invested in the ancestry market Question 2: Let p denote the equalizer of investors who would be more likely to invest in stocks p-head = 0.48, n = 692 SE (p head) = solid root (0.48 * 0.52/ 692) = 0.019 ME = z* X 0.018 = 1.96 X 0.019 = 0.03724 95% confidence interval is (0.44276, 0.51724) We are 95% sure t hat (0.44276, 0.51724) will be covering the ! population proportion of investors who would be more likely to invest in stocks b) Question 1: ME = z* SE = 1.96 * SE = 0.03 SE = 0.0153 = Square root ( 0.65 * 0.35 / n ) N = 972 Question 2: ME = 1.96 * SE = 0.03 SE = 0.0153 = Square root (0.48 *0.52/n) N = 1066 Question 4 Condition Check: np= 25 >0 nq =75 >0 Assumptions: The numerate of...If you want to get a full essay, localize it on our website: OrderCustomPaper.com
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