ASSIGNMENT 4
Due 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
Question 3
a)Question 1:
let p denote the proportion of garmentors who currently call for at least $10,000 invested in the stock 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 proportion of investors who currently have at least $10,000 invested in the stock market
Question 2:
Let p denote the proportion of investors who would be more likely to invest in stocks
p-head = 0.48, n = 692
SE (p head) = Square 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 that (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 number of...If you want to get a full essay, localize it on our website: Orderessay
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