Statistical Analysis of Estimators and Hypothesis Testing, Assignments of Statistics

Download Statistical Analysis of Estimators and Hypothesis Testing and more Assignments Statistics in PDF only on Docsity! Homework 3 8.6 (a) E(θ̂3) = aE(θ̂1) + (1− a)E(θ̂2) = aθ + (1− a)θ = θ (b) V (θ̂3) = a2V (θ̂1) + (1 − a)2V (θ̂2) = a2σ21 + (1 − a)2σ22 , since it was assumed that θ̂1 and θ̂2 are independent. To minimize V (θ̂3), we can take the first derivative (with respect to a), set it equal to zero, to find a = σ22 σ21 + σ 2 2 (One should verify that the second derivative test shows that this is indeed a minimum.) 8.17 It is given that p̂1 is unbiased, and since E(Y ) = np, E(p̂2) = np+1n+2 . (a) B(p̂2) = np+1n+2 − p = 1−2p n+2 (b) Since p̂1 is unbiased, MSE(p̂1) = V (p̂1) = p(1−p) n . MSE(p̂2) = V (p̂2) +B(p̂2) = np(1−p)+(1−2p)2 (n+2)2 . (c) Considering the inequality np(1− p) + (1− 2p)2 (n+ 2)2 > p(1− p) n , this can be written as (8n+ 4)p2 − (8n+ 4)p+ n > 0. Solving for p using the quadratic formula, p will be away from .5 by at least √ n+1 8n+4 . 8.28 The point estimate is given by the difference of the sample proportions: .70− .54 = .16 and an error bound is 2 √ .7(.3) 180 + .54(.46) 100 = .121. 8.90 (a) For the 95% CI for the difference in mean verbal scores, the pooled sample variance is s2p = 14(42)2+14(45)2 28 = 1894.5 and thus 446− 534± 2.048 √ 1894.5( 2 15 ) = −88± 32.55 or (−120.55,−55.45) . (b) For the 95% CI for the difference in mean math scores, the pooled sample variance is s2p = 14(57)2+14(52)2 28 = 2976.5 and thus 548− 517± 2.048 √ 2976.5( 2 15 ) = 31± 40.80 or (−9.8, 71.8) . (c) At the 95% confidence level, there appears to be a difference in the two mean verbal SAT scores achieved by the two groups. However, a difference is not seen in the math SAT scores. (d) We assumed that the sample measurements were independently drawn from normal populations with σ1 = σ2. 8.93 (a) Since the two random samples are assumed to be independent and normally distributed, the quantity 2X̄+ Ȳ is normally distributed with mean 2µ1 +µ2 and variance ( 4n + 3 m )σ 2. Thus, is as σ2 is known, then 2X̄ + Ȳ ± 1.96σ √ 4 n + 3 m is a 95% CI for 2µ1 + µ2. 1