46240HW4_solutionFA10

# 46240HW4_solutionFA10 - Scheaffer Mendenhall Ott Exercises...

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Unformatted text preview: Scheaffer, Mendenhall, Ott Exercises 4.14 p= ∑y i n B = 2 = 25 5 = = 0.83 30 6 pq ⎛ N − n ⎞ (5 / 6)(1 / 6) ⎛ 300 − 30 ⎞ ⎜ ⎟ =2 ⎜ ⎟ =.131 ⎝N⎠ ⎝ 300 ⎠ n −1 29 4.15 B = .05 D = B2 / 4 =(.05)2 / 4 = .000625 From Equation (4.19), we have n= Npq 300 (5 / 6) (1 / 6) = = 127.90 ≈ 128 ( N − 1) D + pq 299 (.000625) + (5 / 6) (1 / 6) 4.16 μ = y = 12.5 B=2 1252 ⎛ 10000 − 100 ⎞ s2 ⎛ N − n ⎞ ⎜ ⎟ =2 ⎜ ⎟ = 7.04 ⎝N⎠ n 100 ⎝ 10000 ⎠ 4.17 τ = Ny = 10000(12.5) = 125,000 ⎛ s2 ⎞ ⎛ N − n ⎞ 2 1252 10000 − 100 B = 2 N 2⎜ ⎟⎜ = 70,412.50 ⎟ = 2 10000 ⎝N⎠ 100 10000 ⎝ n⎠ 4.20 p = 1 1 430 =.430 ∑ yi = n 1000 B = 2 pq ⎛ N − n ⎞ .430(.570) ⎛ 99000 − 1000 ⎞ ⎜ ⎟ =2 ⎜ ⎟ = .0312 n −1⎝ N ⎠ 999 ⎝ 99000 ⎠ 4.21 B = .02, D = B2 / 4 = (.02)2 / 4 = .0001 n= = Npq ( N − 1) D + pq 99000 (.43) (.57) = 23918 ≈ 2392 . 98999 (.0001) +.43 (.57) Scheaffer, Mendenhall, Ott Exercises 4.23 y = 2.1 s =.4 N = 200, n = 20 μ = y = 2.1 B=2 s2 ⎛ N − n ⎞ (.4) 2 ⎜ ⎟ =2 20 n⎝ N ⎠ ⎛ 200 − 20 ⎞ ⎜ ⎟ = .17 ⎝ 200 ⎠ 4.24 σ = 1, B = 1, n= D = B 2 / 4 = 1 / 4 =.25 Nσ 2 200(1) 2 = = 3.94 ≈ 4 2 ( N − 1) D + σ 199 (.25) + 12 ...
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