hw4_sol

# hw4_sol - 11-16 (a) Claim: the scrap rate is less than 7.5%...

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Unformatted text preview: 11-16 (a) Claim: the scrap rate is less than 7.5% H 0 : µ ≥ 7.5 H1 : µ < 7.5 claim x = ∑ xi n = 55.98 8 = 6.9975 , σ = 1.25 , α = 0.10 using a confidence interval zα = z0.10 = 1.282 µ ∈ −∞, x + zα σ µ ∈ [ −∞, 7.564] n 8 µ ∈ −∞, 6.9975 + 1.282 (1.25) Since µ0 ∈ [ −∞, 7.564] , we should accept the null hypothesis (H0). The scrap rate is not less than 7.5%. (b) For δ σ = 1.5 with power = 0.90, δ = µ0 − µ1 , power = 1-β, zβ = z0.10 = 1.282 n= ( zα + z β )2 σ 2 ( µ 0 − µ1 ) 2 = (1.282 + 1.282)2 = 2.92 3 1.52 11-16 (continued) (c) First, we must find x * . x * = µ0 − zα σ n 8 = 6.933 x * = 7.5 − 1.282 (1.25) Then, we calculate β. For δ σ = 2.0 , δ = 2.0(1.25) = 2.50 . δ = µ0 − µ1 ⇒ µ1 = 7.5 − 2.5 = 5.0 β = P [ Accept H 0 | H 0 false] = P X > x * | µ = µ1 x * − µ1 = P Z > σ n 6.933 − 5.0 = P Z > 1.25 8 = P [ Z > 4.37 ] = 0.00 Finally, power = 1-β = 1.00 11-18 Claim: the melting point of the alloy is 1000 H 0 : µ = 1000 claim H1 : µ ≠ 1000 α = 0.05, β = 0.10, σ = 10, µ0 − µ1 = 20 n= ( zα / 2 + z β )2 σ 2 ( µ 0 − µ1 )2 = (1.96 + 1.282) 2102 = 2.62 3 20 2 10-50 n = 25 x = 4.05 s = 0.08 α = 1 − 0.90 = 0.10 t0.10,24 = 1.318 µ ∈ x − tα ,ν s µ ∈ [ 4.029, + ∞ ] n , +∞ 11-14 Claim: the mean titanium content is greater than 9.5% H 0 : µ ≤ 9.5% H1 : µ > 9.5% claim n = 6 x = 10.28% s = 2.55% α = 0.05 tα ,n −1 = 2.015 tobs = x − µ0 10.28 − 9.5 = = 0.75 sn 2.55 6 Since tobs < tα,n-1 , fail to reject H0. The mean titanium content is not grater than 9.5% ...
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## This note was uploaded on 01/04/2011 for the course ISE 31602D taught by Professor Kurtpalmer during the Fall '10 term at USC.

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hw4_sol - 11-16 (a) Claim: the scrap rate is less than 7.5%...

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