2008hw4_solutions

# 2008hw4_solutions - ST 314 4.1 =.7 n = 3 HW#4 Solutions a y...

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ST 314 HW #4 Solutions 4.1 σ = .7 n = 3 a = 5.8 [ ± z α /2 σ / ] where α = .05 α /2 = .025 z .025 = 1.96 [5.8 ± 1.96 (.7) / ] = [5.8 ± .79] = [5.01, 6.59] is a 95% C.I. for the true mean concentricity. b z α /2 σ/ B 1.96 (.7) / .2 1.96 × .7 / .2 n ≥ ( 1.96 × .7 / .2) 2 = 47.06 we must use n = 48 c We assume a sample of 3 measurements is large enough to use the Central Limit Theorem. The baseline data can provide a basis for generating either a stem-and-leaf display or a normal probability plot. These plots can provide insight as to an appropriate minimum sample size in order to assume the Central Limit Theorem. 4.2 σ = 8 n = 4 a = 101.4 [ ± z α /2 σ / ] where α = .01 α /2 = .005 z .005 = 2.58 [101.4 ± 2.58 (8) / ] = [101.4 ± 10.32] = [91.08, 111.72] is a 99% C.I. for the true mean width. b B z α /2 σ / 2 2.58 (8) / 2.58 (8) / 2 n (2.58 (8) / 2) 2 = 106.5 we must use n = 107 c We assume a sample of size 4 parts is large enough to use the Central Limit Theorem. The baseline data can provide a basis for generating either a stem-and-leaf display or a normal probability plot. These plots can provide insight as to an appropriate minimum sample size in order to assume the Central Limit Theorem.

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