IE360%20HW7%20solutions

# IE360%20HW7%20solutions - IE 360 Homework Assignment#7...

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IE 360 Homework Assignment #7 Solutions 10-6 a & b only 10-8 10-16 a (no box plot) & b only 10-18 10-36 a & b only 10-44 10-60 ** Note : several of the Chapter 10 questions ask you to find a P-value – you can skip this part of the question. You do not need to find P-values. 10-6 a) 1) The parameter of interest is the difference in mean burning rate, μ μ 12 2) H 0 : μμ 0 −= or = 3) H 1 : 0 −≠ or 4) α = 0.05 5) The test statistic is 2 2 2 1 2 1 0 2 1 0 ) ( n n x x z σ + Δ = 6) Reject H 0 if |z 0 | > z α /2 = 1.96 7) x 1 = 18 x 2 = 24 σ 1 = 3 σ 2 = 3 n 1 = 20 n 2 = 20 32 . 6 20 ) 3 ( 20 ) 3 ( ) 24 18 ( 2 2 0 = + = z 8) Because | 6.32| > 1.96 reject the null hypothesis and conclude the mean burning rates differ significantly at α = 0.05. b) () 2 2 2 1 2 1 2 / 2 1 2 1 2 2 2 1 2 1 2 / 2 1 n n z x x n n z x x μ α + + + 20 ) 3 ( 20 ) 3 ( 96 . 1 24 18 20 ) 3 ( 20 ) 3 ( 96 . 1 24 18 2 2 2 1 2 2 + + + 14 . 4 86 . 7 2 1 Since 0 not in this interval, we conclude the mean burning rates differ significantly at α = 0.05. Also, since this μ 1 - μ 2 interval is strictly negative we can conclude the mean burn rate of propellant 2 is greater than propellant 1.

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10-8 a) 1) The parameter of interest is the difference in mean batch viscosity before and after the process change, μμ 12 2) H 0 : 10 2 1 = μ or 10 2 1 3) H 1 : 10 2 1 < 4) α = 0.10 5) The test statistic is 2 2 2 1 2 1 0 2 1 0 ) ( n n x x z σ + Δ = 6) Reject H 0 if z 0 < z α where z 0.1 = 1.28 7) x 1 = 750.2 x 2 = 756.88 Δ 0 = 10 σ 1 = 20 σ 2 = 20 n 1 = 15 n 2 = 8 90 . 1 8 ) 20 ( 15 ) 20 ( 10 ) 88 . 756 2 . 750 ( 2 2 0 = + = z 8) Since 1.90 < 1.28 we reject the null hypothesis. We believe the difference in mean batch viscosity is less than 10. b) NOTE: The part b question in the book asks for a ‘90% CI on the difference in mean viscosity’ – this implies a two-sided interval. However, it should be noted that the following two-sided interval is NOT testing the same premise as the hypothesis test in part a) since the hypothesis test was designed to test if the difference in mean batch viscosity is less than 10 (i.e. a one-sided test). The equivalent CI for the part a) hypothesis test would be a 90% upper bound CI, not the two-sided CI shown below. Case 1: Before Process Change Case 2: After Process Change μ 1 = mean batch viscosity before change μ 2 = mean batch viscosity after change x 1 = 750.2 x 2 = 756.88 σ 1 = 2 0 σ 2 = 20 n 1 = 1 5 n 2 = 8 90% confidence on , the difference in mean batch viscosity before and after process change: () 2 2 2 1 2 1 2 / 2 1 2 1 2 2 2 1 2 1 2 / 2 1 n n z x x n n z x x α + + + 8 ) 20 ( 15 ) 20 ( 645 . 1 88 . 756 2 . 750 8 ) 20 ( 15 ) 20 (
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IE360%20HW7%20solutions - IE 360 Homework Assignment#7...

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