HW6_sol - 1. 9‐91 a) 1) The variable of interest is the...

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Unformatted text preview: 1. 9‐91 a) 1) The variable of interest is the form of the distribution for the number of cars passing through the intersection. 2) H0: The form of the distribution is Poisson 3) H1: The form of the distribution is not Poisson 4) α = 0.05 5) The test statistic is Estimated mean = 49.6741 use Poisson distribution with λ=49.674 All expected frequencies are greater than 3. The degrees of freedom are k − p − 1 = 26 − 1 − 1 = 24 χ2 = 0 i =1 ∑ k ( Oi − E i )2 Ei 6) Reject H0 if χ 2 > χ 2.05,24 = 36.42 o 0 7) Estimated mean = 49.6741 χ 02 = 769.5669 8) Since 769.57 >>> 36.42, reject H0. We can conclude that the distribution is not Poisson at α = 0.05. b) P‐value = 0 2. 9‐107. 1. The variable of interest is failures of an electronic component. 2. H0: Type of failure is independent of mounting position. 3. H1: Type of failure is not independent of mounting position. 4. α = 0.01 5. The test statistic is: χ = ∑∑ 2 0 i =1 j =1 r c (O ij − Eij ) E ij 2 6. The critical value is χ 0.01, 3 2 = 11.344 2 7. The calculated test statistic is χ 0 8. χ 0 2 = 10.71 2 > χ 0.01,3 , do not reject H0 and conclude that the evidence is not sufficient to claim that the type of failure is / not independent of the mounting position at α = 0.01. P‐value = 0.013 3. 10‐6 a) 1) The parameter of interest is the difference in mean burning rate, μ1 − μ 2 2) H0 : μ1 − μ 2 = 0 or μ1 = μ 2 3) H1 : μ1 − μ2 ≠ 0 or μ1 ≠ μ 2 4) α = 0.05 5) The test statistic is z0 = ( x1 − x2 ) − Δ 0 2 σ1 σ 2 +2 n1 n2 6) Reject H0 if z0 < −zα/2 = −1.96 or z0 > zα/2 = 1.96 7) x1 = 18 x2 = 24 σ1 = 3 σ2 = 3 n1 = 20 n2 = 20 z0 = (18 − 24) (3) (3) + 20 20 2 2 = −6.32 8) Because −6.32 < −1.96 reject the null hypothesis and conclude the mean burning rates differ significantly at α = 0.05. P‐value = 2(1 − Φ(6.32)) = 2(1 − 1) = 0 b) ( x1 − x2 ) − zα / 2 2 σ1 σ 2 σ2 σ2 + 2 ≤ μ1 − μ 2 ≤ ( x1 − x2 ) + zα / 2 1 + 2 n1 n2 n1 n2 (18 − 24) − 1.96 (3) 2 ( 3) 2 ( 3) 2 (3) 2 + ≤ μ1 − μ 2 ≤ (18 − 24) + 1.96 + 20 20 20 20 −7.86 ≤ μ1 − μ 2 ≤ −4.14 We are 95% confident that the mean burning rate for solid fuel propellant 2 exceeds that of propellant 1 by between 4.14 and 7.86 cm/s. ⎞ ⎛ ⎞ ⎛ ⎟ ⎜ ⎟ ⎜ ⎜ ⎜ Δ − Δ0 ⎟ Δ − Δ0 ⎟ c) β = Φ⎜ zα / 2 − ⎟ − Φ⎜ − zα / 2 − ⎟ 2 2 σ 12 σ 2 ⎟ σ 12 σ 2 ⎟ ⎜ ⎜ + + ⎜ ⎜ n1 n2 ⎟ n1 n2 ⎟ ⎠ ⎝ ⎠ ⎝ ⎛ ⎜ ⎜ = Φ⎜ 1.96 − ⎜ ⎜ ⎝ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ 2.5 . ⎟ − Φ⎜ −196 − 2 2 ( 3) (3) ⎟ ⎜ + ⎜ ⎟ ⎝ 20 20 ⎠ ⎞ ⎟ ⎟ 2.5 ⎟ 2 2 ( 3) ( 3) ⎟ + ⎟ 20 20 ⎠ = Φ(1.96 − 2.64) − Φ( −1.96 − 2.64) = Φ( −0.68) − Φ( −4.6) = 0.24825 − 0 = 0.24825 d) Assume the sample sizes are to be equal, use α = 0.05, β = 1‐power=0.1, and Δ = 4 2 2 2 2 2 n ≅ (zα / 2 + z β ) σ 1 + σ 2 = (1.96 + 1.28) 3 + 3 = 12 2 2 δ ( 4) 2 ( ) ( ) Use n1 = n2 = 12 4. 10‐9 Catalyst 1 Catalyst 2 x1 = 65.22 σ1 = 3 x2 = 68.42 σ2 = 3 n1 = 10 n2 = 10 a) 95% confidence interval on μ1 − μ 2 , the difference in mean active concentration ( x1 − x2 ) − zα / 2 2 σ1 σ 2 σ2 σ2 + 2 ≤ μ1 − μ 2 ≤ ( x1 − x2 ) + zα / 2 1 + 2 n1 n2 n1 n2 ( 65.22 − 68.42) − 1.96 ( 3) 2 (3) 2 ( 3) 2 ( 3) 2 + ≤ μ1 − μ 2 ≤ ( 65.22 − 68.42) + 1.96 + 10 10 10 10 . −583 ≤ μ1 − μ 2 ≤ −0.57 We are 95% confident that the mean active concentration of catalyst 2 exceeds that of catalyst 1 by between 0.57 and 5.83 g/l. p‐value: z0 = ( x1 − x2 ) − Δ 0 σ 12 n1 + 2 σ2 = (65.22 − 68.42) 32 10 + 32 10 = −2.38 n2 Then p‐value=2*0.008656=0.0173 b) Yes, because the 95% confidence interval did not contain the value 0, we would conclude that the mean active concentration depends on the choice of catalyst. c) ⎛ ⎞ ⎛ ⎜ ⎟ ⎜ (5) ⎟ (5) ⎜ ⎜ − Φ⎜ − 1.96 − β = Φ⎜1.96 − 2 ⎟ 3 32 ⎟ 32 32 ⎜ ⎜ +⎟ + ⎜ ⎜ 10 10 ⎠ 10 10 ⎝ ⎝ = Φ (− 1.77 ) − Φ (− 5.69 ) = 0.038364 − 0 = 0.038364 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Power = 1 − β = 1− 0.038364 = 0.9616. d) Based on the power, it appears that the sample sizes are adequate to detect the difference of 5. From the normal probability plots, we see they are approximately normally distributed. 5. 10‐17 a) 1) The parameter of interest is the difference in mean catalyst yield, μ1 − μ 2 , with Δ0 = 0 2) H0 : μ1 − μ2 = 0 or μ1 = μ 2 3) H1 : μ1 − μ 2 < 0 or μ1 < μ 2 4) α = 0.01 5) The test statistic is t0 = ( x1 − x2 ) − Δ 0 sp 1 1 + n1 n2 6) Reject the null hypothesis if t0 < − t α , n1 + n 2 − 2 where − t 0.01, 25 = −2.485 2 ( n1 − 1)s1 + ( n2 − 1)s2 2 n1 + n2 − 2 7) x1 = 86 x2 = 89 sp = s1 = 3 s2 = 2 n1 = 12 n2 = 15 2 2 = 11(3) + 14(2) = 2.4899 25 t0 = (86 − 89) 1 1 2.4899 + 12 15 = −3.11 8) Because −3.11 < −2.485, reject the null hypothesis and conclude that the mean yield of catalyst 2 significantly exceeds that of catalyst 1 at α = 0.01. b) 99% confidence interval: t0.005,25 = 2.787 ( x1 − x2 ) − t α / 2 , n1 + n 2 − 2 ( sp ) 1 1 1 1 + ≤ μ1 − μ 2 ≤ ( x1 − x2 ) + t α / 2 , n1 + n 2 − 2 (sp ) + n1 n2 n1 n2 (86 − 89 ) − 2 . 787 ( 2 . 4899 ) − 5.688 ≤ μ1 − μ 2 ≤ −0.3122 1 1 1 1 + ≤ μ 1 − μ 2 ≤ (86 − 89 ) + 2 .787 ( 2 . 4899 ) + 12 15 12 15 5.688 We are 99% confident that the mean yield of catalyst 2 exceeds that of catalyst 1 by between 0.3122 and 6. 10‐24 a) 1) The parameter of interest is the difference in mean coating thickness, μ1 − μ2 , with Δ0 = 0. 2) H0 : μ1 − μ2 = 0 3) H1 : μ1 − μ 2 > 0 4) α = 0.01 5) The test statistic is t0 = ( x1 − x2 ) − δ 2 s1 s2 +2 n1 n2 6) Reject the null hypothesis if t0 > t 0.01,18 where t 0.01,18 = 2.552 since 2 ⎛ s 12 s2 ⎜ ⎜n +n 2 ⎝1 2 ν= ⎞ ⎟ ⎟ ⎠ 2 2 ⎛ s 12 ⎞ ⎛ s2 ⎞ ⎜ ⎜ ⎜n ⎟ ⎟ ⎜n ⎟ ⎟ ⎝ 1⎠ +⎝ 2⎠ n1 − 1 n2 − 1 18 . 37 ν ≅ 18 (truncated) 7) x1 = 103.5 x2 = 99.7 s1 = 10.2 s2 = 20.1 n1 = 11 n2 = 13 t0 = (103.5 − 99.7) (10.2) 2 (20.1) 2 + 11 13 = 0.597 8) Because 0.597 < 2.552, do not reject the null hypothesis and conclude that increasing the temperature does not significantly reduce the mean coating thickness at α = 0.01. P‐value = P(t > 0.597), 0.25 < P‐value < 0.40 b) If α = 0.01, construct a 99% two‐sided confidence interval on the difference in means. Here, t0.005,19 = 2.878 (x1 − x2 ) − tα / 2 ,ν 2 2 s1 s 2 + ≤ μ 1 − μ 2 ≤ ( x1 − x 2 ) + tα / 2 ,ν n1 n2 2 2 s1 s 2 + n1 n2 (103 . 5 − 99 . 7 ) − 2 . 878 (10 . 2 ) 2 ( 20 . 1) 2 + ≤ μ 1 − μ 2 ≤ (103 . 5 − 99 . 7 ) − 2 . 878 11 13 (10 . 2 ) 2 ( 20 . 1) 2 + 11 13 − 14.52 ≤ μ1 − μ 2 ≤ 22.12 Because the interval contains 0, there is not a significant difference in the mean coating thickness between the two temperatures; that is, raising the process temperature does not significantly reduce the mean coating thickness. ...
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This note was uploaded on 09/02/2010 for the course ISYE 2028 taught by Professor Shim during the Summer '07 term at Georgia Institute of Technology.

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