sol_practicefinal_STA106_fall2018.pdf

# Y 1 0 0735 y 2 0 1905 y 3 0 4600 y 4 0 3655 y 5 0

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Y 1 · = 0 . 0735 , Y 2 · = 0 . 1905 , Y 3 · = 0 . 4600 , Y 4 · = 0 . 3655 , Y 5 · = 0 . 1250 , Y 6 · = 0 . 1515 . (a) Which model shall you use and why? Write down the model equation and model assumptions. (b) Given the total sum of squares being 5.82, derive the ANOVA table. You need to show how you get each number for full credit. (c) Test whether or not all machines of this company have the same mean fill at significance level 0 . 05. State the null and alternative hypotheses, decision rule and the conclusion. (d) Construct a 95% confidence interval for μ · , the overall mean fill for all machines of this company. (e) Construct an approximate 95% confidence interval for σ 2 μ , the machine mean filling variance, by the Satterthwaite procedure. (Hint: Round the degrees of freedom to the nearest integer. For example, if you get a degrees of freedom 1 . 89, then just use 2.) (f) Construct a 90% confidence interval for the error variance σ 2 . Solution: (a) We should use a single-factor random effects model (i.e., single-factor ANOVA model II), since there is only one factor in this study (“machine”) and its six levels are randomly sampled. (b) r = 6, n = 20, n T = rn = 120, SSTO = 5 . 82. Thus, ¯ Y .. = 1 r r X i =1 ¯ Y i. = 0 . 2277 , SSTR = n r X i =1 ( ¯ Y i. - ¯ Y .. ) 2 = 20(0 . 1145) = 2 . 29 , SSE = SSTO - SSTR = 5 . 82 - 2 . 29 = 3 . 53 . The ANOVA table is shown below: Source of Variation SS df MS Between machines 2.29 r - 1 = 5 2.29/5=0.458 Error 3.53 n T - r = 114 3.53/114=0.0310 Total 5.82 n T - 1 = 119 (c) H 0 : σ 2 μ = 0 vs. H a : σ 2 μ > 0. Use the F ratio F * = MSTR MSE , and reject H 0 at level α if F * > F (1 - α ; r - 1 , n T - r ), otherwise accept H 0 . Now r = 6 , n T = 120, α = 0 . 05, and F * = 0 . 458 0 . 0310 = 14 . 77 , 8

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F (0 . 95; 5 , 114) = 2 . 29 . Since 14 . 77 > 2 . 29, we reject H 0 at level 0 . 05 and we conclude that the machines of this company have significantly different mean fills. (d) Note that (1 - α )100% C.I. for μ · : ¯ Y .. ± t (1 - α/ 2; r - 1) s ( ¯ Y .. ), we know r = 6 , n = 20 , ¯ Y .. = 0 . 2277., s ( ¯ Y .. ) = r 0 . 458 6 × 20 = 0 . 0618 , t (0 . 975 , 5) = 2 . 571 95% C.I. for μ · : [0 . 2277 - 2 . 571 × 0 . 0618 , 0 . 2277 + 2 . 571 × 0 . 0618] We are 95% confident that the mean of the deviation of the filling weights across the each cartons is between (0 . 0688 , 0 . 3866) ounces. (e) s 2 μ = MSTR - MSE n = 0 . 458 - 0 . 0310 20 = 0 . 0214 , df = ( ns 2 μ ) 2 MST R 2 r - 1 + MSE 2 n T - r = 0 . 427 2 0 . 458 2 5 + 0 . 0310 2 114 = 4 . 345 . An approximate 95% confidence interval for σ 2 μ : " df × s 2 μ χ 2 (1 - α 2 ; df ) , df × s 2 μ χ 2 ( α 2 ; df ) # 4 . 345(0 . 0214) χ 2 ( . 975; 4) , 4 . 345(0 . 0214) χ 2 ( . 025; 4) = 4 . 345(0 . 0214) 11 . 14 , 4 . 345(0 . 0214) 0 . 484 = [0 . 0083 , 0 . 1921] . (f) Note that (1 - α )100%-C.I for σ 2 h ( n T - r ) MSE χ 2 (1 - α 2 , n T - r ) , ( n T - r ) MSE χ 2 ( α 2 , n T - r ) i Known MSE = 0 . 0310 , n T = 120 , r = 6 with α = 0 . 1, χ 2 ( α 2 , n T - r ) = χ 2 (0 . 05; 114) = 90 . 351 χ 2 (1 - α 2 , n T - r ) = χ 2 (0 . 95; 114) = 139 . 921 Then, 90%-C.I for σ 2 h (120 - 6)0 . 0310 χ 2 (0 . 95 , 114) , (120 - 6)0 . 0310 χ 2 (0 . 05 , 114) i = [0 . 0253 , 0 . 0391] We are 90% confident that the error variance σ 2 is between (0.0253, 0.0391). 9
Table 1: Studentized range distribution table: q (0 . 94; r, ν ) r \ ν 15 16 17 18 19 20 4 3.94 3.91 3.89 3.87 3.85 3.83 5 4.23 4.20 4.17 4.15 4.12 4.10 6 4.45 4.42 4.39 4.36 4.34 4.32 7 4.64 4.60 4.57 4.54 4.52 4.49 Table 2: t table: t ( A ; ν ) A \ ν 5 10 12 14 16 18 20 0.90 1.476 1.372 1.3567 1.345 1.337 1.330 1.325 0.94 1.873 1.700 1.674 1.656 1.642 1.632 1.624 0.975 2.571 2.228 2.179 2.145 2.120 2.101 2.086 0.99 3.365 2.764 2.681 2.624 2.583 2.552 2.528 Table 3: F table: F (0 . 94; ν 1 , ν 2 ) r \ ν 15 16 17 18 19 20 4 2.87 2.83 2.79 2.76 2.73 2.70 5 2.73 2.69 2.65 2.62 2.59 2.56 6 2.63 2.59 2.55 2.52 2.49 2.46 7 2.56 2.52 2.48 2.44 2.41 2.39 10

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Table 4: F table: F (0 . 95; ν 1 , ν 2 ) ν 1 \ ν 2 50 55 60 65 70 114 5 2.400 2.383 2.368 2.356 2.346 2.294 10 2.026 2.008 1.993 1.980 1.969 1.915 12 1.952
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• Spring '13
• Random effects model, SSAB

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