hw3-solution.pdf

# F test for all pairs of factor level means whether or

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f. Test for all pairs of factor level means whether or not they di er: use the Tukey procedure with = . 05. Set up groups of factor levels whose means do not di er. Solution: a. The line plot of the estimated factor level means ¯ Y i . is shown below: This plot suggests that the prior physical fitness has some e ects on the mean time required in therapy, since the three points on the plot are not very close to each other. 4

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b. Since ¯ Y 2 . = 32, s ( ¯ Y 2 . ) = p MS E / n 2 = 1 . 407 and t (1 - / 2; n T - I ) = t ( . 995; 21) = 2 . 831, the 99 percent confidence interval of μ 2 is: ¯ Y 2 . ± s ( ¯ Y 2 . ) t (1 - / 2; n T - I ) = 32 ± 1 . 407(2 . 831) = [28 . 01 , 35 . 99] . c. Since ˆ D 1 = ¯ Y 2 . - ¯ Y 3 . = 8, ˆ D 2 = ¯ Y 1 . - ¯ Y 2 . = 6, s ( ˆ D 1 ) = p MS E (1 / n 2 + 1 / n 3 ) = 2 . 298, s ( ˆ D 2 ) = p MS E (1 / n 1 + 1 / n 2 ) = 2 . 111, and B = t (1 - 2 g ; n T - I ) = t (1 - . 05 2(2) ; 21) = 2 . 414, the 95 percent family confidence intervals of D 1 and D 2 are: ˆ D 1 ± s ( ˆ D 1 ) · B = 8 ± 2 . 298(2 . 414) = [2 . 45 , 13 . 55] , ˆ D 2 ± s ( ˆ D 2 ) · B = 6 ± 2 . 111(2 . 414) = [ . 90 , 11 . 10] . The interval estimate for D 1 is positive providing evidence that above average physical fitness ac- celerates rehabilitation time when compared to avg. physical fitness. The interval estimate for D 2 is positive providing evidence that average physical fitness will have quicker rehabilitation time when compared to below avg. physical fitness. d. T = 1 p 2 q (1 - ; I , n T - I ) = 1 p 2 q ( . 95; 3 , 21) = 2 . 52. Since T = 2 . 52 > 2 . 414 = B , the Tukey procedure is less e ffi cient to use in part(c). If you are using the table B.9, the tukey multiplier is 1 p 2 q ( . 95; 3 , 24) < T < 1 p 2 q ( . 95; 3 , 20) = 2 . 5 < T < 2 . 53 q ( . 95; 3 , 21) since it is not precisely at 2 = 21. Still T > B . e. The B multiple need to be modified, but not the T multiple in the Tukey procedure. f. Let D 1 = μ 2 - μ 3 , D 2 = μ 1 - μ 2 , D 3 = μ 1 - μ 3 . Since ˆ D 3 = ¯ Y 1 . - ¯ Y 3 . = 14, s ( ˆ D 3 ) = p MS E (1 / n 1 + 1 / n 3 ) = 2 . 404, T = 2 . 52, and the rest statistics are calculated in part (c). The 95 percent simultaneous Tukey confidence intervals of D 1 , D 2 and D 3 are: ˆ D 1 ± s ( ˆ D 1 ) · T = 8 ± 2 . 298(2 . 52) = [2 . 21 , 13 . 79] , ˆ D 2 ± s ( ˆ D 2 ) · T = 6 ± 2 . 111(2 . 52) = [ . 68 , 11 . 32] , ˆ D 3 ± s ( ˆ D 3 ) · T = 14 ± 2 . 404(2 . 52) = [7 . 94 , 20 . 06] .
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• Spring '13
• RaissaD'Souza
• Statistics, Student's t-distribution, Credible interval

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