# hw4_solution - Statistics HW 4 solution 14.27 1 Burning...

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Statistics HW4 solution 14.27 μ 1 : Burning time of flare of brand A μ 2 : Burning time of flare of brand B μ 3 : Burning time of flare of brand C μ 4 : Burning time of flare of brand D (a) 2 1 0 : H 3 = 4 : 1 H At least two means differ. Required Conditions: (1) Assume the four population ~ND (2) 2 2 1 2 0 : H 3 2 = 4 2 : 1 H At least two variances differ. As the p-value = 0.27 > 0.05, we couldn’t reject the null hypothesis. That is, we could conclude with 95% confidence that there is insufficient evidence to say the variances of burning time of flare differ. (3) Observations of the four brands are independent. ANOVA 變源 SS 自由度 MS F P- 臨界值 組間 662.675 3 220.8917 3.556873 0.023642 2.866266 組內 2235.7 36 62.10278 總和 2898.375 39 F = 3.56, p-value = .0236. p-value < reject H0. There is enough evidence to infer that differences exist between the flares with respect to burning times. (b) LSD method: C = 4(3)/2 = 6, E = .05, C / E .0083 794 . 2 t t 36 , 0042 . k n , 2 / (from Excel)

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LSD = j i k n , 2 / n 1 n 1 MSE t = 85 . 9 10 1 10 1 11 . 62 794 . 2 The means of flares C and D differ. (c) Tukey’s method: ) , k ( q 05 . q (4, 36) 3.79 = 3.79 10 11 . 62 = 9.45 Multiple Comparisons LSD Omega Treatment Treatment Difference Alpha = 0.0083 Alpha = 0.05 Flare A Flare B 4.3 9.845272985 9.444842 Flare C -0.2 9.845272985 9.444842 Flare D 9.8 9.845272985 9.444842 Flare B Flare C -4.5 9.845272985 9.444842 Flare D 5.5 9.845272985 9.444842 Flare C Flare D 10 9.845272985 9.444842 The means of flares A and D, and C and D differ. 14.29 (a) μ 1 : Average time that engineer can receive a promotion in small firms μ 2 : Average time that engineer can receive a promotion in median firms μ 3 : Average time that engineer can receive a promotion in large firms 2 1 0 : H 3 : H 1 At least two means differ.
Required conditions Except large firms , its approximate ~ND 2 2 1 2 0 : H 3 2 : 1 H At least two variances differ. 2 7 12 8 1 0 0 5 10 15 30 45 60 75 90 其他 頻率 time small firms 2 5 10 8 4 1 0 0 2 4 6 8 10 12 30 40 50 60 70 80 其他 time medium firms 1 2 6 9 12 0 0 5 10 15 20 30 40 50 60 其他 time large firms

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As the p-value = 0.55 > 0.05, we couldn’t reject the null hypothesis. That is, we could conclude
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## This note was uploaded on 01/06/2012 for the course MANAGEMENT 000 taught by Professor 游啟璋 during the Fall '11 term at National Taiwan University.

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hw4_solution - Statistics HW 4 solution 14.27 1 Burning...

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