12+Analysis+of+variance

# 5 73 b c 143 35 b d 63 171 c d 152

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Unformatted text preview: − µC , µ A − µD , µB − µC , µ B − µD , µ C − µD . Calculate s = 907.1/36 = 5.02, use α = 0.01, α/c = 0.05/6 = 0.0167, tα/(2c) = t0.0167 ≈ 2.42. The Bonferroni intervals are (with the probability P = 1−α/c ≈ 0.9833) µA − µB ∈ [−15.2, −4.4], µA − µC ∈ [−24.1, −13.2], µA − µD ∈ [−3.5, 7.3], µB − µC ∈ [−14.3, −3.5], µB − µD ∈ [6.3, 17.1], µC − µD ∈ [15.2, 26.0]. All intervals contain respective diﬀerences of µ with α = 0.1. We see that only interval for µA − µD contains 0 and does not support the conclusion that the brand’s mean distances diﬀer. One factor model as well as multifactor models are usually analyzed on a computer using, say, SP SS software. Randomized block design. One-way ANOVA tests for diﬀerence in the treatments, splitting the variation into two parts: due to treatments and due to errors. However, 6 A. Zhensykbaev ANOVA there are likely to be sources of variation other than the treatments. If another source of variation is removed, perhaps the treatments would now be seen to have a signiﬁcant eﬀect. This question is for two-way ANOVA. It allows a source of variation to be isolated before testing for treatments. First source of variation is referred to as the treatments, the second one as the block. The matched sets of experimental units are called blocks. As early we present these data in the table Treatment 1 Treatment 2 . . . Treatment k Mean Block 1 x11 x21 ... xk 1 xB1 ¯ Block 2 x12 x22 ... xk 2 xB2 ¯ . . . . . . . . . . . . . . . . . . Block b x1b x2b ... xkb xBb ¯ Mean xT1 ¯ xT2 ¯ ... xTk ¯ x ¯ ANOVA F test to compare treatment means for randomized block design. Assumptions: 1. The probability distribution of observations associated with all blocktreatment combinations are approximately normal with mean =0. 2. The variances of block-treatment distributions are equal. 3. The blocks are randomly selected and all treatments are applied to each block. Like before, the null hypothesis is H0 = {µ1 = · · · = µk }, alternative hypothesis is Ha = {at least two means diﬀer}, test statistics is F= M ST = SST , k−1 M ST , M SE M SE = 7 S...
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