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12+Analysis+of+variance

# The unbiased estimator of 2 being 1 nk ni k 2 xij

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Unformatted text preview: III-XI (McClave and Benson) with (k − 1) numerator degree of freedom and df = n − k for denominator. The unbiased estimator of σ 2 being 1 σ= ˆ n−k ni k 2 (xij − xi )2 . ¯ i=1 j =1 Example 2. Refer to the example 1. There are k = 4 groups of measurements for 4 level (brands of golf balls), ni = 10 (i = 1 : 4). Results are shown in the Table. Set up the test to compare the mean distances for the four brands. Use α = 0.05. 4 A. Zhensykbaev ANOVA # Brand A Brand B Brand C Brand D x1i x2i x 3i x4i 1 251.2 263.2 269.7 251.6 2 245.1 262.9 263.2 248.6 3 248.0 265.0 277.5 249.4 4 251.1 254.5 267.4 242.0 5 265.5 264.3 270.5 246.5 6 250.0 257.0 265.5 251.3 7 253.9 262.8 270.7 262.8 8 244.6 264.4 272.9 249.0 9 254.6 260.6 275.6 247.1 10 248.8 255.9 266.5 245.9 Mean x1 = 251.3 x2 = 261.1 x3 = 270 x4 = 249.4 ¯ ¯ ¯ ¯ Calculate means and than test statistic. x1 = 251.3, x2 = 261.1, x3 = 270, x4 = 249.4, x = 257.95, ¯ ¯ ¯ ¯ ¯ 4 10(¯i − x)2 = 2724.5, x ¯ SST = i=1 4 10 (xij − xi )2 = 907.1, ¯ SSE = i=1 j =1 F= 2724.5/3 SST /(k − 1) = = 36. SSE/(n − k − 1) 907.1/36 The quantity Fα = F0.05 ≈ 2.88 is deﬁned from the Table IX (ν1 = k − 1 = 3, ν2 = n − k = 36). Since 36>2.88 we reject H0 (the brand is essential). The Bonferroni procedure. This is a procedure to make multiple comparisons of a set of treatment means µ1 , ..., µk . This is usually done by constructing the set of conﬁdence intervals for all possible diﬀerences µi − µj of the overall conﬁdence α. 5 A. Zhensykbaev ANOVA So each conﬁdence interval will be of conﬁdence α/c, where c is the total number of all possible diﬀerences µi − µj . For each diﬀerence µi − µj the conﬁdence interval of the level α/c is (¯i − xj ) ± t 2αc · s · x ¯ 1 1 +, ni nj (i = j, i, j = 1 : k ), where tα/(2c) is deﬁned from Table VI with df = n − k , s= S SE/(n − k ). Example 3. Refer to the example 1. There are exactly c = 6 pairs of treatment means: µA − µB , µ A...
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