12+Analysis+of+variance

Ss total sse sst ssb i1 j 1 the reject region

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Unformatted text preview: SE , (k − 1)(b − 1) A. Zhensykbaev ANOVA where b - number of blocks, the sum of squares for error SSE = SS (Total) − SST − SSB, the sum of squares for treatments k SST = b(¯Ti − x), x ¯ i=1 the sum squares for blocks SSB is b (¯Bi − x)2 , x ¯ SSB = k i=1 the total sum of squares, SS (Total),is the sum of squared differences of each measurement from over-all mean (total number of measurement is n = b · k) k b (xij − x)2 . ¯ SS (Total) = SSE + SST + SSB = i=1 j =1 The reject region is RR = {F > Fα }, where Fα is defined from the Tables VIII-XI (McClave and Benson) with (k − 1) numerator degree of freedom and df = kb − k − b +1 = (k − 1)(b − 1) for denominator. Example 4. Let five observations on crop yield be taken from a different plot of land and that observation i always referred to plot i for all treatments. It is likely that the nature of the plot (slope, drainage, etc.) would affect the crop yield. Possibly the effect is such that the effect of the treatments is masked. Let us given the following data (on crop yield): Plot 1 Plot 2 Plot 3 Plot 4 Plot 5 Mean Treatment 1 Treatment 2 Treatment 3 Mean 4 3 7 4.67 7 2 5 4.67 5 4 6 5.0 6 8 4 6.0 3 3 8 4.67 5 4 6 Total=5.0 8 A. Zhensykbaev ANOVA Calculate test statistic. The sum squares for blocks: x = 5, xi are ¯ ¯ presented in last column of the table and SST = 5 · ((5 − 5)2 + (4 − 5)2 + (6 − 5)2 ) = 10, SSB = 3 · (3 · (4.67 − 5)2 + (5.0 − 5)2 + (6.0 − 5)2 ) = 4, The total sum of squares is 3 5 (xij − x)2 = (4 − 5)2 + (3 − 5)2 + · · · + (8 − 5)2 = 52, ¯ SS (Total) = i=1 j =1 SSE = SS (Total) − SST − SSB = 52 − 10 − 4 = 38. The results of an ANOVA can be summarized in a tabular format Variation Treatment df k−1=2 SS SST=10 Block b−1=4 SSB=4 Error Total (k − 1)(b − 1) SSE=38 =8 n − 1 = 14 SS (Total) =52 MS F M ST = M ST /M SE = 10/2=5 5/4.75 = 1.05 M SB = M SB/M SE = 4/4 = 1 1/4.75 = 0.21 M SE = 38/8 = 4.75 Let α = 0.05, than tα = t0.05 ≈ 4.46 (ν1 = 2, ν2 = 8). Since F = 1.05 < 4.4 = F0.05 we do not reject H0 (the fertilizers are not essential, even if one takes into account effects of blocks). Example 5. Teaching material for economics School 1 School 2 School 3 School 4 School 5 School 6 Mean Material A Material B Material C Mean 64 61 55 60 69 63 63 65 72 68 70 70 58 65 60 61 64 60 59 61 63 61 59 61 65 63 61 Total=63 9 A. Zhensykbaev ANOVA The results of an ANOVA we summarize in the table Variation Treatment df k−1=2 SS SST=48 Block b−1=5 SSB=222 Error Total (k − 1)(b − 1) SSE=74 =10 n − 1 = 17 SS (Total) =344 MS F M ST = M ST /M SE = 48/2=24 24/7.4 = 3.24 M SB = M SB/M SE = 222/5 = 44.4 44.4/7.4 = 6.0 M SE = 74/10 = 7.4 Let α = 0.05, than Fα = F0.05 ≈ 4.1 (ν1 = 2, ν2 = 10). Since F = 3.24 < 4.1 = F0.05 we do not reject H0 (no significant differences between the effects of the teaching materials, even if to take into account the possible difference between schools). 10...
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This note was uploaded on 02/11/2014 for the course MATH 1390 taught by Professor Christopherstocker during the Spring '13 term at Marquette.

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