19SplitLME - ANOVA ANALYSIS OF A BALANCED SPLIT-PLOT...

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Unformatted text preview: ANOVA ANALYSIS OF A BALANCED SPLIT-PLOT EXPERIMENT I Example: the corn genotype and fertilization response study I Main plots: genotypes, in blocks I Split plots: fertilization I 2 way factorial treatment structure I split plot variability nested in main plot variability nested in blocks. c 2011 Dept. Statistics (Iowa State University) Stat 511 section 19 1 / 22 Field Block 1 Block 2 Block 3 Block 4 Genotype A Genotype B Genotype C Genotype A Genotype B Genotype C Genotype A Genotype B Genotype C Genotype A Genotype B Genotype C 50 100 150 50 100 150 150 100 50 150 100 50 100 50 150 100 50 150 100 150 50 50 100 150 50 100 150 150 50 100 150 100 50 50 150 100 Plot Split Plot or Sub Plot c 2011 Dept. Statistics (Iowa State University) Stat 511 section 19 2 / 22 I A model: y ijk = μ ij + b k + w ik + e ijk μ ij =Mean for Genotype i , Fertilizer j Genotype i = 1 , 2 , 3 , Fertilizer j = 1 , 2 , 3 , 4 , Block k = 1 , 2 , 3 , 4 u = b 1 . . . b 4 w 11 w 21 . . . w 34 = b w ∼ N , σ 2 b I σ 2 w I u ∼ N , G σ 2 e I c 2011 Dept. Statistics (Iowa State University) Stat 511 section 19 3 / 22 Source DF Blocks 4- 1 = 3 Genotypes 3- 1 = 2 Blocks × Geno ( 4- 1 )( 3- 1 ) = 6 whole plot × u Fert 4- 1 = 3 Geno × Fert ( 3- 1 )( 4- 1 ) = 6 Block × Fert ( 4- 1 )( 4- 1 ) + Blocks × Geno × Fert +( 4- 1 )( 3- 1 )( 4- 1 ) 27 split plot × u C . Total 48- 1 47 c 2011 Dept. Statistics (Iowa State University) Stat 511 section 19 4 / 22 Source Sum of Squares Blocks ∑ 3 i = 1 ∑ 4 j = 1 ∑ 4 k = 1 (¯ y .. k- ¯ y ... ) 2 Geno ∑ 3 i = 1 ∑ 4 j = 1 ∑ 4 k = 1 (¯ y i ..- ¯ y ... ) 2 Blocks × Geno ∑ 3 i = 1 ∑ 4 j = 1 ∑ 4 k = 1 (¯ y i . k- ¯ y i ..- ¯ y .. k + ¯ y ... ) 2 Fert ∑ 3 i = 1 ∑ 4 j = 1 ∑ 4 k = 1 (¯ y . j .- ¯ y ... ) 2 Geno × Fert ∑ 3 i = 1 ∑ 4 j = 1 ∑ 4 k = 1 (¯ y ij .- ¯ y i ..- ¯ y . j . + ¯ y ... ) 2 error ∑ 3 i = 1 ∑ 4 j = 1 ∑ 4 k = 1 (¯ y ijk- ¯ y i . k- ¯ y ij . + ¯ y i .. ) 2 C . Total ∑ 3 i = 1 ∑ 4 j = 1 ∑ 4 k = 1 (¯ y ijk- ¯ y ... ) 2 c 2011 Dept. Statistics (Iowa State University) Stat 511 section 19 5 / 22 E ( MS GENO ) = FB G- 1 ∑ G i = 1 E (¯ y i ..- ¯ y ... ) 2 = FB G- 1 ∑ G i = 1 E (¯ μ i .- ¯ μ.. + ¯ w i .- ¯ w .. + ¯ e i ....
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This note was uploaded on 02/11/2012 for the course STAT 511 taught by Professor Staff during the Spring '08 term at Iowa State.

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19SplitLME - ANOVA ANALYSIS OF A BALANCED SPLIT-PLOT...

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