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multcompf2010 - Statistics 514 Compare Treatment Means...

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Statistics 514: Compare Treatment Means Lecture 5: Comparing Treatment Means Montgomery: Sections 3.3-5 Page 1

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Statistics 514: Compare Treatment Means Linear Combinations of Treatment Means ANOVA Model: y ij = μ + τ i + ij ( τ i : treatment effect) = μ i + ij ( μ i : treatment mean) Linear combination with given coefficients c 1 , c 2 , . . . , c a : L = c 1 μ 1 + c 2 μ 2 + . . . + c a μ a = a X i =1 c i μ i , Want to test: H 0 : L = c i μ i = L 0 Examples: 1. Pairwise comparison: μ i - μ j = 0 for all possible i and j . 2. Compare treatment vs control: μ i - μ 1 = 0 when treatment 1 is a control and i = 2 , ..., a are new treatments. 3. General cases such as μ 1 - 2 μ 2 + μ 3 = 0 , μ 1 + 3 μ 2 - 6 μ 3 = 0 , etc. Page 2
Statistics 514: Compare Treatment Means Estimate of L : ˆ L = X c i ˆ μ i = X c i ¯ y i. Var ( ˆ L ) = X c 2 i Var y i. ) = σ 2 X c 2 i n i = σ 2 n X c 2 i Standard Error of ˆ L S . E . ˆ L = s MSE X c 2 i n i Test statistic t 0 = ( ˆ L - L 0 ) S.E. ˆ L t ( N - a ) under H 0 Page 3

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Statistics 514: Compare Treatment Means Example: Lambs Diet Experiment Recall there are three diets and their treatment means are denoted by μ 1 , μ 2 and μ 3 . Suppose one wants to consider L = μ 1 + 2 μ 2 + 3 μ 3 = 6 μ + τ 1 + 2 τ 2 + 3 τ 3 and test H 0 : L = 60 . data lambs; input diet [email protected]@; cards; 1 8 1 16 1 9 2 9 2 16 2 21 2 11 2 18 3 15 3 10 3 17 3 6 ; proc glm; class diet; model wtgain=diet; means diet; estimate ’l1’ intercept 6 diet 1 2 3; run; Page 4
Statistics 514: Compare Treatment Means Example: Lambs Diet Experiment SAS output Level of ------------wtgain----------- diet N Mean Std Dev 1 3 11.0000000 4.35889894 2 5 15.0000000 4.94974747 3 4 12.0000000 4.96655481 Dependent Variable: wtgain Standard Parameter Estimate Error t Value Pr > |t| l1 77.0000000 8.88506862 8.67 <.0001 t 0 = (77 . 0 - 60) / 8 . 89 = 1 . 91 P - value = P ( t ≤ - 1 . 91 or t 1 . 91 | t (12 - 3)) = . 088 Fail to reject H 0 : μ 1 + 2 μ 2 + 3 μ 3 = 60 at α = 5% . Page 5

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Statistics 514: Compare Treatment Means Contrasts Γ = a i =1 c i μ i is a contrast if a i =1 c i = 0 . Equivalently, Γ = a i =1 c i τ i . Examples 1. Γ 1 = μ 1 - μ 2 = μ 1 - μ 2 + 0 μ 3 + 0 μ 4 , c 1 = 1 , c 2 = - 1 , c 3 = 0 , c 4 = 0 Comparing μ 1 and μ 2 . 2. Γ 2 = μ 1 - 0 . 5 μ 2 - 0 . 5 μ 3 = μ 1 - 0 . 5 μ 2 - 0 . 5 μ 3 + 0 μ 4 c 1 = 1 , c 2 = - 0 . 5 , c 3 = - 0 . 5 , c 4 = 0 Comparing μ 1 and the average of μ 2 and μ 3 . Estimate of Γ : C = a i =1 c i ¯ y i. Page 6
Statistics 514: Compare Treatment Means Test: H 0 : Γ = 0 t 0 = C S . E . C t ( N - a ) t 2 0 = ( c i ¯ y i. ) 2 MSE c 2 i n i = ( c i ¯ y i. ) 2 / c 2 i /n i MSE = SS C / 1 MSE Under H 0 , t 2 0 F 1 ,N - a . Contrast Sum of Squares SS C = X c i y i. 2 / X ( c 2 i /n i ) SS C represents the amount of variation attributable Γ . Page 7

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Statistics 514: Compare Treatment Means SAS Code (cont.sas) Tensile Strength Example options ls=80; title1 ’Contrast Comparisons’; data one; infile ’c:\saswork\data\tensile.dat’; input percent strength time; proc glm data=one; class percent; model strength=percent; contrast ’C1’ percent 0 0 0 1 -1; contrast ’C2’ percent 1 0 1 -1 -1; contrast ’C3’ percent 1 0 -1 0 0; contrast ’C4’ percent 1 -4 1 1 1; Page 8
Statistics 514: Compare Treatment Means ___________________________________________________________ Dependent Variable: STRENGTH Sum of Mean Source DF Squares Square F Value Pr > F Model 4 475.76000 118.94000 14.76 0.0001 Error 20 161.20000 8.06000 Corrected Total 24 636.96000 Source DF

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