On Synchronized Permutation tests in Two-way ANOVA Dario Basso 1 , - PowerPoint PPT Presentation

On Synchronized Permutation tests in Two-way ANOVA Dario Basso 1 , Luigi Salmaso 1 , Fortunato Pesarin 2 1 : Department of Management and Engineering, University of Padova 2 : Department of Statistics, University of Padova dario@stat.unipd.it mODa8, June 3 - 8, 2007 - Almagro – p.1/29

The I × J Balanced Design In I × J replicated designs, the responses are assumed to follow the linear model: y ijk = µ + α i + β j + γ ij + ε ijk where: α i is the effect of the i th level of Factor A, i = 1 , . . . , I ; β j is the effect of the j th level of Factor B, j = 1 , . . . , J ; γ ij is the effect of the ij th level of interaction AB; ε ijk ( k = 1 , . . . , n ) are i.i.d. random errors with: X X X X E [ ε ijk ] = 0 ; V ar [ ε ijk ] = σ 2 . The effects are assumed to satisfy the side-conditions: ∀ j, ∀ i. α i = β j = 0; γ ij = 0 γ ij = 0 i j i j – p.2/29

Representation of the response’s structure Factor B . . . B j . . . B h . . . . . . ... ... . . . . . . . . . A i . . . µ + α i + β j + γ ij + ε ijk . . . µ + α i + β h + γ ih + ε ihk . . . . . . ... ... ... . . . Factor A . . . A s . . . µ + α s + β j + γ sj + ε sjk . . . µ + α s + β h + γ sh + ε shk . . . . . . ... ... . . . . . . . . . – p.3/29

Hypotheses of Interest H 0 A : α i = 0 ∀ i vs. H 1 A : ∃ α i � = 0 ∀ j H 1 B : ∃ β j � = 0 H 0 B : β j = 0 vs. H 0 AB : γ ij = 0 ∀ i, j H 1 AB : ∃ γ ij � = 0 vs. The aim is to find three uncorrelated test statistics (saparately) for each null hypothesis. – p.4/29

Normality Assumption: The Parametric Solution P P P P If ε ijk ∼ N (0 , σ 2 ) , define: � � 2 /IJ ( n − 1) P P P y ··· ] 2 / ( I − 1) MS A = i [¯ y i ·· − ¯ y ··· ] 2 / ( J − 1) y · j · − ¯ MS B = i [¯ y ··· ] 2 / ( I − 1)( J − 1) MS AB = j [¯ y ij · − ¯ y i ·· − ¯ y · j · + ¯ i MS ε = y ijk − ¯ y ij · i j k F A = MS A /MS ε ∼ F I − 1 ,IJ ( n − 1) if H 0 A is true. F B = MS B /MS ε ∼ F J − 1 ,IJ ( n − 1) if H 0 B is true. F AB = MS AB /MS ε ∼ F ( I − 1)( J − 1) ,IJ ( n − 1) if H 0 AB is true. Can the Normality if errors be assumed? The two-way ANOVA test statistics are positively correlated . – p.5/29

Correlation Among Two-Way ANOVA Test Statistics The test statistics for main factors: F A = MS A /MS ε F B = MS B /MS ε are positively correlated. In fact: COV [ F A , F B ] = E [ F A · F B ] − E [ F A ] E [ F B ] E [ MS A ] E [ MS B ] E [ MS − 2 = ] + ε ] 2 E [ MS A ] E [ MS B ] E [ MS − 1 − ε E [ MS A ] E [ MS B ] V ar [ MS − 1 = ] > 0 . ε – p.6/29

Correlation Among Two-Way ANOVA Test Statistics p It can be proved that the correlation between F A and F B is: � α 2 + [( I − 1) + 2 α ][( J − 1) + IJ ( n − 1) − 2] � ; ρ ( F A , F B ) = σ 4 [( I − 1) + α ][( J − 1) + β ] � � σ 2 ( F A )˜ σ 2 ( F B ) ˜ P where: P σ 2 ( F A ) = σ 4 ˜ β 2 + [( J − 1) + 2 β ][( I − 1) + IJ ( n − 1) − 2] σ 2 ( F B ) = σ 4 ˜ . i α 2 i /σ 2 α = j β 2 j /σ 2 β = – p.7/29

Error Exchangeability Assumption: Nonparametric Solution A jointly sufficient statistic for the three null hypotheses is: Y = [ y 11 , y 12 , . . . , y ij , . . . , y IJ ] where y ij is the vector of n units in block A i B j . ⇒ Units are not exchangeable between blocks, unless H 0 A ∩ H 0 B ∩ H 0 AB is true. ⇒ To separately test for factor A effects we need that exchangeability is satisfied under H 0 A regardless H 0 B ∪ H 0 AB is true or not. Therefore.... – p.8/29

Permuting within columns Factor B . . . B j . . . B h . . . . . . ... ... . . . . . . . . . A i . . . y ij 1 y ij 2 . . . y ijn . . . y ih 1 y ih 2 . . . y ihn . . . . . � ν ∗ � ν ∗ � is | j /n � is | h /n � Factor A . A s . . . y sj 1 y sj 2 . . . y sjn . . . y sh 1 y sh 2 . . . y shn . . . . . . ... ... . . . . . . . . . If ν ∗ is | j � = ν ∗ is | h no proper solution is possible. – p.9/29

Testing Factor A X X Suppose that ν ∗ is | j units are exchanged between block A i B j and A s B j , j = 1 , . . . , J . X X The totals of blocks A i B j and A s B j have the following structure: n n y ∗ ijk = ν ∗ is | j [ µ + α s + β j + γ sj ] + ( n − ν ∗ ε ∗ is | j ) [ µ + α i + β j + γ ij ] + ijk , k k n n X X y ∗ sjk = ν ∗ is | j [ µ + α i + β j + γ ij ] + ( n − ν ∗ ε ∗ is | j ) [ µ + α s + β j + γ sj ] + sjk . k k Let us consider the structure of the difference of these totals, to compare α i and α s at level B j : X X a T ∗ y ∗ y ∗ sjk = ( n − 2 ν ∗ is | j )[ α i − α s + γ ij − γ sj ] + ε ∗ ijk − is | j = is | j k k Similarly, at level B j : a T ∗ y ∗ y ∗ shk = ( n − 2 ν ∗ is | h )[ α i − α s + γ ih − γ sh ] + ε ∗ is | h = ihk − is | h k k – p.10/29

Testing Factor A X X X Because of the side-conditions on γ ’s, the sum over all columns: a T ∗ ( n − 2 ν ∗ ε ∗ is | j = is | j )[ α i − α s + γ ij − γ sj ] + is | j , j j j X X only depends on effects α i and α s and on exchangeable errors if and only if ν ∗ is | j = ν ∗ is , ∀ j . We have to synchronize the permutations with respect to the levels of factor B. When ν ∗ is | j = ν ∗ is , ∀ j , X X a T ∗ is | j = ( n − 2 ν ∗ ε ∗ is ) J [ α i − α s ] + is | j , j j and (when ν ∗ is = ν ∗ ) the test statistic: a T ∗ a T ∗ is | j ] 2 , A = [ i<s j only depends on effects of factor A and on exchangeable errors, hence it is a separate exact test on factor A. – p.11/29

Permuting within rows Factor B . . . B j . . . B h . . . . . . ... ... . . . ⇐ ⇒ . . . A i . . . y ij 1 y ij 2 . . . y ijn ⇐ ⇒ y ih 1 y ih 2 . . . y ihn . . . ν ∗ jh | i /n . . . ... ... . . . ⇐ ⇒ Factor A . . . A s . . . y sj 1 y sj 2 . . . y sjn ⇐ ⇒ y sh 1 y sh 2 . . . y shn . . . ν ∗ jh | s /n . . . ... ... . . . . . ⇐ ⇒ . If ν ∗ jh | i � = ν ∗ jh | s no proper solution is possible. – p.12/29

Testing Factor B P P Let: b T ∗ jh | i = ( n − 2 ν ∗ k ε ∗ k ε ∗ jh | i ) [ β j − β h + γ ij − γ ih ] + ijk − ihk " # be the difference between the totals of blocks A i B j and A i B h after we have X X X X exchanged ν ∗ jh | i units (out of n ); if ν ∗ jh | i = ν ∗ jh ∀ i , because of the side-conditions on γ ’s, the sum: I I b T ∗ jh | i = ( n − 2 ν ∗ ε ∗ ε ∗ jh ) I [ β j − β h ] + ijk − ihk " # 2 i =1 X X i =1 k k Similarly, (when ν ∗ jh = ν ∗ ) by interchanging indeces and by applying synchronized permutations within rows, the test statistic: b T ∗ b T ∗ B = jh | i i j<h allows for nonparametric exact testing on Factor B. – p.13/29

Testing the Interaction Consider the synchronized permutations within columns; recall: a T ∗ ( n − 2 ν ∗ is )[ α i − α s + γ ij − γ sj ] + ε ∗ = is | j is | j a T ∗ ( n − 2 ν ∗ is )[ α i − α s + γ ih − γ sh ] + ε ∗ = is | h . is | h The difference: h i 2 X X a T ∗ is | j − a T ∗ is | h = ( n − 2 ν ∗ is )[ γ ij − γ sj − γ ih + γ sh ] + ε ∗ is | j − ε ∗ is | h only depends on the interaction effects, therefore (when ν ∗ is = ν ∗ ), the test statistic: a T ∗ a T ∗ is | j − a T ∗ AB = is | h h i 2 X X i<s j<h allows for nonparametric exact testing on Interaction. By interchanging indeces (and by synchronizing within rows), there is another test statistic: b T ∗ b T ∗ jh | i − b T ∗ AB = jh | s j<h i<s which allows for nonparametric exact testing on Interaction. – p.14/29

Constrained Synchronized Permutations (CSPs) Factor B . . . B j . . . B h . . . . . . ... ... . . . . . . . . . A i . . . y ij 1 y ij 2 . . . y ijn . . . y ih 1 y ih 2 . . . y ihn . . . . . . . . . . . � ν ∗ � ν ∗ � ν ∗ � ν ∗ � ν ∗ Factor A . A s . . . y sj 1 y sj 2 . . . y sjn . . . y sh 1 y sh 2 . . . y shn . . . . . . . . . . . . ... ... . . . . . . . . . Note that ν ∗ is itself a random variable. In this example ν ∗ = 1 . ν ∗ units from the same original positions are exchanged between each pair of blocks. – p.15/29

Unconstrained Synchronized Permutations (USPs) Factor B . . . B j . . . B h . . . . . . ... ... . . . . . . . . . A i . . . y ij 1 y ij 2 . . . y ijn . . . y ih 1 y ih 2 . . . y ihn . . . . . . . . . . . � ν ∗ � ν ∗ � ν ∗ � ν ∗ � ν ∗ Factor A . A s . . . y sj 1 y sj 2 . . . y sjn . . . y sh 1 y sh 2 . . . y shn . . . . . . . . . . . . ... ... . . . . . . . . . Note that ν ∗ is itself a random variable. In this example ν ∗ = 1 . ν ∗ units are exchanged between each pair of blocks (regardless their original positions). – p.16/29

Cardinality of CSPs and USPs � 2 n � CSPs depend on the rearrangement made in the 1 st pair of blocks ( A i B 1 and A s B 1 ), hence the cardinality of CSPs is: C CSP = � n � 2 × J × I ( I − 1) / 2 n X USPs depend on all possible ways to exchange ν ∗ units between pairs of blocks for all columns and all possible pairs of rows, hence: n C USP = ν ∗ ν ∗ =0 – p.17/29