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1、(a) k = two tables of explanatory variables (b) k = three tables of explanatory variables (c) k = four tables of explanatory variables abc d Table 1Table 2 Y = ab c g fe d Table 1Table 2 Table 3 h Y = ab c gf e d Table 1Table 2 Table 3 hi j k l mn Table 4 o p The contributions of Table 4 are represe
2、nted by the two rectangles Y = Partitioning Y among k = 4 tables of explanatory variables Number of fractions: Total Resid. No Intersections between inters. 2 3 4 tables Cr k r0 = k 4! 0!4! - - 4! 1!3! - - 4! 2!2! - - 4! 3!1! - - 4! 4!0! - -+= 16 3 regression/canonical analyses and 3 subtraction equ
3、ations are needed to estimate the 4 (= 22) fractions. a and c and pairs containing a and c can be tested for significance (3 canonical analyses per permutation). Fraction b cannot be tested singly. 7 regression/canonical analyses and 10 subtraction equations are needed to estimate the 8 (= 23) fract
4、ions. a to c and subsets containing a to c can be tested for significance (4 canonical analyses per permutation to test a to c). Fractions d to g cannot be tested singly. 15 regression/canonical analyses and 27 subtraction equations are needed to estimate the 16 (= 24) fractions. a to d and subsets
5、containing a to d can be tested for significance (5 canonical analyses per permutation to test a to d). Fractions e to o cannot be tested singly. fractions (Pascal triangle) Partitioning diagrams Diagrams describing the partitioning of the variation of a response data table Y (rectangles) by (a) two
6、, (b) three, and (c) 4 tables of explanatory variables. The fraction names a to p in the output of programs Partition2, Partition3, and Partition4 follow the nomenclature in these Venn diagrams. Page Page 1 1 of of 1 1Dalmore:Programmes Dalmore:Programmes labo:Partitioning labo:Partitioning among.:U
7、sers among.:Users guide:Varpart2-3 guide:Varpart2-3 procedures procedures (65%).txt(65%).txt Printed: Printed: Lundi Lundi 28 28 novembre novembre 2005 2005 16:54:3016:54:30 Variation partitioning for two explanatory data tables - Table 1 with m1 explanatory variables, Table 2 with m2 explanatory va
8、riables Number of fractions: 4, called a . d indicates the 3 regression or canonical analyses that have to be computed. # Partial canonical analyses are only computed if tests of significance or biplots are needed. ComputeFittedResidualsDerived fractionsDegrees of freedom, numerator of F Y.1a+bc+d (
9、1)df(a+b) = m1 Y.2b+ca+d (2)df(b+c) = m2 Y.1,2a+b+cd(3)df(a+b+c) = m3 m1+m2 (there may be collinearity) # Y.1|2addf(a) = m3-m2 # Y.2|1cddf(c) = m3-m1 Partial analyses(4) a = a+b+c - b+cdf(a) = m3-m2* controlling for 1 table X (5) c = a+b+c - a+bdf(c) = m3-m1* (6) b = a+b + b+c - a+b+cdf(b) = m1+m2-(
10、m1+m2) = 0 (7) d = residuals = 1 - a+b+cdf2(d) = n-1-m3 for denominator of F * Calculation of d.f. for difference between nested models: see Sokal & Rohlf (1981, 1995) equation 16.14. Tests of significance - F(a+b) = (a+b/m1)/(c+d/(n-1-m1) F(b+c) = (b+c/m2)/(a+d/(n-1-m2) F(a+b+c) = (a+b+c/m3)/(d/(n-
11、1-m3) F(a) = (a/(m3-m2)/(d/(n-1-m3) F(c) = (c/(m3-m1)/(d/(n-1-m3) The only testable fractions are those that can be obtained directly by regression or canonical analysis. The non-testable fraction is b. That fraction cannot be obtained directly by regression or canonical analysis. - Variation partit
12、ioning for three explanatory data tables - Table 1 with m1 explanatory variables, Table 2 with m2 explanatory variables, Table3 with m3 explanatory variables Number of fractions: 8, called a . h indicates the 7 regression or canonical analyses that have to be computed. # Partial canonical analyses a
13、re only computed if tests of significance or biplots are needed. ComputeFittedResidualsDerived fractionsDegrees of freedom, numerator of F Direct canonical analysis Y.1a+d+f+gb+c+e+h (1)df(a+d+f+g) = m1 Y.2b+d+e+ga+c+f+h (2)df(b+d+e+g) = m2 Y.3c+e+f+ga+b+d+h (3)df(c+e+f+g) = m3 Y.1,2a+b+d+e+f+gc+h (
14、4)df(a+b+d+e+f+g) = m4 m1+m2 (collinearity?) Y.1,3a+c+d+e+f+gb+h (5)df(a+c+d+e+f+g) = m5 m1+m3 (collinearity?) Y.2,3b+c+d+e+f+ga+h (6)df(b+c+d+e+f+g) = m6 m2+m3 (collinearity?) Y.1,2,3 a+b+c+d+e+f+gh (7)df(a+b+c+d+e+f+g) = m7 m1+m2+m3 (collinearity?) # Y.1|2a+fc+hdf(a+f) = m4-m2 # Y.1|3a+db+hdf(a+d)
15、 = m5-m3 # Y.2|1b+ec+hdf(b+e) = m4-m1 # Y.2|3b+da+hdf(b+d) = m6-m3 # Y.3|1c+eb+hdf(c+e) = m5-m1 # Y.3|2c+fa+hdf(c+f) = m6-m2 # Y.1|2,3 ahdf(a) = m7-m6 # Y.2|1,3 bhdf(b) = m7-m5 # Y.3|1,2 chdf(c) = m7-m4 Partial analyses(8) a = a+b+c+d+e+f+g - b+c+d+e+f+gdf(a) = m7-m6 controlling for two tables X(9) b = a+b+c+d+e+f+g - a+c+d+e+f+gdf(b) = m7-m5 (10) c = a+b+c+d+e+f+g - a+b+d+e+f+gdf(c) = m7-m4 controlling for one table X
