Derivation of partitioning of total variability

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The discussion focuses on deriving the equation for the sum of squares in ANOVA (Analysis of Variance). A key point mentioned is the identity -\bar{Y}_{i.} + \bar{Y}_{i.} = 0, which can be manipulated without altering the equation's integrity. Participants seek clarification on how to construct a working equation to reach the sum of squares identity. The conversation emphasizes the importance of understanding this derivation in the context of statistical analysis. Mastering these concepts is essential for accurate data interpretation in ANOVA.
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it is a sum of squares in anova(analysis of data). how can i derive this equation?tnx..
 

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-\bar{Y}_{i.} + \bar{Y}_{i.} = 0

So you can place that anywhere and you change nothing, it is a common trick.
 
viraltux said:
-\bar{Y}_{i.} + \bar{Y}_{i.} = 0

So you can place that anywhere and you change nothing, it is a common trick.
what is my working equation so i can arrive at the sum of squares identity?
 

Thread 'Formal derivation of statement from Peano Arithmetic system'

I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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