Applied Statistics and Probability for Engineers

13-2.7 Technical Details about the Analysis of Variance (CD Only)

Derivation of the ANOVA Identity

The proof of the fundamental ANOVA identity in Equation 13-5 is straightforward. Note that

we may write

or

Note that the cross-product term in the previous equation is zero, since

Therefore, we have shown that Equation 13-5 is correct.

Expected Mean Squares

In the text we state that

We can prove this directly by apply the expected value operator. Since

we will initially work with the treatment sum of squares. Now

and from the model Yijiijwe have

and

since. Substituting for and in the expression for SSTreatmentsyields

E cn (^) a
a
i 1
^2 in (^) a
a
i 1
(^2) i.an 2 .. 2 n (^) a
a
i 1
i (^) i. 2 n .. (^) a
a
i 1
i 2 n .. (^) a
a
i 1
(^) i.d
E 1 SSTreatments 2 E cn (^) a
a
i 1
1 ii. .. 22 d
g Yi. Y..
a
i 1 i^0
Y....
Yi.ii.
E 1 SSTreatments 2 E cn (^) a
a
i 1
1 Yi. Y.. 22 d
MSTreatments
SSTreatments
a 1
E 1 MSTreatments 2 ^2 
n (^) a
a
i 1

2
i
a 1
a
n
j 1
1 yij^ yi.^2 yi.^ nyi.yi.^ n^1 yi.n^2 ^0
 (^2) a
a
i 1 a
n
j 1
1 yi. y.. 21 yij yi. 2
a
a
i 1
(^) a
n
j 1
1 yij y.. 22 na
a
i 1
1 yi.^ y..^2
(^2) 
a
a
i 1 a
n
j 1
1 yij yi. 22
a
a
i 1 a
n
j 1
1 yij y.. 22  a
a
i 1 a
n
j 1
31 yi. y.. 2  1 yij yi. 242
13-6
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