Re: One way anova

"anne001" <anne@xxxxxxxxxxxxxxx> schreef in bericht
news:1125315673.651382.199020@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> so you are saying that
> E(SST) = (n-J) *var(alfa) +(n-1) var(e) is correct
> and
> var(y) = var(alfa) + var(e) is correct
>
> but that Var(y) is not E(SST)/n-1.
>
> How can that be. SST is the sum of all (yij-y..)^2/n-1. That is the
> definition of the variance.

In calculating the expected value of SST/(n-1) you have to account for
the fact that the observations are not independent: each alfa(i) is
part of J observations. This is correctly performed by first
decomposing SST in SSE and SSM, resulting in the value of E(SST) as
stated above.

> I see that you have decomposed E(SST) into two terms. But I did not
> understand the last point.

No. What I did was decomposing the expected value of the sum of
squares of all observations [Sum yij^2] in the expected value of SST
and that of the "correction term" [(Sum yij)^2 / n], in order to show
the complete decomposition of the Sum of Squares:

correction term df=1 expected value= var(e) +
J*var(alfa) + n*mu^2
betw. groups (SSM) df=I-1 expected value= (I-1)*var(e) +
(I-1)J*var(alfa)
within groups (SSE) df=I(J-1) expected value= I(J-1)*var(e)

total sum of squares df=IJ=n expected value= n*var(e) + n*var(alfa)
+ n*mu^2

Jos Jansen


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