#### The "Sum of Squares Total" in ANOVA

Many of the computer implementations of ANOVA, including the one
in Excel, print out two values that are not used in the later
steps of ANOVA: the sum of the *SSE* (the sum of the squared
deviations within samples from the sample averages) and the *SSG*
(the sum of the squared deviations of the sample averages from
the overall average, weighted by the size of the samples), and the
sum of the *DFE* and the *DFG*. The first of these is
often denoted *SST* and called the "total squared deviation
(from the average)", because it is also equal to the sum of the squared
deviations of all the data values from the grand average. And the
second, denoted *DFT,* is called the total degrees of freedom.
It is easy to see that

*DFT* = *DFE* + *DFG* =
*(N* - *I)* + *(I* - 1) = *N* - 1,
and this
is a reasonable quantity to call the "total degrees of freedom". But
there are two ways of interpreting *SST*,
on the one hand as the sum of *SSE* and *SSG*, and
on the other hand as the sum
of the squares of the differences of the data values from the overall
average; and it is not so obvious that the two interpretations
give the same value. Their equality would imply the following equation
(which I must render as a graphic because of the limitations of HTML):

At first glance, it looks reasonable that the two end expressions
and are equal. But in
general, it is not true that

**(A - B)**^{2} + (B - C)^{2}
= (A - C)^{2}
-- because of the squaring, there are several "middle
terms" in these expressions that do not "cancel out". Do they really
cancel out in this case, so that the two interpretations of SST are
equivalent? To see why it does work here, we note first that the definitions
of *AV*_{g} and *AV* give us some substitutions to use:

Using these and some familiar facts about
summations, and starting with
the more complicationed expression , we have:

where the second and third terms in the expression
labelled add to zero. In a similar way,
but working on the expression , we have:

This is the same expression as we got from , so
they are indeed equal.