# Squared deviations

{{#invoke:see also|seealso}} In probability theory and statistics, the definition of variance is either the expected value (when considering a theoretical distribution), or average value (for actual experimental data), of squared deviations from the mean. Computations for analysis of variance involve the partitioning of a sum of squared deviations. An understanding of the complex computations involved is greatly enhanced by a detailed study of the statistical value:

$\operatorname {E} (X^{2}).$ $\sigma ^{2}=\operatorname {E} (X^{2})-\mu ^{2}$ Therefore

$\operatorname {E} (X^{2})=\sigma ^{2}+\mu ^{2}.$ From the above, the following are easily derived:

$\operatorname {E} \left(\sum \left(X^{2}\right)\right)=n\sigma ^{2}+n\mu ^{2}$ $\operatorname {E} \left(\left(\sum X\right)^{2}\right)=n\sigma ^{2}+n^{2}\mu ^{2}$ ## Sample variance

{{#invoke:main|main}} The sum of squared deviations needed to calculate sample variance (before deciding whether to divide by n or n − 1) is most easily calculated as

$S=\sum x^{2}-{\frac {\left(\sum x\right)^{2}}{n}}$ From the two derived expectations above the expected value of this sum is

$\operatorname {E} (S)=n\sigma ^{2}+n\mu ^{2}-{\frac {n\sigma ^{2}+n^{2}\mu ^{2}}{n}}$ which implies

$\operatorname {E} (S)=(n-1)\sigma ^{2}.$ This effectively proves the use of the divisor n − 1 in the calculation of an unbiased sample estimate of σ2.

## Partition — analysis of variance

In the situation where data is available for k different treatment groups having size ni where i varies from 1 to k, then it is assumed that the expected mean of each group is

$\operatorname {E} (\mu _{i})=\mu +T_{i}$ and the variance of each treatment group is unchanged from the population variance $\sigma ^{2}$ .

Under the Null Hypothesis that the treatments have no effect, then each of the $T_{i}$ will be zero.

It is now possible to calculate three sums of squares:

Individual
$I=\sum x^{2}$ $\operatorname {E} (I)=n\sigma ^{2}+n\mu ^{2}$ Treatments
$T=\sum _{i=1}^{k}\left(\left(\sum x\right)^{2}/n_{i}\right)$ $\operatorname {E} (T)=k\sigma ^{2}+\sum _{i=1}^{k}n_{i}(\mu +T_{i})^{2}$ $\operatorname {E} (T)=k\sigma ^{2}+n\mu ^{2}+2\mu \sum _{i=1}^{k}(n_{i}T_{i})+\sum _{i=1}^{k}n_{i}(T_{i})^{2}$ Under the null hypothesis that the treatments cause no differences and all the $T_{i}$ are zero, the expectation simplifies to

$\operatorname {E} (T)=k\sigma ^{2}+n\mu ^{2}.$ Combination
$C=\left(\sum x\right)^{2}/n$ $\operatorname {E} (C)=\sigma ^{2}+n\mu ^{2}$ ### Sums of squared deviations

Under the null hypothesis, the difference of any pair of I, T, and C does not contain any dependency on $\mu$ , only $\sigma ^{2}$ .

$\operatorname {E} (I-C)=(n-1)\sigma ^{2}$ total squared deviations aka total sum of squares
$\operatorname {E} (T-C)=(k-1)\sigma ^{2}$ treatment squared deviations aka explained sum of squares
$\operatorname {E} (I-T)=(n-k)\sigma ^{2}$ residual squared deviations aka residual sum of squares

The constants (n − 1), (k − 1), and (n − k) are normally referred to as the number of degrees of freedom.

### Example

In a very simple example, 5 observations arise from two treatments. The first treatment gives three values 1, 2, and 3, and the second treatment gives two values 4, and 6.

$I={\frac {1^{2}}{1}}+{\frac {2^{2}}{1}}+{\frac {3^{2}}{1}}+{\frac {4^{2}}{1}}+{\frac {6^{2}}{1}}=66$ $T={\frac {(1+2+3)^{2}}{3}}+{\frac {(4+6)^{2}}{2}}=12+50=62$ $C={\frac {(1+2+3+4+6)^{2}}{5}}=256/5=51.2$ Giving

Total squared deviations = 66 − 51.2 = 14.8 with 4 degrees of freedom.
Treatment squared deviations = 62 − 51.2 = 10.8 with 1 degree of freedom.
Residual squared deviations = 66 − 62 = 4 with 3 degrees of freedom.

## Two-way analysis of variance

The following hypothetical example gives the yields of 15 plants subject to two different environmental variations, and three different fertilisers.

Extra CO2 Extra humidity
No fertiliser 7, 2, 1 7, 6
Nitrate 11, 6 10, 7, 3
Phosphate 5, 3, 4 11, 4

Five sums of squares are calculated:

Finally, the sums of squared deviations required for the analysis of variance can be calculated.

Factor Sum $\sigma ^{2}$ Total Environment Fertiliser Fertiliser × Environment Residual
Individual 641 15 1 1
Fertiliser × Environment 556.1667 6 1 −1
Fertiliser 525.4 3 1 −1
Environment 519.2679 2 1 −1
Composite 504.6 1 −1 −1 −1 1
Squared deviations 136.4 14.668 20.8 16.099 84.833
Degrees of freedom 14 1 2 2 9