Squared deviations

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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 ).

It is well known that for a random variable X with mean \mu and variance \sigma^2:

\sigma^2 = \operatorname{E}(  X ^ 2 ) - \mu^2[1]

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

If \hat{Y} is a vector of n predictions, and Y is the vector of the true values, then the SSE of the predictor is: SSE=\frac{1}{2}\sum_{i=1}^n(\hat{Y_i} - Y_i)^2

Sample variance[edit]

Main article: Sample variance

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[edit]

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_iT_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[edit]

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[edit]

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[edit]

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:

Factor Calculation Sum \sigma^2
Individual 7^2+2^2+1^2 + 7^2+6^2 + 11^2+6^2 + 10^2+7^2+3^2 + 5^2+3^2+4^2 + 11^2+4^2 641 15
Fertiliser × Environment \frac{(7+2+1)^2}{3} + \frac{(7+6)^2}{2} + \frac{(11+6)^2}{2} + \frac{(10+7+3)^2}{3} + \frac{(5+3+4)^2}{3} + \frac{(11+4)^2}{2} 556.1667 6
Fertiliser \frac{(7+2+1+7+6)^2}{5} + \frac{(11+6+10+7+3)^2}{5} + \frac{(5+3+4+11+4)^2}{5} 525.4 3
Environment \frac{(7+2+1+11+6+5+3+4)^2}{8} + \frac{(7+6+10+7+3+11+4)^2}{7} 519.2679 2
Composite \frac{(7+2+1+11+6+5+3+4+7+6+10+7+3+11+4)^2}{15} 504.6 1

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

See also[edit]

References[edit]

  1. ^ Mood & Graybill: An introduction to the Theory of Statistics (McGraw Hill)