Squared deviations

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The definition of variance is either the expected value (when considering a theoretical distribution), or average (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 $μ$ and variance $σ 2$ :

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

Therefore

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

From the above, the following are readly derived:

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

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

1 Sample variance
2 Partition — analysis of variance
- 2.1 Sums of squared deviations
- 2.2 Example
3 Two-way analysis of variance
4 See also
5 References

[edit] Sample variance

The sum of squared deviations needed to calculate variance (before deciding whether to divide by n or n − 1) is most easily calculated as

$S = \sum x ^ 2 - \left(\sum x\right)^2/n$

From the two derived expectations above the expected value of this sum is

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

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 $σ 2$

[edit] Partition — analysis of variance

In the situation where data is available for k different treatment groups having size n_i 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 $σ 2$ .

Under the Null Hyporthesis 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

$C = \left(\sum x\right)^2/n$

$\operatorname{E}(C) = \sigma^2 + n\mu^2$

[edit] Sums of squared deviations

Under the null hypothesis, the difference of any pair of I, T, and C does not contain any dependency on $μ$ , only $σ 2$ .

$\operatorname{E}(I - C) = (n - 1)\sigma^2$ Total Squared Deviations

$\operatorname{E}(T - C) = (k - 1)\sigma^2$ Treatment Squared Deviations

$\operatorname{E}(I - T) = (n - k)\sigma^2$ Residual Squared Deviations

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

[edit] 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.

[edit] Two-way analysis of variance

The following hypothetical example gives the yields of 15 plants subject to two environmental variations, and three 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	$σ 2$
Individual	$72 + 2 2 + 1 2 + 7 2 + 6 2 + 11 2 + 6 2 + 10 2 + 7 2 + 3 2 + 5 2 + 3 2 + 4 2 + 11 2 + 4 2$	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	$σ 2$	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

[edit] See also

Variance decomposition

[edit] References

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

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Squared deviations

From Wikipedia, the free encyclopedia

Contents

[edit] Sample variance

[edit] Partition — analysis of variance

[edit] Sums of squared deviations

[edit] Example

[edit] Two-way analysis of variance

[edit] See also

[edit] References

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