- Squared deviations
In

probability theory andstatistics , the definition ofis either thevariance expected value (when considering a theoretical distribution), or average (for actual experimental data) of**squared deviations**from the mean. Computations forinvolve the partitioning of a sum ofanalysis of variance **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 $sigma^2$:: $sigma^2\; =\; operatorname\{E\}(\; X\; ^\; 2\; )\; -\; mu^2$ [

*Mood & Graybill: "An introduction to the Theory of Statistics" (McGraw Hill)*]Therefore

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

From the above, the following are readly derived:

: $operatorname\{E\}left(\; sumleft(\; X\; ^\; 2\; ight)\; ight)\; =\; nsigma^2\; +\; nmu^2$

: $operatorname\{E\}left(\; left(sum\; X\; ight)^\; 2\; ight)\; =\; nsigma^2\; +\; n^2mu^2$

**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

: $S\; =\; sum\; x\; ^\; 2\; -\; left(sum\; x\; ight)^2/n$

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

: $operatorname\{E\}(S)\; =\; nsigma^2\; +\; nmu^2\; -\; (nsigma^2\; +\; n^2mu^2)/n$

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 $sigma^2$**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 $sigma^2$.

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

:$I\; =\; sum\; x^2$

:$operatorname\{E\}(I)\; =\; nsigma^2\; +\; nmu^2$

;Treatments

:$T\; =\; sum\_\{i=1\}^k\; left(left(sum\; x\; ight)^2/n\_i\; ight)$

:$operatorname\{E\}(T)\; =\; ksigma^2\; +\; sum\_\{i=1\}^k\; n\_i(mu\; +\; T\_i)^2$

:$operatorname\{E\}(T)\; =\; ksigma^2\; +\; nmu^2\; +\; 2mu\; sum\_\{i=1\}^k\; (n\_iT\_i)\; +\; sum\_\{i=1\}^k\; n\_i(T\_i)^2$

Under the null hypothesis that the treatments cause no differences and all the $T\_i$ are zero, the expectation simplifies to

:$operatorname\{E\}(T)\; =\; ksigma^2\; +\; nmu^2.$

;Combination

:$C\; =\; left(sum\; x\; ight)^2/n$

:$operatorname\{E\}(C)\; =\; sigma^2\; +\; nmu^2$

**ums 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

:$operatorname\{E\}(T\; -\; C)\; =\; (k\; -\; 1)sigma^2$ 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.

**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.

**ee also***

Variance decomposition

*Errors and residuals in statistics **References**

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