ANOVA

``Analysis of Variance.'' A Statistical Test for heterogeneity of Means by analysis of group Variances. To apply the test, assume random sampling of a variate $y$ with equal Variances, independent errors, and a Normal Distribution. Let $n$ be the number of Replicates (sets of identical observations) within each of $K$ Factor Levels (treatment groups), and $y_{ij}$ be the $j$th observation within Factor Level $i$. Also assume that the ANOVA is ``balanced'' by restricting $n$ to be the same for each Factor Level.

Now define the sum of square terms

$${\rm SST} \equiv \sum_{i=1}^k \sum_{j=1}^n (y_{ij}-\bar{\bar y})^2$$ (1)

$$= \sum_{i=1}^k \sum_{j=1}^n {y_{ij}}^2-{\left({\sum_{i=1}^k \sum_{j=1}^n y_{ij}}\right)^2\over Kn}$$ (2)

$${\rm SSA} \equiv {1\over n}\sum_{i=1}^k \left({\sum_{j=1}^n y_{ij}}\right)^2 -{1\over Kn}\left({\sum_{i=1}^k \sum_{j=1}^n y_{ij}}\right)^2$$ (3)

$${\rm SSE} \equiv \sum_{i=1}^k \sum_{j=1}^n (y_{ij}-\bar{y_i})^2$$ (4)

$$= {\rm SST}-{\rm SSA},$$ (5)

which are the total, treatment, and error sums of squares. Here, $\bar y_i$ is the mean of observations within Factor Level $i$, and $\bar{\bar y}$ is the ``group'' mean (i.e., mean of means). Compute the entries in the following table, obtaining the P-Value corresponding to the calculated F-Ratio of the mean squared values

$$F={{\rm MSA}\over{\rm MSE}}.$$ (6)

Category	SS	°Freedom	Mean Squared	F-Ratio
Treatment	SSA	$K-1$	${\rm MSA}\equiv{{\rm SSA}\over K-1}$	${{\rm MSA}\over{\rm MSE}}$
Error	SSE	$K(n-1)$	${\rm MSE}\equiv{{\rm SSE}\over K(n-1)}$
Total	SST	$Kn-1$	${\rm MST}\equiv{{\rm SST}\over Kn-1}$

If the P-Value is small, reject the Null Hypothesis that all Means are the same for the different groups.

See also Factor Level, Replicate, Variance