   #### STAT 330: 95-3 The sum of squares decomposition in one example The data consist of blood coagulation times for 24 animals fed one of 4 different diets. In the following I write the data in a table and decompose the table into a sum of several tables. The 4 columns of the table correspond to Diets A, B, C and D. Later in the course we will do matrix linear algebra and then want to think of stacking up thes 24 values into a single column vector but the tables save space. The sums of squares of the entries of each of these arrays are intervals for differences between the 4 population means. On the left hand side . This is the uncorrected total sum of squares. The first term on the right hand side gives . This term is sometimes put in ANOVA tables as the Sum of Squares due to the Grand Mean but it is usually subtracted from the total to produce the Total Sum of Squares we usually put at the bottom of the table and often called the Corrected (or Adjusted) Total Sum of Squares. In this case the corrected sum of squares is the squared length of the table which is 340. The second term on the right hand side of the equation has squared length (which is the Treatment Sum of Squares produced by SAS). The formula for this Sum of Squares is but I want you to see that the formula is just the squared length of the vector of individual sample means minus the grand mean. The last vector of the decomposition is called the residual vector and has squared length . Corresponding to the decomposition of the total squared length of the data vector is a decomposition of its dimension, 24, into the dimensions of subspaces. For instance the grand mean is always a multiple of the single vector all of whose entries are 1; this describes a one dimensional space. The second vector, of deviations from a grand mean lies in the three dimensional subspace of tables which are constant in each column and have a total equal to 0. Similarly the vector of residuals lies in a 20 dimensional subspace -- the set of all tables whose columns sum to 0. This decomposition of dimensions is the decomposition fo degrees of freedom. So 24 = 1+3+20 and the degrees of freedom for treatment and error are 3 and 20 respectively. The vector whose squared length is the Corrected Total Sum of Squares lies in the 23 dimensional subspace of vectors whose entries sum to 1; this produces the 23 total degrees of freedom in the usual ANOVA table.   