Block Contrasts in Balanced Designs

Why Blocks Create a Useful Decomposition

In a balanced complete-block design, every treatment is observed in every block. Blocks can account for systematic variation among experimental conditions, patients, batches, or locations, while treatment comparisons are made within those blocks. Balance matters algebraically: each treatment occurs the same number of times, and the treatment and block contrast spaces are orthogonal. This lets us separate their sums of squares cleanly.

Consider \(a\) treatments and \(b\) blocks, with one observation in each treatment-block combination. Write the observation as \(Y_{ij}\), where \(i=1,\ldots,a\) indexes treatments and \(j=1,\ldots,b\) indexes blocks. For the exact distributional statements below, assume the observations are independent and \(Y_{ij}\sim N(\mu_{ij},\sigma^2)\), with common variance \(\sigma^2>0\). The means \(\mu_{ij}\) need not be additive unless that condition is stated.

Orthogonal Spaces and Sums of Squares

Arrange the \(ab\) observations in a vector \(Y\), using any fixed ordering of the treatment-block cells. The inner product on this space is the usual Euclidean inner product. Four mutually orthogonal spaces describe the decomposition: the grand-mean space, the treatment-contrast space, the block-contrast space, and the interaction space. The interaction space consists of arrays whose row and column sums are all zero.

For an array \(x=(x_{ij})\), the treatment component is constant across blocks within each treatment, and its treatment-level values sum to zero. The block component is constant across treatments within each block, with block-level values summing to zero. An interaction array has zero sums within every treatment row and every block column. These spaces are orthogonal: for example, the inner product of a treatment array \(u_i\) repeated across blocks and a block array \(v_j\) repeated across treatments is \(\sum_i\sum_j u_i v_j=(\sum_i u_i)(\sum_j v_j)=0\). The same zero-sum conditions show orthogonality with the interaction space.

Define the treatment means \(\bar Y_{i.}=b^{-1}\sum_jY_{ij}\), block means \(\bar Y_{.j}=a^{-1}\sum_iY_{ij}\), and grand mean \(\bar Y_{..}=(ab)^{-1}\sum_i\sum_jY_{ij}\). The resulting sums of squares have direct projection interpretations.

To verify the identity, decompose each centered observation as \(Y_{ij}-\bar Y_{..}=(\bar Y_{i.}-\bar Y_{..})+(\bar Y_{.j}-\bar Y_{..})+(Y_{ij}-\bar Y_{i.}-\bar Y_{.j}+\bar Y_{..})\). The three arrays on the right belong to the treatment, block, and interaction spaces. Their pairwise inner products vanish by the zero-sum conditions, so squaring the norm gives the sum of their squared norms. Repeating each treatment deviation across \(b\) blocks gives the factor \(b\) in \(SS_{\mathrm{Tr}}\); repeating each block deviation across \(a\) treatments gives the factor \(a\) in \(SS_{\mathrm{Bl}}\).

For the dimensions, the grand-mean space has dimension \(1\). The treatment and block contrast spaces have dimensions \(a-1\) and \(b-1\). An interaction array is determined by its first \(a-1\) rows and first \(b-1\) columns: the final entries are forced by the zero row and column sums, giving dimension \((a-1)(b-1)\). These dimensions add to \(ab-1\), the dimension of the centered observation space. This also verifies that the spaces exhaust all corrected variation.

Worked Examples

When Are the Sums of Squares Independent?

Orthogonality gives an algebraic partition for every data array. Independence is a distributional statement and requires assumptions. Under the independent normal model with common variance, the observation vector is multivariate normal with covariance matrix \(\sigma^2 I\). Projections onto orthogonal spaces are then independent. Their sums of squares are central or noncentral according to whether the mean vector has a component in the corresponding space.

To prove independence, take any vectors \(s_r\) in the range of \(P_r\). The joint normal vector formed by the projections has cross-covariance \(\operatorname{Cov}(P_rY,P_tY)=P_r(\sigma^2I)P_t=\sigma^2P_rP_t=0\) for \(r\ne t\), because the ranges are orthogonal. Jointly normal vectors with zero cross-covariance are independent; hence the projections, and therefore their squared norms, are independent.

For the distribution, choose an orthonormal basis of the range of \(P_r\), and let \(U\) be the matrix with those \(q_r\) basis vectors as columns. Then \(P_r=UU'\), and \(Y'P_rY/\sigma^2=\|U'Y/\sigma\|^2\). The coordinates of \(U'Y/\sigma\) are independent unit-variance normal variables with mean vector \(U'\mu/\sigma\). Their squared norm is noncentral chi-square with \(q_r\) degrees of freedom and noncentrality \(\|U'\mu/\sigma\|^2=\mu'P_r\mu/\sigma^2\). This noncentrality is zero exactly when \(P_r\mu=0\). \(\square\)

For the treatment, block, and interaction projections, independence therefore holds under the stated Gaussian spherical model. Under an additive mean structure \(\mu_{ij}=\mu+\tau_i+\beta_j\), with \(\sum_i\tau_i=0\) and \(\sum_j\beta_j=0\), the interaction projection of \(\mu\) is zero. Thus \(SS_{\mathrm{E}}/\sigma^2\) is central chi-square with \((a-1)(b-1)\) degrees of freedom. Treatment and block components can be noncentral when their corresponding effects are nonzero. If the mean has interaction, the residual component is generally noncentral instead.

Why Balance Matters and a Common Pitfall

Balance makes a treatment contrast orthogonal to every block contrast because each treatment contrast is replicated equally over blocks, and vice versa. In an unbalanced layout, those repeated contributions need not cancel. The simple projection formulas and their orthogonality cannot then be assumed without checking the actual design spaces.

• Wrong reasoning: “The three sums of squares add to the total, so they must be independent.” A numerical identity alone does not establish independence.
• Correction: Independence here follows from orthogonal projections of a jointly normal vector with covariance \(\sigma^2I\). Without normality or spherical covariance, orthogonality alone does not generally imply independence.
• Wrong reasoning: “The residual sum of squares is always central.” It is central only when the mean has no interaction-space component. Under nonadditive cell means, its distribution is noncentral.
• Correction: Check whether the mean array is additive before calling the interaction component central.

The decomposition is useful because it separates treatment comparisons from block-to-block variation while keeping track of the remaining treatment-by-block pattern. Its exact Gaussian distributions also make explicit which assumptions support independence and which mean structures make a component central.