Why Incomplete Blocks Need an Adjusted Analysis
Blocking is useful when experimental units differ in ways that may affect the response. In a complete block design, every block contains every treatment. An incomplete block contains only some treatments, so treatment totals can reflect both treatment differences and the particular blocks in which each treatment appeared. A balanced incomplete-block design arranges those appearances so that every treatment is replicated equally and every pair of treatments occurs together equally often.
The analysis must account for block differences when comparing treatments. This tutorial works through a small hypothetical experiment, showing how to check the design, calculate the block and treatment sums of squares, and report the treatment test. The ANOVA model assumes additive treatment and block effects; its errors are independent, normally distributed, and have a common variance.
For the standard additive model, write the response to treatment \(i\) in block \(j\) as \(y_{ij}=\mu+\tau_i+\beta_j+\epsilon_{ij}\), for observed treatment-block combinations only. Here \(\mu\) is the overall mean, \(\tau_i\) is the treatment effect, and \(\beta_j\) is the block effect. Use the constraints \(\sum_i\tau_i=0\) and \(\sum_j\beta_j=0\) to define the effects uniquely. The errors \(\epsilon_{ij}\) are assumed independent with distribution \(N(0,\sigma^2)\). The additive model assumes there is no treatment-by-block interaction.
The treatment test asks whether all treatment effects are equal, \(H_0:\tau_1=\cdots=\tau_t=0\), against the alternative that at least one differs. The test adjusts for block effects; it does not compare unadjusted treatment averages as though the blocks were irrelevant.
Worked Analysis From Raw Observations
Consider a hypothetical study with three treatments, A, B, and C, arranged in three blocks. Each block contains two treatments: AB, AC, and BC. Each treatment occurs in two blocks, and each pair occurs together once. Thus \(t=3\), \(b=3\), \(k=2\), \(r=2\), and \(\lambda=1\). The balance checks are \(bk=3(2)=6=tr\) and \(\lambda(t-1)=1(2)=2=r(k-1)=2(1)\).
| Block | Treatment | Response (units) | Block total |
|---|---|---|---|
| 1 | A | 8 | 18 |
| 1 | B | 10 | 18 |
| 2 | A | 9 | 21 |
| 2 | C | 12 | 21 |
| 3 | B | 11 | 24 |
| 3 | C | 13 | 24 |
Worked Example: Checking the Raw Totals and Block Sum of Squares
First calculate the treatment totals, block totals, and corrected total sum of squares for these six hypothetical responses. The treatment totals are \(T_A=8+9=17\), \(T_B=10+11=21\), and \(T_C=12+13=25\). The block totals are \(B_1=18\), \(B_2=21\), and \(B_3=24\), giving grand total \(G=63\).
The correction factor is \(G^2/N=63^2/6=661.5\), where \(N=6\). Therefore, the corrected total sum of squares is
Since each block contains \(k=2\) observations, the block sum of squares is
This is the block sum of squares entered before the treatment sum of squares in the decomposition used here. The observations are assigned to the stated blocks, each block has the same size, and there is one observation for each treatment appearing in a block. The model assumptions about independent errors, common variance, normality, and additivity are assumptions about how responses arise; they cannot be verified from these six values alone.
Adjusting Treatment Comparisons for Blocks
Let \(N\) be the treatment-by-block incidence matrix: its entry is 1 when a treatment occurs in a block and 0 otherwise. Let \(T\) be the vector of treatment totals and \(B\) the vector of block totals. The adjusted treatment totals are
In words, for each treatment, subtract the average of the totals of the blocks containing it. These \(Q_i\) values remove the contribution of block totals from the treatment totals. They are not treatment means and need not be read as effects in the original response units.
For a balanced incomplete-block design, the treatment sum of squares adjusted for blocks is
Worked Example: Calculating Adjusted Treatment Totals
For the three-block dataset, treatment A occurs in blocks 1 and 2, B in blocks 1 and 3, and C in blocks 2 and 3. Thus
The adjusted totals sum to zero, as expected. Using \(k=2\), \(\lambda=1\), and \(t=3\), the treatment sum of squares adjusted for blocks is
The adjusted treatment effects under the sum-to-zero constraint are \(\hat{\tau}_i=\frac{k}{\lambda t}Q_i\). They are \(-1.667\), \(0\), and \(1.667\) response units for A, B, and C. For example, the adjusted estimate of the A-minus-C difference is \(-1.667-1.667=-3.333\) response units. This comparison accounts for the block pattern rather than simply subtracting the raw treatment means.
Why the Adjustment Formula Works
A short derivation clarifies the role of balance. When block effects are fitted, the treatment information matrix after accounting for blocks is \(C=rI-\frac{1}{k}NN'\), where \(I\) is the identity matrix. In a balanced design, two treatments share \(r\) blocks with themselves and \(\lambda\) blocks with each other. Therefore, \(NN'=(r-\lambda)I+\lambda J\), where \(J\) is the all-ones matrix. On treatment contrasts, which sum to zero, multiplication by \(J\) gives zero. The balance identity \(r(k-1)=\lambda(t-1)\) then gives the contrast-space information coefficient \(\lambda t/k\).
The adjusted totals \(Q\) sum to zero, so they lie in this contrast space. The treatment sum of squares is the quadratic form \(Q'C^{-}Q\), where \(C^{-}\) is the inverse on the contrast space. There \(C^{-}\) multiplies by \(k/(\lambda t)\), yielding the stated formula. This argument explains why the balance conditions matter: they give a single common adjustment for all treatment contrasts.
Worked Example: Completing the Treatment F-Test
For the same dataset, the residual sum of squares follows from the decomposition:
The degrees of freedom are 5 in total, 2 for blocks, and 2 for treatments. The residual degrees of freedom are \(6-3-3+1=1\). Thus \(MS_E=0.1667/1=0.1667\) square response units, and the treatment mean square is \(8.3333/2=4.1667\) square response units. The treatment statistic is
Under \(H_0\) and the stated model assumptions, this statistic has an \(F\) distribution with 2 and 1 degrees of freedom. Its upper-tail p-value is approximately \(0.1400\). At significance level \(\alpha=0.05\), the result is not statistically significant: the data do not provide sufficient evidence that the three treatment means differ after accounting for blocks. This conclusion is not proof that the treatment means are equal. The very low residual degrees of freedom also make this small design’s test imprecise.
Reading the ANOVA Table and Avoiding a Common Error
The completed table separates the block variation, the treatment variation adjusted for blocks, and the residual variation. The block row is included to show the decomposition; the treatment row supplies the test of interest.
| Source | df | SS | MS | F | p-value |
|---|---|---|---|---|---|
| Blocks | 2 | 9.000 | 4.500 | ||
| Treatments adjusted for blocks | 2 | 8.333 | 4.167 | 25.00 | 0.1400 |
| Error | 1 | 0.1667 | 0.1667 | ||
| Total | 5 | 17.50 |
A common mistake is to use the ordinary, unadjusted treatment sum of squares, calculated from treatment totals as if all treatments had appeared in the same blocks. That calculation does not isolate treatment variation when treatments occur in different subsets of blocks. Here, use the adjusted totals \(Q_i\) and the balanced-design formula for \(SS_{\text{Treatments adjusted}}\). Also keep the specified order of this decomposition: the block sum of squares is followed by the treatment sum of squares adjusted for blocks. An adjusted block sum of squares from a different order of fitting cannot simply be substituted into this table.
The F-test is an omnibus test: a small p-value would indicate evidence that at least one treatment mean differs, not identify which pair differs. Any follow-up comparisons need to account for the adjustment for blocks and, if several comparisons are made, the multiple-testing procedure used. The unadjusted treatment means alone are not a substitute for adjusted comparisons.
Check Your Understanding
Use the definitions, calculations, and assumptions in this worked analysis to answer these questions.
- For a design with \(t=4\), \(b=6\), \(k=2\), \(r=3\), and \(\lambda=1\), check both balance identities.
- In the worked dataset, explain why the raw treatment totals alone do not provide a block-adjusted comparison.
- What is the residual degrees of freedom for a balanced incomplete-block design with \(N=12\), \(b=6\), and \(t=4\)?
- State the null hypothesis tested by the treatment F-statistic and identify its numerator and denominator degrees of freedom for the worked dataset.
- Why is a nonsignificant treatment test not evidence that all treatment effects are exactly equal?