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A randomized complete block design groups similar experimental units into experimental blocks and places every treatment once in each experimental block, independently randomizing treatment allocation within experimental blocks. Blocking removes background between-block variation from within-block treatment comparisons. A balanced incomplete block design has $v$ treatments and $b$ experimental blocks of size $k<v$, with each treatment occurring in $r$ experimental blocks and every distinct treatment pair together in $\lambda$ experimental blocks. Counting treatment occurrences and co-occurrences gives $vr=bk$ and $r(k-1)=\lambda (v-1)$.

A Latin square of order $q$ places each of $q$ treatment symbols once in every row and every column of a $q\times q$ array. This controls two nuisance directions. A Graeco-Latin square superposes two mutually orthogonal Latin squares: each square is Latin, and each ordered pair of symbols from the two squares occurs exactly once. The two symbol factors can then be fitted together with row and column effects. Random permutations of row labels, column labels and treatment symbols provide a suitable randomized allocation without destroying these balances.

For the operator experiment, fit the additive normal linear modelHere $r_i,c_j$ account for the two fabric directions, and ${\tau }_o$ is the operator effect. Additivity, independent equal-variance errors and the normal assumption supply the exact F-test; there is only one observation per cell, so an unrestricted interaction model is not available.

In the analysis of variance for a Latin square, the three centered factor spaces each have dimension $4-1=3$. Together with the intercept the model has rank ten, leaving $16-10=6$ residual statistical degrees of freedom. The corrected total has $16-1=15$. The completed calculation isUnder the null hypothesis ${\tau }_1=\cdots ={\tau }_4=0$, the operator F-statistic isThe upper five-percent critical value for $F_{3,6}$ is $4.76$, so the operator effect is not significant at five percent in this model.

The decision is independent of the order of fitting rows, columns and operators. To prove the required orthogonality, let $u_i$ be any centered row coefficients and $w_o$ any centered operator coefficients. Their observation-space inner product is ${\sum }_{i,j}{u}_i{w}_{o(i,j)}={\sum }_i{u}_i{\sum }_o{w}_o=0$, since every operator occurs once in each row. The same calculation applies to columns and operators, and all row-column pairs occur once as well. Thus the three centered spaces are mutually orthogonal; their projections commute and sequential sums of squares equal adjusted sums of squares in every fitting order.

With machines included, the layout is a Graeco-Latin square: the machine symbols are Latin, and all sixteen operator-machine pairs occur exactly once. Fitin addition to the earlier constraints. The machine space contributes three new statistical degrees of freedom and is orthogonal to the previous factor spaces, so the operator sum of squares remains $18.25$. The residual sum of squares and statistical degrees of freedom becomeHenceThis exceeds $F_{3,3}(0.05)=9.28$ but not $F_{3,3}(0.01)=29.46$. Therefore the operator effect is now significant at five percent, but not at one percent. Adjustment for machines has removed substantial residual variation without altering the operator numerator. This illustrates why a balanced nuisance factor can change precision and significance even when it does not change the estimated treatment contrasts.

Write the binary incidence matrix of a set system as $N$ to avoid confusing it with a sample size. Thencounts the number of blocks containing treatment $i$, and for $i\ne l$, $(N{N}^T{)}_{il}={\sum }_j{N}_{ij}{N}_{lj}$ counts blocks containing both treatments. The displayed structure therefore makes every treatment replication equal to $r$ and every pair concurrence equal to $\lambda$.

Assume the usual nonempty block design, with $r>0$ and $k>0$. If $r=\lambda$, any block containing one treatment contains every other treatment: otherwise that treatment's replication would exceed its concurrence with the omitted one. Every block is consequently complete, so $k=t$ and $r=b$. With random allocation within blocks this is a randomized complete block design. If $\lambda <r$, the design is a balanced incomplete block design. Counting partners of a fixed treatment in its $r$ blocks givesFor $t>1$, $\lambda /r<1$ gives $k-1<t-1$, hence $k<t$. Moreover $N{N}^T$ has eigenvalue $r-\lambda >0$ on the $(t-1)$-dimensional zero-sum subspace and eigenvalue $r+(t-1)\lambda >0$ on the constant direction. Thus $\mathrm{rank}(N{N}^T)=t$, while $\mathrm{rank}(N{N}^T)\le \mathrm{rank}(N)\le b$. This proves $b\ge t$, the Fisher's inequality for block designs. The case $\lambda =0$ allows singleton blocks; their lack of connectedness prevents general treatment comparison after block adjustment, but does not invalidate the rank inequality.

First ignore the day information and use a one-way normal linear model with five group means and independent errors of common variance. There are $20$ observations. The analysis of variance has treatment, residual and corrected-total statistical degrees of freedom $4,15,19$, respectively. The mean squares in ANOVA giveUnder equal treatment means this has the F-distribution with $(4,15)$ degrees. It exceeds the given $5\%$ upper critical value $3.06$, so the unblocked analysis rejects equality at $5\%$.

The actual day allocation omits a different treatment on each of the five days. Hence it is a balanced incomplete block design withEvery pair occurs on exactly the three days omitting neither member. In the additive two-way analysis of variance, use an intercept, four independent treatment contrasts and four day contrasts. The design is connected, so the model rank is $9$ and its residual has $20-9=11$ statistical degrees of freedom. Day effects and treatment effects are not orthogonal, because a different treatment is absent each day; the unadjusted treatment sum of squares cannot be used for the adjusted test.

Enter days first and then treatments. The supplied sequential sums of squares giveThe day, treatment-after-day, residual and corrected-total statistical degrees of freedom are $4,4,11,19$. Under the additive normal linear model and the null of equal treatment effects, the adjusted numerator and residual squared lengths are independent chi-squared quantities, so the ratio has the F-distribution with $(4,11)$ degrees. It is below even the given $10\%$ critical value $2.54$. After accounting for days there is no significant treatment difference at either $10\%$ or $5\%$. This inference assumes independent homoscedastic normal errors and an additive day effect; with one observation per observed treatment–day cell, unrestricted interactions cannot also be fitted.

Use a randomized complete block design, with day as block and each biological cell as an experimental unit. Each day should contain two biological cells at each of the five dose levels. Each level then has ten replicates overall and occurs equally often in every day: the resulting block design is an orthogonal block design.

This spreads any day-to-day changes in preparation, temperature or measurement across all levels instead of confounding dose with day. Treat the zero level as a control with otherwise comparable handling. Keep the treatment-to-measurement delay standardized because the response variable is a diffusion rate measured immediately after preparation.

A completely randomized design uses $2m$ eligible animals, randomly selecting $m$ to receive $P$ and assigning the remainder to $Q$. Use comparable follow-up and assess the same response variable in both groups. Its advantage is simplicity and freedom from previous-treatment carryover effects; its disadvantage is that between-animal variation enters the residual and can make a treatment contrast imprecise.

A randomized complete block design with matched pairs first forms $m$ pairs using pre-treatment characteristics such as initial disease severity, age or breed. Independently choose which animal in each pair receives $P$, with its partner receiving $Q$. The experimental units are animals; pairs are blocks. The average within-pair difference estimates the treatment contrast, and positive within-pair similarity can reduce its variance. Its advantage is control of known heterogeneity; its disadvantage is the need for useful matching, with fewer residual statistical degrees of freedom and little gain if the matching variables are uninformative. Do not construct pairs using outcomes observed after assignment.

A two-period crossover design randomly assigns half the animals to sequence $PQ$ and half to $QP$. Each animal receives both treatments in separate periods, with a scientifically justified interval between them and the same outcome assessment after each period. Animal blocks remove persistent between-animal differences, while the two sequences balance treatment against period. Its advantage is potentially high precision from within-animal comparisons; its disadvantage is vulnerability to carryover effects, changing disease state and irreversible effects. It is suitable only when comparing the treatments in both periods remains meaningful and residual effects of the first treatment are adequately controlled. A two-period crossover design does not by itself disentangle arbitrary treatment-specific carryover effects.

A randomized complete block design places every treatment in every block, with a common replication pattern, and uses independent within-block randomization. The usual basic version has one occurrence of each treatment per block; equal repeated occurrences give a replicated complete-block version. Such equal-proportion designs are orthogonal block designs.