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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.