= Solution
Write the binary <incidence matrix of a set system> as $N$ to avoid confusing it with a sample size. Then
$$
(NN^T)_{ii}=\sum_jN_{ij}
$$
counts the number of blocks containing treatment $i$, and for $i\ne l$, $(NN^T)_{il}=\sum_jN_{ij}N_{lj}$ 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 \b[$k=t$ 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 gives
$$
r(k-1)=\lambda(t-1),\qquad tr=bk.
$$
For $t>1$, $\lambda/r<1$ gives $k-1<t-1$, hence $k<t$. Moreover $NN^T$ 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 $\operatorname{rank}(NN^T)=t$, while $\operatorname{rank}(NN^T)\leq\operatorname{rank}(N)\leq b$. This proves \b[$b\geq t$], 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> give
$$
\boxed{F_{\mathrm{unblocked}}=
\frac{42.4/4}{46.7/15}=3.4047.}
$$
Under equal treatment means this has the <F-distribution> with $(4,15)$ degrees. It exceeds the given $5\%$ upper critical value $3.06$, so \b[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> with
$$
\boxed{t=b=5,\qquad r=k=4,\qquad\lambda=3.}
$$
Every 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> give
$$
\mathrm{SS}_{E}=89.1-28.4-18.3=42.4,\qquad
\boxed{F_{\mathrm{adjusted}}=
\frac{18.3/4}{42.4/11}=1.1869.}
$$
The 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 $2.54$. \b[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.
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