For the initial normal linear model, the design matrix has rows $(1,x_1,x_2,x_3)$. The eight factorial corners have balanced signs and orthogonal coordinate columns; the six centre points contribute only to the intercept. ConsequentlyUsing the least-squares normal equations gives the least-squares estimatorIt is a linear transformation of independent normal responses, so its sampling distribution isThe four estimated coefficients are independent, since they are jointly normal with diagonal covariance matrix.

Adding the axial settings produces a central composite design: eight factorial corners, six axial points and six replicated centre points. The axial distance $1.682$ is the rounded value of $8^{1/4}=1.6817928\dots$. With that exact distance it is a rotatable design for a full quadratic model. Indeed, sign symmetry annihilates odd moments, all coordinates have the same second moment, and the fourth moments satisfyRotational symmetry requires the first to be three times the second, giving ${\alpha }^4=8$. Using the printed rounded distance makes rotatability approximate, rather than mathematically exact.

The full quadratic regression mean has ten coefficients:The expanded design matrix has rank ten. To verify this rather than just count parameters, suppose such a quadratic vanishes at every setting. Its centre value sets its constant term to zero; subtracting the values at opposite axial points sets every linear coefficient to zero, and adding them then sets every squared-coordinate coefficient to zero. On the factorial corners, orthogonality of the three pair-product columns forces their coefficients to zero as well. Thus all ten columns are independent, leaving $20-10=10$ residual statistical degrees of freedom.

There are $15$ distinct settings: eight corners, six axial points and the centre. The six centre replicates give the only pure error, with $6-1=5$ degrees and sum of squares $859.33$. The saturated setting-mean model has $15$ parameters, so the lack of fit beyond the quadratic model has $15-10=5$ degrees. For any setting $r$ with $n_r$ replicates, the identitysplits the residual sum of squares into pure error and lack of fit; the cross term vanishes because the deviations from the setting mean sum to zero. Summing givesUnder a correct quadratic mean with independent errors of constant normal variance, the pure-error and lack-of-fit projections are orthogonal, hence their sums of squares are independent with respective distributions ${\sigma }^2{\chi }_5^2$ and ${\sigma }^2{\chi }_5^2$. The Lack-of-fit F-test therefore usesThis is below the given $10\%$ and $5\%$ critical values $3.45$ and $5.05$. There is no significant lack of fit at either level. That is failure to reject this quadratic mean, not proof that it is exact or that a maximum has already been located. The pure-error denominator also assumes the error variance at the centre represents the common variance throughout the experiment.

Code each two-level factor as a sign $x_j\in \{-1,1\}$. A factorial contrast indexed by a factor subset $A$ has sign column ${\chi }_A(x)={\prod }_{j\in A}{x}_j$, with the empty subset giving the intercept. Construct a regular fractional factorial design by choosing $k$ independent words $G_1,\dots ,G_k$ and retaining the settings satisfying ${\chi }_{G_i}(x)=s_i$, where each $s_i\in \{-1,1\}$. In binary exponent notation these are $k$ independent linear equations over ${\mathbb{F}}_2$, leaving $m-k$ freely chosen coordinates and exactly $2^{m-k}$ settings. Dependence of the chosen words would fail to give the stated fraction size.

The words generated by multiplying the $G_i$ form the defining contrast subgroup $\mathcal{H}$, of order $2^k$. Multiplication cancels a repeated factor, so word multiplication is symmetric difference of subsets. On the fraction, ${\chi }_H=s_H$ is constant for every $H\in \mathcal{H}$, and henceTo see that these are all the aliases, average ${\chi }_A{\chi }_B={\chi }_{AB}$ over the fraction. If $AB\notin \mathcal{H}$, this is a nonconstant character of its freely varying binary coordinates; pairing settings that reverse that character makes the average zero. If $AB\in \mathcal{H}$, the product is constant. Thus the aliasing in a fractional factorial design classes are exactly the cosets of $\mathcal{H}$. For a half fraction, $\mathcal{H}=\{I,G\}$ with $G\ne I$, and every contrast column is aliased with exactly the other word $AG$, with a plus or minus sign according to the chosen fraction. This includes the defining word paired with the intercept.

For the six-factor experiment, take the 32-run half fractionIts resolution of a fractional factorial design is VI. Every main effect is aliased with a five-factor interaction, and every two-factor interaction with a four-factor interaction. In particular, no two required low-order effects are aliased with one another. Under the stated negligible-higher-interaction assumption, all six main effects and the fourteen required two-factor interactions are estimable before introducing blocks.

To form four equal days, classify runs by the two signsTheir product is $x_B{x}_C{x}_F$. The three nonconstant block contrasts therefore coincide with the treatment-column alias classesOnly $AF$ among the main effects and two-factor interactions is lost to block confounding in a factorial design; the other two block directions involve only negligible three-factor interactions.

Here is an explicit construction of every run without an ambiguous generator convention. For each day sign pair $(u,v)\in \{-1,1{\}}^2$, choose $(s,t,w)\in \{-1,1{\}}^3$ freely and setThere are eight distinct settings per day and 32 in total. Direct multiplication gives $ABCDEF=1$, $AF=u$ and $ABC=v$. Conversely those three equations solve $C,E,F$ uniquely from $A,B,D$, so there are no missing or repeated settings. Assign the four sign pairs randomly to the four days, and randomize the eight runs within each day.

Distinct nonaliased factorial contrasts are orthogonal over the fraction. None of the 20 required low-order contrast columns belongs to a block class, so all are also orthogonal to the three block-effect columns. Their joint design matrix, including an intercept, therefore has rank $1+3+6+14=24$. The analysis of variance partition isThe eight residual directions are the remaining unconfounded three-factor alias classes. They estimate error only under the negligible-higher-interaction assumption; they are not a separate replicated pure error stratum. This provides the requested comparisons within exactly four days of eight runs.

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.