= Solution
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. Consequently
$$
X^TX=\operatorname{diag}(14,8,8,8).
$$
Using the <least-squares normal equations> gives the <least-squares estimator>
$$
\boxed{\begin{aligned}
\widehat\beta_0&=\frac1{14}\sum_{i=1}^{14}y_i,\\
\widehat\beta_1&=\frac{-y_1-y_2-y_3-y_4+y_5+y_6+y_7+y_8}{8},\\
\widehat\beta_2&=\frac{-y_1-y_2+y_3+y_4-y_5-y_6+y_7+y_8}{8},\\
\widehat\beta_3&=\frac{-y_1+y_2-y_3+y_4-y_5+y_6-y_7+y_8}{8}.
\end{aligned}}
$$
It is a linear transformation of independent normal responses, so its <sampling distribution> is
$$
\boxed{\widehat\beta\sim N_4\!\left(\beta,\sigma^2
\operatorname{diag}(1/14,1/8,1/8,1/8)\right).}
$$
The 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 $8^{1/4}=1.6817928\ldots$. 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 satisfy
$$
\sum x_j^4=8+2\alpha^4,\qquad
\sum x_j^2x_l^2=8\quad(j\ne l).
$$
Rotational 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:
$$
y=\beta_0+\sum_{j=1}^3\beta_jx_j+
\sum_{j=1}^3\beta_{jj}x_j^2+
\beta_{12}x_1x_2+\beta_{13}x_1x_3+\beta_{23}x_2x_3+\varepsilon.
$$
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 identity
$$
\sum_j(y_{rj}-\widehat y_r)^2
=\sum_j(y_{rj}-\bar y_r)^2+n_r(\bar y_r-\widehat y_r)^2
$$
splits 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 gives
$$
\mathrm{SS}_{\mathrm{LOF}}=1860.98-859.33=1001.65.
$$
Under 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 $\sigma^2\chi^2_5$ and $\sigma^2\chi^2_5$. The <Lack-of-fit F-test> therefore uses
$$
\boxed{F=\frac{1001.65/5}{859.33/5}=1.1656,\qquad F\sim F_{5,5}\text{ under the null}.}
$$
This is below the given $10\%$ and $5\%$ critical values $3.45$ and $5.05$. \b[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.
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