Solution. (a) Using the table of signs, \begin {align*} \text {Contrast}_{AB} &= 10 - 16 - 18 + 26 + 22 - 16 - 32 + 20 = -4 , \\ \text {Contrast}_{AC} &= 10 - 16 + 18 - 26 - 22 + 16 - 32 + 20 = -32 , \\ \text {Contrast}_{ABC} &= -10 + 16 + 18 - 26 + 22 - 16 - 32 + 20 = -8 , \end {align*}

and dividing by \(4\): \(AB = -1\), \(AC = -8\), \(ABC = -2\).

(b) The ordered effects \(-8\) (\(AC\)), \(-2\) (\(ABC\)), \(-1\), \(-1\), \(-1\) (\(A\), \(AB\), \(BC\)), \(5\) (\(C\)), \(8\) (\(B\)) are plotted against the quantiles \(-1.47\), \(-0.79\), \(-0.37\), \(0\), \(0.37\), \(0.79\), \(1.47\). The four effects near zero lie on a line through the origin; \(B\), \(C\) and \(AC\) fall well off it.

(c) Each sum of squares is \(2\hat E^2\): \(A\): \(2\), \(B\): \(128\), \(C\): \(50\), \(AC\): \(128\). Keeping \(A\) because \(AC\) is in the model, the error is \(AB + BC + ABC = 2 + 2 + 8 = 12\) on \(3\) degrees of freedom, \(MSE = 4\), and \(SST = 320\).

With \(F_{0.05,1,3} = 10.13\), \(B\), \(C\) and \(AC\) are significant.

(d) The mean is \(160/8 = 20\) and each coefficient is half the effect: \[ \hat y = 20 - 0.5x_1 + 4x_2 + 2.5x_3 - 4x_1x_3 . \] The strong negative \(AC\) interaction means that \(C\) helps at the low level of \(A\) (by \(2.5 + 4 = 6.5\) per unit of \(x_3\)) but not at the high level (\(2.5 - 4 = -1.5\)).

Solution. (a) The contrasts are \(A\): \(8\), \(B\): \(68\), \(C\): \(22\), \(AB\): \(52\), \(AC\): \(-54\), \(BC\): \(-10\), \(ABC\): \(-10\), and dividing by \(4\) the effects are \begin {gather*} A = 2 , \quad B = 17 , \quad C = 5.5 , \quad AB = 13 , \\ AC = -13.5 , \quad BC = -2.5 , \quad ABC = -2.5 . \end {gather*}

(b) Ordered: \(-13.5\) (\(AC\)), \(-2.5\), \(-2.5\) (\(BC\), \(ABC\)), \(2\) (\(A\)), \(5.5\) (\(C\)), \(13\) (\(AB\)), \(17\) (\(B\)). The effects \(A\), \(BC\), \(ABC\) and perhaps \(C\) lie near a line through the origin; \(B\), \(AB\) and \(AC\) fall well off it and appear significant.

(c) Each sum of squares is \(2\hat E^2\), and \(SST = 1374\). With \(A\) (\(8\)) and \(ABC\) (\(12.5\)) as error, \(SSE = 20.5\) on \(2\) degrees of freedom and \(MSE = 10.25\):

With \(F_{0.05,1,2} = 18.51\), \(B\), \(AB\) and \(AC\) are significant; \(C\) and \(BC\) are not.

(d) \(\Var (\hat E) = \sigma ^2/2^{3-2} = \sigma ^2/2\), estimated by \(10.25/2\), so \(\operatorname {se}(\widehat {BC}) = 2.264\). With \(t_{0.025,2} = 4.303\) the interval is \(-2.5 \pm 4.303(2.264) = -2.5 \pm 9.74\), that is \((-12.24,\ 7.24)\). It contains zero, consistent with (c).

(e) Keeping \(A\) and \(C\) for hierarchy, since they appear in the significant interactions, and using half-effects with \(\bar y = 34.5\), \[ \hat y = 34.5 + 1.0x_1 + 8.5x_2 + 2.75x_3 + 6.5x_1x_2 - 6.75x_1x_3 . \]

(f) At \(x_1 = -1\), \(x_2 = -1\), \(x_3 = +1\), \[ \hat y = 34.5 - 1.0 - 8.5 + 2.75 + 6.5 + 6.75 = 41.0 , \] which equals the observed value \(41\) at that combination.

Solution. An effect estimate is \(\bar y_+ - \bar y_-\), the difference of two independent means of \(N/2 = n2^{k-1}\) observations each, so its variance is \(2\sigma ^2/(n2^{k-1}) = \sigma ^2/(n2^{k-2})\). For \(k = 3\) the standard error is \(\sigma /\sqrt {2n}\), which is at most \(0.5\sigma \) when \(2n \ge 4\): \(n = 2\) replicates, \(16\) runs.

Solution. The column of an effect is the product of the main-effect columns of its letters. Multiplying two such columns, each main-effect column that appears in both is multiplied by itself, giving a column of \(+1\)’s, so only the letters in exactly one of the two remain: for example \(ABC \cdot BCD = AB^2C^2D = AD\). If the two effects differ, at least one letter remains, and the product is the column of an effect, which has equal numbers of \(+\) and \(-\) signs because each main-effect column does and the levels of different factors are crossed. So the inner product of the two columns, the sum of the product column, is zero.