Solution. (a) In a \(2^k\) design confounded in blocks, the principal block is the block containing the treatment combination \((1)\); equivalently, the runs for which every defining contrast is \(0\). It is closed under multiplication of treatment combinations, and the other blocks are obtained from it by multiplication. (b) A replicated design in which the same effect (or effects) is confounded with blocks in every replicate, so that no information on it is available.

Solution. The generalised interaction is \(ABC \cdot BCD = AB^2C^2D = AD\), also confounded. With \(L_1 = x_1 + x_2 + x_3\) and \(L_2 = x_2 + x_3 + x_4 \pmod 2\), the principal block is \(\{(1), bc, abd, acd\}\) (check: \(bc \cdot abd = ab^2cd = acd\)). Multiplying by \(a\), by \(b\) and by \(d\):

Solution. (a) Block \(1\) is \(\{ab, ac, (1), bc\}\), the runs with an even number of letters, so the \(ABC\) column is \(-\) throughout block \(1\) and \(+\) throughout block \(2\): \(ABC\) is confounded.

(b) Both are estimable, since only \(ABC\) is confounded. The \(A\) column is \(+\) for runs \(1\), \(2\), \(5\), \(6\), and the \(BC\) column is \(+\) for runs \(3\), \(4\), \(5\), \(6\), so \begin {gather*} \hat A = \tfrac 14(y_1 + y_2 + y_5 + y_6 - y_3 - y_4 - y_7 - y_8) , \\ \widehat {BC} = \tfrac 14(y_3 + y_4 + y_5 + y_6 - y_1 - y_2 - y_7 - y_8) . \end {gather*} Each column has two \(+\) signs in each block, so the block difference cancels.

(c) In block \(3\), the runs \(a\), \(bc\), \((1)\), \(abc\) all have \(B\) and \(C\) at the same level, so the \(BC\) column is \(+\) throughout; in block \(4\) it is \(-\): \(BC\) is confounded. From blocks \(3\) and \(4\) alone, \(A\) is estimable (its column has two \(+\) signs in each block), as \(\hat A = \frac 14\bigl (a + abc + ac + ab - bc - (1) - c - b\bigr )\), but \(BC\) is not. Taken together, the two replicates partially confound \(ABC\) and \(BC\).

Solution. Choose \(ABC\) and \(CDE\): their generalised interaction is \(ABC^2DE = ABDE\), of order four, so the confounded effects are \(ABC\), \(CDE\) and \(ABDE\). (Choosing, say, \(ABCD\) and \(BCDE\) would confound \(AE\).) The principal block satisfies \(x_1 + x_2 + x_3 \equiv 0\) and \(x_3 + x_4 + x_5 \equiv 0\): \[ \{(1),\ ab,\ de,\ abde,\ acd,\ ace,\ bcd,\ bce\} . \] It contains \(8\) runs and is closed under multiplication, for example \(acd \cdot bce = abde\).

Solution. With \(N = 2^k\) runs and block totals \(B_1\), \(B_2\) of \(N/2\) runs each, \(SS_{\mathrm {Blocks}} = (B_1^2 + B_2^2)/(N/2) - (B_1 + B_2)^2/N = (B_2 - B_1)^2/N\). The contrast of the confounded effect is \(\pm (B_2 - B_1)\), and its sum of squares is \(\text {contrast}^2/N\), the same.