Solution. In \(SSE(x) = \sum y^2 - \frac 1p\bigl (\sum R^2 + \sum C^2 + \sum T^2\bigr ) + \frac {2}{p^2}G^2\), the missing value enters \(\sum y^2\) as \(x^2\), one row total, one column total and one treatment total as \(R' + x\), \(C' + x\) and \(T' + x\), and the grand total as \(G' + x\). Setting the derivative to zero, \[ 2x - \frac 2p\bigl (R' + C' + T' + 3x\bigr ) + \frac {4}{p^2}(G' + x) = 0 . \] Multiplying by \(p^2/2\): \(x(p^2 - 3p + 2) = p(R' + C' + T') - 2G'\), and \(p^2 - 3p + 2 = (p - 1)(p - 2)\).

For the fuel data with \(y = 27.2\) removed, \(R' = 93.2 - 27.2 = 66.0\), \(C' = 92.4 - 27.2 = 65.2\), \(T' = 92.6 - 27.2 = 65.4\) and \(G' = 341.4 - 27.2 = 314.2\), so \[ \hat x = \frac {4(66.0 + 65.2 + 65.4) - 2(314.2)}{3 \times 2} = \frac {786.4 - 628.4}{6} = 26.33 , \] against the observed \(27.2\).

Solution. The totals are: days \(37\), \(38\), \(38\); operators \(36\), \(40\), \(37\); methods \(A\): \(32\), \(B\): \(39\), \(C\): \(42\); grand total \(113\), and \(\sum y^2 = 1441\). The correction factor is \(113^2/9 = 1418.78\), so \(SST = 22.22\), \(SS_{\mathrm {Days}} = 0.22\), \(SS_{\mathrm {Operators}} = 2.89\), \(SS_{\mathrm {Methods}} = 17.56\) and \(SSE = 1.56\) on \((2)(1) = 2\) degrees of freedom. Then \(F_0 = (17.56/2)/(1.56/2) = 11.29\), against \(F_{0.05,2,2} = 19.0\), so the difference between methods is not significant (\(p = 0.08\)), even though the methods account for nearly \(80\%\) of the total variation. A \(3 \times 3\) Latin square has only \(2\) error degrees of freedom, and its test has very little power; the experiment should be replicated, with further squares.

Solution. With \(L_1(i, j) = i + j\) and \(L_2(i, j) = 2i + j \pmod 5\), writing the symbols \(0, \dots , 4\) as \(A, \dots , E\) for \(L_1\) and \(\alpha , \dots , \varepsilon \) for \(L_2\):

Each row and column contains each Latin and each Greek letter once, and each of the \(25\) pairs occurs once.

Solution. (a) \(bk = tr\) gives \(r = 8 \times 3/6 = 4\), and then \(\lambda = r(k - 1)/(t - 1) = 8/5\), not an integer. No.

(b) \(\lambda (t - 1) = r(k - 1)\) is \(5\lambda = 2r\), so \(r\) must be a multiple of \(5\); the smallest is \(r = 5\), \(\lambda = 2\), and then \(b = tr/k = 10\). (A design with these parameters exists.)

(c) \(t = b = 7\), \(k = r = 3\): each treatment appears in three blocks (treatment \(1\) in \(\{1,2,4\}\), \(\{5,6,1\}\), \(\{7,1,3\}\), and similarly for the others by the cyclic construction). Each block contains three pairs, giving \(21 = \binom 72\) pairs in all, and checking shows that no pair is repeated, so each pair occurs exactly once: \(\lambda = 1\). Indeed \(\lambda (t - 1) = 6 = r(k - 1)\). This is the Fano plane, and with \(b = t\) it attains Fisher’s bound.

Solution. \(Q_i = y_{i\cdot } - \frac 1k\sum _jn_{ij}y_{\cdot j}\). The total \(y_{i\cdot }\) has variance \(r\sigma ^2\); the block totals are independent with variance \(k\sigma ^2\), and each block containing \(i\) shares one observation with \(y_{i\cdot }\). So \begin {align*} \Var (Q_i) &= r\sigma ^2 - \frac 2k\sum _jn_{ij}\sigma ^2 + \frac {1}{k^2}\sum _jn_{ij}k\sigma ^2 \\ &= r\sigma ^2 - \frac {2r\sigma ^2}{k} + \frac {r\sigma ^2}{k} = \frac {r(k - 1)}{k}\sigma ^2 . \end {align*}

For the Fano plane, \(E = \lambda t/(rk) = 7/9 = 0.78\): about \(22\%\) of the information about treatment differences is lost to the blocks, compared with a complete block design of the same replication.