Solution. With \(t = 5\) and \(k = 3\), Proposition 6.3.2 requires \(3b = 5r\) and \(4\lambda = 2r\). So \(r = 2\lambda \) is even and \(r\) is a multiple of \(3\), which makes \(r\) a multiple of \(6\). The smallest choice is \(r = 6\), \(\lambda = 3\), \(b = 10\), and all \(\binom 53 = 10\) triples of catalysts achieve it. By Theorem 6.3.4, \[ \Var (\hat \tau _i - \hat \tau _l) = \frac {2k\sigma ^2}{\lambda t} = \frac {6\sigma ^2}{15} = 0.4\sigma ^2 . \] In a randomised block design with \(r = 6\) the variance would be \(2\sigma ^2/6 = 0.333\sigma ^2\), so the incomplete design keeps a fraction \(\lambda t/(rk) = 15/18 = 0.83\) of the information. This assumes the same \(\sigma ^2\); in practice blocks of five runs would be less homogeneous than blocks of three, so the true loss is smaller, and may be a gain.

Solution. (a) The diet is applied to a pen, not to a pig: the pigs in a pen share a feeder, and nothing assigns the diet to one pig independently of the other. The pen is the experimental unit, and the pig is a subsample (an observational unit). The model is a two-stage nested design with pens random within fixed diets, \[ y_{ijk} = \mu + \tau _i + \beta _{j(i)} + \varepsilon _{ijk} , \quad i = 1, 2, 3 , \quad j = 1, \dots , 4 , \quad k = 1, 2 , \] with \(\sum \tau _i = 0\), \(\beta _{j(i)} \sim N(0, \sigma ^2_\beta )\) the pen effects, \(\varepsilon _{ijk} \sim N(0, \sigma ^2)\) the pig-to-pig variation, and all random terms independent.

(b) By Theorem 12.1.2, \[ E(MS_{\mathrm {Diets}}) = \sigma ^2 + 2\sigma ^2_\beta + 4\sum \tau _i^2 , \quad E(MS_{\mathrm {Pens(Diets)}}) = \sigma ^2 + 2\sigma ^2_\beta , \] so diets are tested against pens within diets. The sums of squares are \(SS_{\mathrm {Diets}} = 19.163\), \(SS_{\mathrm {Pens(Diets)}} = 22.095\) and \(SSE = 6.220\), adding to \(SST = 47.478\):

For diets, \(F_0 = 9.582/2.455 = 3.90 < F_{0.05, 2, 9} = 4.26\) (\(P = 0.060\)): the difference between diets is not significant at the \(5\%\) level. Pens differ (\(F_0 = 4.74\), \(P = 0.007\)), with \(\hat \sigma ^2_\beta = (2.455 - 0.518)/2 = 0.968\), larger than the pig-to-pig variance \(\hat \sigma ^2 = 0.518\). The same \(F\) test results from a one-way analysis of the twelve pen means, which is the simplest correct analysis.

(c) Pooling pens and pigs gives an error mean square \((22.095 + 6.220)/21 = 1.348\) on \(21\) degrees of freedom, so \(F_0 = 9.582/1.348 = 7.11\) with \(P = 0.004\), a highly significant difference. The pooled mean square estimates \(\sigma ^2 + \frac {18}{21}\sigma ^2_\beta \) instead of \(\sigma ^2 + 2\sigma ^2_\beta \): it leaves out most of the pen-to-pen variation that the diet means actually contain, and it claims \(21\) degrees of freedom when there are only \(12\) independent units. This error is called pseudo-replication. Weighing more pigs per pen reduces \(\sigma ^2/2\), but not \(\sigma ^2_\beta \); only more pens make the diet comparison more precise.

Solution. (a) The design has resolution IV, with aliases \(A = BCD\), \(B = ACD\), \(C = ABD\), \(D = ABC\), \(AB = CD\), \(AC = BD\), \(AD = BC\). The runs are the full \(2^3\) in \(A\), \(B\), \(C\) with \(D = ABC\), so each effect is the contrast in the signs of the \(2^3\) table divided by \(4\). With \(\bar y = 61.875\), \begin {gather*} A = 9.75 , \quad B = 0.25 , \quad C = 8.25 , \quad D = 1.75 , \\ AB + CD = 0.25 , \quad AC + BD = -6.75 , \quad AD + BC = 0.75 . \end {gather*}

(b) Three effects stand apart: \(A\), \(C\) and the string \(AC + BD\); the other four are below \(2\) in absolute value. Since neither \(B\) nor \(D\) has an effect, the heredity principle attributes the string to \(AC\). If that were in doubt, the alternate fraction \(I = -ABCD\) (eight more runs) would separate \(AC\) from \(BD\).

(c) Dropping \(B\) and \(D\) projects the design into a \(2^2\) in \(A\) and \(C\) with two replicates, and the four negligible effects give an error sum of squares \(SSE = 8(0.25^2 + 1.75^2 + 0.25^2 + 0.75^2)/4 = 7.5\) on \(4\) degrees of freedom (an effect estimate \(E\) has \(SS = 2E^2\) here):

With \(F_{0.05, 1, 4} = 7.71\), all three are significant (\(P \le 0.002\)). In coded units \[ \hat y = 61.875 + 4.875x_1 + 4.125x_3 - 3.375x_1x_3 , \] which predicts \(49.5\), \(66.0\), \(64.5\) and \(67.5\) at \((x_1, x_3) = (-1, -1)\), \((1, -1)\), \((-1, 1)\) and \((1, 1)\); these equal the cell means, as they must for a saturated model in a \(2^2\). Yield is highest with both \(A\) and \(C\) high. The interaction says that either factor at its high level gives most of the gain: raising \(A\) adds \(16.5\) when \(C\) is low but only \(3\) when \(C\) is high. A further experiment beyond the high levels might be worthwhile.

Solution. The blocks are distinguished by the sign of some effect \(W\), which is confounded with blocks, and with \(W\) goes its alias \(W \cdot ABCDE\). The letters of \(W\) and of its alias are complementary subsets of \(\{A, B, C, D, E\}\), so their lengths add to \(5\) and one of them has length at most \(2\). Hence blocks are confounded with either a main effect (when \(W\) has length \(1\) or \(4\)) or a two-factor interaction (length \(2\) or \(3\)). The better choice is a two-factor interaction believed negligible. To confound \(DE = ABC\), take the sixteen runs of the half fraction (the full \(2^4\) in \(A\), \(B\), \(C\), \(D\) with \(E = ABCD\)) and divide them by the sign of \(ABC\):

The other main effects and the other nine two-factor interactions remain estimable, each aliased only with a three- or four-factor interaction.

Solution. Every effect has been tested against the error mean square, which is correct only for a fixed effects model. In the restricted mixed model of Theorem 11.4.2, with \(a = 3\), \(b = 4\), \(n = 2\), \[ E(MS_A) = \sigma ^2 + 2\sigma ^2_{\tau \beta } + 4\sum \tau _i^2 , \qquad E(MS_{AB}) = \sigma ^2 + 2\sigma ^2_{\tau \beta } , \] so machines must be tested against the interaction: \(F_0 = 40/10 = 4.0\) on \(2\) and \(6\) degrees of freedom, below \(F_{0.05, 2, 6} = 5.14\) (\(P = 0.079\)). The machines are not shown to differ. The reason is the significant interaction (\(F_0 = 5.0\), correctly against error): machine differences change from operator to operator, and with only four operators the average difference over the population of operators is poorly determined. The test against error answers a different question, about these four operators only.

Operators are tested against error in the restricted model (\(F_0 = 6.0\), \(P = 0.010\)), and against the interaction in the unrestricted one (\(F_0 = 1.2\), \(P = 0.39\)); the report should say which model it uses. The restricted estimates are \(\hat \sigma ^2_{\tau \beta } = (10 - 2)/2 = 4\) and \(\hat \sigma ^2_\beta = (12 - 2)/6 = 1.67\). Comparisons of machine means use the interaction mean square too: their standard error is \(\sqrt {2 \cdot 10/8} = 1.58\), not \(\sqrt {2 \cdot 2/8} = 0.71\).

Solution. By Theorem 11.2.2, \(MS_{\mathrm {Treatments}} = (\sigma ^2 + r\sigma ^2_\tau )U/(t - 1)\) and \(MSE = \sigma ^2V/(N - t)\), where \(U \sim \chi ^2_{t-1}\) and \(V \sim \chi ^2_{N-t}\) are independent. A \(\chi ^2_\nu \) variable has variance \(2\nu \), so a mean square \(M = \mathcal E\chi ^2_\nu /\nu \) with expectation \(\mathcal E\) has \(\Var (M) = 2\mathcal E^2/\nu \). By independence, \[ \Var (\hat \sigma ^2_\tau ) = \frac 1{r^2}\bigl (\Var (MS_{\mathrm {Treatments}}) + \Var (MSE)\bigr ) , \] which is the stated formula. For the estimator, \(E(M^2) = \Var (M) + \mathcal E^2 = \mathcal E^2(\nu + 2)/\nu \), so \(M^2/(\nu + 2)\) is unbiased for \(\mathcal E^2/\nu \), and \[ \widehat {\Var }(\hat \sigma ^2_\tau ) = \frac 2{r^2}\biggl (\frac {MS_{\mathrm {Treatments}}^2}{t + 1} + \frac {MSE^2}{N - t + 2}\biggr ) \] is unbiased. In Example 11.2.4, \(t = 6\), \(r = 4\), \(N = 24\), \(MS_{\mathrm {Treatments}} = 9.743\) and \(MSE = 1.024\): \[ \widehat {\Var }(\hat \sigma ^2_\tau ) = \frac 2{16}\biggl (\frac {9.743^2}7 + \frac {1.024^2}{20}\biggr ) = 1.70 , \] a standard error of \(1.30\) for the estimate \(2.18\). Almost all of it comes from the first term: a variance component between lots is estimated from only six lots, and its precision is governed by \(t - 1 = 5\) degrees of freedom however many tablets are measured. (A simulation of \(200000\) experiments with \(\sigma ^2 = 1\), \(\sigma ^2_\tau = 2\) gives a variance of \(2.030\) against the formula’s \(2.032\).)

Solution. By the pivot of Section 11, \(F_0/(1 + r\theta ) \sim F_{t-1, N-t}\) with \(\theta = \sigma ^2_\tau /\sigma ^2\), so the power is \[ P\bigl (F_{t-1, N-t} > F_{\alpha , t-1, N-t}/(1 + r\theta )\bigr ) , \] computed from a central \(F\) distribution. With \(r = 4\) and \(\theta = 1\) the power for \(t = 6, 7, 8, 9, 10\) lots is \(0.733\), \(0.791\), \(0.836\), \(0.872\) and \(0.901\), so ten lots (\(40\) tablets) are needed. With \(40\) tablets in all, the divisions \(5 \times 8\), \(10 \times 4\) and \(20 \times 2\) (lots \(\times \) tablets) give power \(0.880\), \(0.901\) and \(0.768\). Few lots leave few numerator degrees of freedom; many lots with few tablets make \(1 + r\theta \) small. Ten lots of four is the best of the three, and in general sampling more lots matters more than measuring more within each, once \(r\theta \) is moderately large.