7.6 Practice problems
Problem 7.6.1. Define a factorial experiment, and explain why the main effect of a factor can be misleading when the factor interacts with another.
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Solution. In a factorial experiment each complete replicate contains all combinations of the levels of the factors studied, so that the factors are varied together. A main effect is the average effect of a factor over the levels of the others. When the factor interacts with another, its effect differs from one level of the other factor to the next, and the average may describe none of them; effects of opposite sign may cancel to give a main effect of zero, as for the adhesives in Example 7.2.4.
Problem 7.6.2. In the two-factor model, prove directly that \[ E(SS_{AB}) = (a - 1)(b - 1)\sigma ^2 + n\sum _i\sum _j(\tau \beta )_{ij}^2 , \] using \(SS_{AB} = SS_{\mathrm {Subtotals}} - SS_A - SS_B\).
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Solution. \(SS_{\mathrm {Subtotals}}\) is the treatment sum of squares of a one-way model with \(ab\) treatments of \(n\) replicates and means \(\mu _{ij}\), so by Theorem 3.3.2 its expectation is \((ab - 1)\sigma ^2 + n\sum \sum (\mu _{ij} - \mu )^2\). Similarly \(E(SS_A) = (a - 1)\sigma ^2 + bn\sum \tau _i^2\) and \(E(SS_B) = (b - 1)\sigma ^2 + an\sum \beta _j^2\). Since \(\mu _{ij} - \mu = \tau _i + \beta _j + (\tau \beta )_{ij}\) and the side conditions make the three terms orthogonal when summed over \(i\) and \(j\), \[ \sum _i\sum _j(\mu _{ij} - \mu )^2 = b\sum \tau _i^2 + a\sum \beta _j^2 + \sum \sum (\tau \beta )_{ij}^2 . \] Subtracting, \(E(SS_{AB}) = (ab - 1 - a + 1 - b + 1)\sigma ^2 + n\sum \sum (\tau \beta )_{ij}^2 = (a - 1)(b - 1)\sigma ^2 + n\sum \sum (\tau \beta )_{ij}^2\).
Problem 7.6.3. Three machines and four operators were used in a factorial experiment with one observation per combination:
| machine | operator \(1\) | operator \(2\) | operator \(3\) | operator \(4\) |
| \(1\) | \(12\) | \(15\) | \(14\) | \(18\) |
| \(2\) | \(14\) | \(17\) | \(17\) | \(21\) |
| \(3\) | \(11\) | \(13\) | \(12\) | \(16\) |
Assuming no interaction, test the main effects at \(\alpha = 0.05\). Then check the assumption with Tukey’s one-degree-of-freedom test.
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Solution. The totals are \(59\), \(69\), \(52\) for machines and \(37\), \(45\), \(43\), \(55\) for operators; \(y_{\cdot \cdot } = 180\) and \(\sum y^2 = 2794\). The correction factor is \(2700\), so \(SST = 94\), \(SS_{\mathrm {Machines}} = (59^2 + 69^2 + 52^2)/4 - 2700 = 36.5\), \(SS_{\mathrm {Operators}} = (37^2 + 45^2 + 43^2 + 55^2)/3 - 2700 = 56.0\), and the residual, used as error, is \(1.5\) on \(6\) degrees of freedom (\(MSE = 0.25\)). Then \(F_{\mathrm {Machines}} = 18.25/0.25 = 73.0\) and \(F_{\mathrm {Operators}} = 18.67/0.25 = 74.7\), far beyond \(F_{0.05,2,6} = 5.14\) and \(F_{0.05,3,6} = 4.76\): both factors matter.
For Tukey’s test, with \(c_i\) the machine deviations and \(a_j\) the operator deviations, \(\sum \sum c_ia_jy_{ij} = 11.333\), \(\sum c_i^2 = 9.125\) and \(\sum a_j^2 = 18.667\), so \(SS_N = 11.333^2/(9.125 \times 18.667) = 0.754\), and \(F_0 = 0.754/\bigl ((1.5 - 0.754)/5\bigr ) = 5.05\) on \(1\) and \(5\) degrees of freedom (\(p = 0.074\)). There is no significant evidence of non-additivity at the \(5\%\) level, though the \(p\)-value is small enough that the assumption deserves a second look in further work.
Problem 7.6.4. A factorial experiment has four factors with \(2\), \(3\), \(3\) and \(4\) levels and \(n = 2\) replicates. Give the degrees of freedom for each main effect, for the interaction of the last three factors, for the four-factor interaction, and for error.
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Solution. The main effects have \(1\), \(2\), \(2\) and \(3\) degrees of freedom. The three-factor interaction \(BCD\) has \(2 \times 2 \times 3 = 12\), the four-factor interaction \(ABCD\) has \(1 \times 2 \times 2 \times 3 = 12\), and error has \(abcd(n - 1) = 72\). In all there are \(abcdn - 1 = 143\) degrees of freedom; the \(71\) for the treatment combinations split into the \(15\) main effects and interactions of all orders.
Problem 7.6.5. For the adhesive experiment, suppose the interaction had been ignored and an additive two-factor model fitted. Compute the resulting \(F\) statistic for adhesives, and explain why the conclusion about adhesives would be wrong even though the arithmetic is right.
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Solution. The interaction sum of squares joins the error: \(SSE = 54.00 + 118.89 = 172.89\) on \(22\) degrees of freedom, \(MSE = 7.86\), and \(F_{\mathrm {Adhesives}} = 1.44/7.86 = 0.18\). The test says the adhesives do not differ on average over the three temperatures, which is true. But the additive model claims that this average describes every temperature, which is false: the adhesives differ by up to \(6\) MPa at \(160\,^\circ \)C and by up to \(4.3\) MPa at \(120\,^\circ \)C, in opposite directions. The error mean square is inflated by the omitted interaction, and the model answers a question nobody should be asking.
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