All sixteen are worked below, each set out as 1. Problem, 2. Aim, 3. Formula, 4. Calculation, 5. Result. Practicals 5, 9 and 10 analyse the same \(2^{3}\) data three ways — unconfounded, totally confounded and partially confounded — so each result checks the others.
| # | Practical | Worked here | Also in |
|---|---|---|---|
| 1 | ANOVA for two-way classification, \(m\) observations per cell | Practical 1 | Unit 1, Example 1.1 |
| 2 | ANOVA for two-way classification, \(n_{ij}\) observations per cell | Practical 2 | Unit 1, Examples 1.2 and 1.3 |
| 3 | Analysis of covariance, one-way classification | Practical 3 | — |
| 4 | Analysis of covariance, two-way classification | Practical 4 | Unit 1, section 5 (method) |
| 5 | ANOVA for \(2^{3}\) and \(2^{4}\) factorial experiments | Practical 5 | Unit 2, Example 2.1 |
| 6 | ANOVA for \(3^{2}\) factorial experiments | Practical 6 | Unit 2, Example 2.2 |
| 7 | Identification of confounded terms in \(2^{3}\), \(2^{4}\), \(3^{2}\) | Practical 7 | — |
| 8 | Construction of a design with a specified effect confounded | Practical 8 | Unit 3, section 1 |
| 9 | ANOVA for total confounding of \(2^{3}\), \(2^{4}\) | Practical 9 | Unit 3, Example 3.1 |
| 10 | ANOVA for partial confounding of \(2^{3}\), \(2^{4}\) | Practical 10 | Unit 3, section 2 (method) |
| 11 | ANOVA for a one-half fraction of \(2^{4}\) | Practical 11 | Unit 3, section 3 (method) |
| 12 | ANOVA for a one-quarter fraction of \(2^{5}\) | Practical 12 | Unit 3, section 3 (method) |
| 13 | ANOVA for a split-plot design | Practical 13 | Unit 3, section 4 (method) |
| 14 | Analysis of a balanced incomplete block design | Practical 14 | Unit 3, Example 3.2 |
| 15 | Analysis of a Youden square design | Practical 15 | Unit 4, section 3 (method) |
| 16 | Analysis of a partially balanced incomplete block design | Practical 16 | Unit 4, Example 4.1 (design) |
A has \(a = 3\) levels and B has \(b = 2\), with \(m = 3\) observations per cell:
| B₁ | B₂ | Row total \(A_i\) | |
|---|---|---|---|
| A₁ | 7, 9, 8 (24) | 12, 10, 14 (36) | 60 |
| A₂ | 10, 12, 11 (33) | 15, 17, 16 (48) | 81 |
| A₃ | 6, 8, 7 (21) | 9, 11, 13 (33) | 54 |
| Column total \(B_j\) | 78 | 117 | 195 |
Analyse the data and test the interaction and the main effects at 5%.
To analyse a two-way classification with several observations per cell, separating the interaction from the error.
With \(m\) observations per cell the divisors are \(bm\), \(am\) and \(m\). The variation within a cell estimates \(\sigma^{2}\) on its own, so the interaction can be separated and tested.
Applying it:
| Source | SS | df | MS | \(F\) | \(F_{0.05}\) |
|---|---|---|---|---|---|
| A | 67.0 | 2 | 33.5 | 16.75 | 3.885 |
| B | 84.5 | 1 | 84.5 | 42.25 | 4.747 |
| AB | 1.0 | 2 | 0.5 | 0.25 | 3.885 |
| Error | 24.0 | 12 | 2.0 | ||
| Total | 176.5 | 17 |
The interaction is nowhere near significant (\(F = 0.25\)), so the main effects can be read as they stand: both A (\(F = 16.75\)) and B (\(F = 42.25\)) are significant. The cell means are 8, 12 for A₁, 11, 16 for A₂ and 7, 11 for A₃: the B₂ − B₁ gap is 4, 5, 4, nearly constant, which is additivity (worked also in Unit 1, Example 1.1).
(a) Proportional frequencies: analyse
| B₁ | B₂ | \(n_{i\cdot}\) | \(A_i\) | |
|---|---|---|---|---|
| A₁ | 5, 7, 6, 6 (total 24) | 9, 11 (20) | 6 | 44 |
| A₂ | 8, 10, 9, 9, 11, 7 (54) | 13, 15, 14 (42) | 9 | 96 |
| \(n_{\cdot j}\), \(B_j\) | 10, 78 | 5, 62 | 15 | 140 |
(b) Disproportionate frequencies: analyse
| B₁ | B₂ | \(n_{i\cdot}\) | \(A_i\) | |
|---|---|---|---|---|
| A₁ | 5, 7, 6, 6 (mean 6) | 9, 11 (mean 10) | 6 | 44 |
| A₂ | 8, 10, 9 (mean 9) | 13, 15, 14, 12, 16, 14 (mean 14) | 9 | 111 |
| \(n_{\cdot j}\), \(B_j\) | 7, 51 | 8, 104 | 15 | 155 |
To analyse a two-way classification with unequal cell frequencies: directly when they are proportional, and by fitting constants when they are not.
The frequencies are proportional when \(n_{ij} = n_{i\cdot}n_{\cdot j}/N\) for every cell; then the formulae above apply. Otherwise the design is not orthogonal, and A and B must each be adjusted for the other by fitting the additive model: \(R(A \mid \mu, B) = R(\mu, A, B) - R(\mu, B)\), \(R(B \mid \mu, A) = R(\mu, A, B) - R(\mu, A)\), with \(SS_{AB} = SS_{\text{cells}} - R(\mu, A, B)\).
Applying it:
(a) \(6 \times 10/15 = 4 = n_{11}\), \(6 \times 5/15 = 2\), \(9 \times 10/15 = 6\), \(9 \times 5/15 = 3\): proportional. \(\checkmark\) \(\text{CF} = 140^{2}/15 = 1306.666667\); \(SS_{\text{total}} = 1434 - 1306.666667 = 127.333333\);
\[ SS_A = \frac{44^{2}}{6} + \frac{96^{2}}{9} - \text{CF} = 40, \quad SS_B = \frac{78^{2}}{10} + \frac{62^{2}}{5} - \text{CF} = 70.533333, \quad SS_{\text{cells}} = 144 + 200 + 486 + 588 - \text{CF} = 111.333333, \] \[ SS_{AB} = 0.8, \qquad SS_E = 16 \ \text{ on } 11 \text{ d.f.}, \ MS_E = 1.454545; \qquad F_A = 27.50, \quad F_B = 48.49, \quad F_{AB} = 0.55 \ (F_{0.05;\,1,\,11} = 4.844). \](b) \(6 \times 7/15 = 2.8 \ne 4\): not proportional. With \(\text{CF} = 155^{2}/15 = 1601.666667\), \(SS_{\text{total}} = 177.333333\), \(SS_A = 90\), \(SS_B = 121.904762\) and \(SS_{\text{cells}} = 161.333333\), the "interaction by subtraction" would be \(161.333333 - 211.904762 = -50.571429\) — impossible. Fitting \(\mu, \alpha, \beta\) (with \(\alpha_2 = \beta_2 = 0\)) by weighted least squares on the cell means:
\[ 15\mu + 6\alpha + 7\beta = 155, \qquad 6\mu + 6\alpha + 4\beta = 44, \qquad 7\mu + 4\alpha + 7\beta = 51, \] \[ \mu = \frac{208}{15} = 13.866667, \quad \alpha = -\frac{52}{15} = -3.466667, \quad \beta = -\frac{23}{5} = -4.6, \]with fitted cell means 5.8, 10.4, 9.266667, 13.866667, and
\[ R(\mu, A, B) = 1762.2 - 1601.666667 = 160.533333, \quad R(A \mid \mu, B) = 160.533333 - 121.904762 = 38.628571, \] \[ R(B \mid \mu, A) = 160.533333 - 90 = 70.533333, \quad SS_{AB} = 161.333333 - 160.533333 = 0.8, \quad SS_E = 16. \]| Source | SS | df | MS | \(F\) |
|---|---|---|---|---|
| A adjusted for B | 38.628571 | 1 | 38.628571 | 26.557143 |
| B adjusted for A | 70.533333 | 1 | 70.533333 | 48.491667 |
| AB | 0.800000 | 1 | 0.800000 | 0.550000 |
| Error | 16.000000 | 11 | 1.454545 |
(a) A and B are significant, the interaction is not; the three components add to \(SS_{\text{cells}}\), which is the orthogonality. (b) Again A and B are significant and AB is not, but A's sum of squares falls from 90 unadjusted to 38.63 adjusted: more than half of what A appeared to explain was really B's. Only the adjusted figure can be reported, and the three terms need not add to \(SS_{\text{cells}}\) (worked also in Unit 1, Examples 1.2 and 1.3).
Three treatments, four units each. For every unit a covariate \(x\) was measured before the treatment was applied, and the response \(y\) after.
| Treatment | \((x, y)\) pairs | \(\sum x\) | \(\sum y\) |
|---|---|---|---|
| T₁ | (2, 8), (3, 10), (4, 11), (5, 13) | 14 | 42 |
| T₂ | (3, 12), (4, 13), (5, 15), (6, 17) | 18 | 57 |
| T₃ | (4, 11), (5, 13), (6, 14), (7, 16) | 22 | 54 |
| Total | 54 | 153 |
Test the treatments after adjusting for the covariate, and find the adjusted treatment means.
To compare treatments after removing, by regression, the part of the response explained by a covariate.
The slope comes from the error line because that is the within-treatment relation between \(x\) and \(y\), the one the model assumes common to all treatments.
Applying it:
The raw sums are \(\sum x^{2} = 266\), \(\sum y^{2} = 2023\), \(\sum xy = 725\), so \(T_{xx} = 23\), \(T_{yy} = 72.25\), \(T_{xy} = 36.5\). From the treatment totals,
\[ R_{xx} = 251 - 243 = 8, \qquad R_{yy} = 1982.25 - 1950.75 = 31.5, \qquad R_{xy} = 700.5 - 688.5 = 12. \]| Source | df | \(S_{xx}\) | \(S_{xy}\) | \(S_{yy}\) |
|---|---|---|---|---|
| Treatments | 2 | 8.00 | 12.00 | 31.50 |
| Error | 9 | 15.00 | 24.50 | 40.75 |
| Total | 11 | 23.00 | 36.50 | 72.25 |
| Treatment | \(\bar x_i\) | \(\bar y_i\) | \(\bar y'_i\) |
|---|---|---|---|
| T₁ | 3.50 | 10.50 | 12.133333 |
| T₂ | 4.50 | 14.25 | 14.250000 |
| T₃ | 5.50 | 13.50 | 11.866667 |
Ignoring the covariate: \(F = (31.5/2)/(40.75/9) = 3.478528\), \(p = 0.076\).
After adjustment the treatments differ highly significantly (\(F = 74.1\)); without it they did not (\(F = 3.48\)). The error collapsed — the covariate explained 98.2% of what had been called error — and the ranking reversed: unadjusted, T₃ (13.50) beats T₁ (10.50); adjusted, T₁ (12.13) beats T₃ (11.87), because T₃'s units simply started higher. Reporting the unadjusted means would have been reporting the covariate.
Three treatments are compared in a randomised block design with four blocks. The covariate \(x\) (initial weight) was measured before treatment and the response \(y\) (final weight) after:
| Block | T₁ \((x, y)\) | T₂ \((x, y)\) | T₃ \((x, y)\) | Block totals \((x, y)\) |
|---|---|---|---|---|
| 1 | (10, 16) | (12, 22) | (11, 18) | (33, 56) |
| 2 | (14, 23) | (13, 24) | (15, 25) | (42, 72) |
| 3 | (9, 13) | (11, 19) | (10, 15) | (30, 47) |
| 4 | (13, 21) | (15, 27) | (14, 22) | (42, 70) |
| Treatment totals | (46, 73) | (51, 92) | (50, 80) | (147, 245) |
Test the treatments after adjusting for the covariate, and find the adjusted treatment means.
To carry the analysis of covariance into a randomised block design, with one more line in the table.
Two assumptions are part of the answer: the slope is the same for every treatment (parallel regressions), and the covariate is unaffected by the treatment.
Applying it:
\(N = 12\); \(\sum x = 147\), \(\sum y = 245\), \(\sum x^{2} = 1847\), \(\sum y^{2} = 5203\), \(\sum xy = 3093\).
\[ \text{CF}_{xx} = \frac{147^{2}}{12} = 1800.75, \qquad \text{CF}_{yy} = \frac{245^{2}}{12} = 5002.083333, \qquad \text{CF}_{xy} = \frac{147 \times 245}{12} = 3001.25. \] \[ \text{Blocks: } B_{xx} = \frac{33^{2}+42^{2}+30^{2}+42^{2}}{3} - 1800.75 = 38.25, \quad B_{yy} = 140.916667, \quad B_{xy} = \frac{33(56)+42(72)+30(47)+42(70)}{3} - 3001.25 = 72.75. \] \[ \text{Treatments: } R_{xx} = \frac{46^{2}+51^{2}+50^{2}}{4} - 1800.75 = 3.5, \quad R_{yy} = 46.166667, \quad R_{xy} = 11.25. \]| Source | df | \(S_{xx}\) | \(S_{xy}\) | \(S_{yy}\) |
|---|---|---|---|---|
| Blocks | 3 | 38.25 | 72.75 | 140.916667 |
| Treatments | 2 | 3.50 | 11.25 | 46.166667 |
| Error | 6 | 4.50 | 7.75 | 13.833333 |
| Total | 11 | 46.25 | 91.75 | 200.916667 |
With \(\bar x = 12.25\):
| Treatment | \(\bar x_i\) | \(\bar y_i\) | \(\bar y'_i\) |
|---|---|---|---|
| T₁ | 11.50 | 18.25 | 19.541667 |
| T₂ | 12.75 | 23.00 | 22.138889 |
| T₃ | 12.50 | 20.00 | 19.569444 |
Ignoring the covariate: \(F = (46.166667/2)/(13.833333/6) = 23.083333/2.305556 = 10.01\) on (2, 6) d.f.
Adjusted for initial weight, the treatments differ significantly (\(F = 74.0 > 5.786\)), and T₂ is best (adjusted mean 22.14). The covariate cut the error mean square from 2.31 to 0.097. It also changed the comparison of T₁ and T₃: unadjusted, T₃ leads T₁ by 1.75; adjusted, they are almost equal (19.54 and 19.57), because T₃'s animals started heavier. The analysis assumes a common slope in every treatment, which should be checked by fitting separate slopes.
A \(2^{3}\) factorial is run in \(r = 2\) replicates (one complete replicate per block):
| Combination | Replicate I | Replicate II | Total |
|---|---|---|---|
| (1) | 9 | 11 | 20 |
| a | 16 | 16 | 32 |
| b | 12 | 14 | 26 |
| ab | 21 | 23 | 44 |
| c | 11 | 13 | 24 |
| ac | 18 | 20 | 38 |
| bc | 15 | 15 | 30 |
| abc | 25 | 27 | 52 |
| Replicate total | 127 | 139 | 266 |
Estimate every effect by Yates's algorithm and complete the analysis of variance. State how the method extends to a \(2^{4}\).
To estimate the main effects and interactions of a two-level factorial by Yates's algorithm, and test them.
A factor in the effect contributes \((\text{letter} - 1)\), a factor not in it \((\text{letter} + 1)\). Yates's algorithm takes \(k\) passes of sums of pairs then differences of pairs on the totals in standard order. Checks: the first entry of the last column is the grand total; the effect sums of squares add to the treatment sum of squares. For a \(2^{4}\) the method is the same with sixteen totals and four passes, giving fifteen effects.
Applying it:
| Combination | Total | (1) | (2) | (3) | Effect | SS |
|---|---|---|---|---|---|---|
| (1) | 20 | 52 | 122 | 266 = \(G\) | ||
| a | 32 | 70 | 144 | 66 = [A] | 8.25 | 272.25 |
| b | 26 | 62 | 30 | 38 = [B] | 4.75 | 90.25 |
| ab | 44 | 82 | 36 | 14 = [AB] | 1.75 | 12.25 |
| c | 24 | 12 | 18 | 22 = [C] | 2.75 | 30.25 |
| ac | 38 | 18 | 20 | 6 = [AC] | 0.75 | 2.25 |
| bc | 30 | 14 | 6 | 2 = [BC] | 0.25 | 0.25 |
| abc | 52 | 22 | 8 | 2 = [ABC] | 0.25 | 0.25 |
Check: \([A] = (32 + 44 + 38 + 52) - (20 + 26 + 24 + 30) = 166 - 100 = 66\). \(\checkmark\) The effect sums of squares total \(407.75\). With \(\text{CF} = 266^{2}/16 = 4422.25\) and raw sum of squares 4842:
\[ SS_{\text{total}} = 419.75, \qquad SS_{\text{rep}} = \frac{127^{2}+139^{2}}{8} - \text{CF} = 9, \qquad SS_E = 419.75 - 9 - 407.75 = 3 \ \text{ on } 7 \text{ d.f.}, \ MS_E = 0.428571. \]| Source | SS | df | \(F\) |
|---|---|---|---|
| Replicates | 9.00 | 1 | |
| A | 272.25 | 1 | 635.25 |
| B | 90.25 | 1 | 210.58 |
| C | 30.25 | 1 | 70.58 |
| AB | 12.25 | 1 | 28.58 |
| AC | 2.25 | 1 | 5.25 |
| BC | 0.25 | 1 | 0.58 |
| ABC | 0.25 | 1 | 0.58 |
| Error | 3.00 | 7 | |
| Total | 419.75 | 15 |
\(F_{0.05;\,1,\,7} = 5.591\).
A, B, C and the AB interaction are significant; AC (\(F = 5.25\)) is marginal and BC and ABC are not. Read AB from the data: the effect of A is 6.5 at low B and 10 at high B; half the difference, 1.75, is the AB effect. Negligible three-factor interactions are what make the confounded and fractional designs of Practicals 9–12 worth having (worked also in Unit 2, Example 2.1).
A \(3^{2}\) factorial has \(r = 2\) replicates; the two observations in each cell differ by 2. The cell totals are:
| b₀ | b₁ | b₂ | \(A_i\) | |
|---|---|---|---|---|
| a₀ | 10 | 14 | 18 | 42 |
| a₁ | 16 | 22 | 26 | 64 |
| a₂ | 20 | 24 | 30 | 74 |
| \(B_j\) | 46 | 60 | 74 | 180 |
Analyse the experiment, splitting each main effect and the interaction into single degrees of freedom.
To analyse a three-level factorial and split its effects into linear and quadratic components.
For a main-effect component on marginal totals, the divisor is \(r\sum c_i^{2}\) times the three levels of the other factor.
Applying it:
\(SS_E = 9 \times 2 = 18\) on 9 d.f., \(MS_E = 2\). With \(\text{CF} = 180^{2}/18 = 1800\) and raw sum of squares 1974:
\[ SS_A = \frac{11336}{6} - 1800 = 89.333333, \quad SS_B = \frac{11192}{6} - 1800 = 65.333333, \quad SS_{\text{cells}} = \frac{3912}{2} - 1800 = 156, \quad SS_{AB} = 1.333333. \] \[ C_{A_L} = -42 + 74 = 32, \ SS = \frac{1024}{12} = 85.333333; \qquad C_{A_Q} = 42 - 128 + 74 = -12, \ SS = \frac{144}{36} = 4; \] \[ C_{B_L} = -46 + 74 = 28, \ SS = \frac{784}{12} = 65.333333; \qquad C_{B_Q} = 46 - 120 + 74 = 0. \] \[ C_{A_QB_Q} = 10 - 28 + 18 - 32 + 88 - 52 + 20 - 48 + 30 = 6, \qquad SS_{A_QB_Q} = \frac{36}{72} = 0.5. \]| Component | Contrast | Divisor | SS | \(F\) (1, 9) |
|---|---|---|---|---|
| \(A_L\) | 32 | 12 | 85.333333 | 42.666667 |
| \(A_Q\) | −12 | 36 | 4.000000 | 2.000000 |
| \(B_L\) | 28 | 12 | 65.333333 | 32.666667 |
| \(B_Q\) | 0 | 36 | 0.000000 | 0.000000 |
| \(A_LB_L\) | 2 | 8 | 0.500000 | 0.250000 |
| \(A_LB_Q\) | 2 | 24 | 0.166667 | 0.083333 |
| \(A_QB_L\) | −2 | 24 | 0.166667 | 0.083333 |
| \(A_QB_Q\) | 6 | 72 | 0.500000 | 0.250000 |
The four interaction components sum to 1.333333 \(= SS_{AB}\), and all eight to 156 \(= SS_{\text{cells}}\). \(\checkmark\) \(F_{0.05;\,1,\,9} = 5.117\).
Almost everything is in \(A_L\) and \(B_L\): the response rises nearly linearly in both factors, with no useful curvature and no interaction (B is exactly linear: 46, 60, 74). Testing \(A_L\) alone (\(F = 42.7\)) is nearly twice as sensitive as the 2-d.f. test of A (\(F = 22.3\)), because it is aimed at the alternative that actually holds (worked also in Unit 2, Example 2.2).
Identify the confounded effects in each arrangement.
(a) A \(2^{3}\) in two blocks: Block 1: \((1), ab, c, abc\); Block 2: \(a, b, ac, bc\).
(b) A \(2^{4}\) in four blocks of four, whose principal block is \((1), ac, abd, bcd\).
(c) A \(3^{2}\) in three blocks, one of which is \((0,0), (1,1), (2,2)\).
To identify which effects a given blocked arrangement of a factorial has confounded with blocks.
The principal block (the one containing \((1)\)) must be a subgroup: closed under multiplication with squares reduced. With \(2^{p}\) blocks, \(2^{p} - 1\) effects are confounded: the generators and their generalised interactions.
Applying it:
(a) \(ab \cdot c = abc\), \(ab \cdot abc = c\), \(c \cdot abc = ab\): closed. \(\checkmark\)
| Effect | (1) | ab | c | abc | All even? |
|---|---|---|---|---|---|
| \(A\) | 0 | 1 | 0 | 1 | no |
| \(B\) | 0 | 1 | 0 | 1 | no |
| \(C\) | 0 | 0 | 1 | 1 | no |
| \(AB\) | 0 | 2 | 0 | 2 | yes |
| \(AC\) | 0 | 1 | 1 | 2 | no |
| \(BC\) | 0 | 1 | 1 | 2 | no |
| \(ABC\) | 0 | 2 | 1 | 3 | no |
Modulo 2, block 1 is \((000), (110), (001), (111)\), which satisfy \(x_1 + x_2 \equiv 0\) — the defining equation of \(AB\).
(b) \(ac \cdot abd = bcd\), \(ac \cdot bcd = abd\), \(abd \cdot bcd = ac\): closed. \(\checkmark\) The letters shared with each member \((1), ac, abd, bcd\):
| Effect | counts | All even? | Effect | counts | All even? |
|---|---|---|---|---|---|
| \(A\) | 0, 1, 1, 0 | no | \(BC\) | 0, 1, 1, 2 | no |
| \(B\) | 0, 0, 1, 1 | no | \(BD\) | 0, 0, 2, 2 | yes |
| \(C\) | 0, 1, 0, 1 | no | \(CD\) | 0, 1, 1, 2 | no |
| \(D\) | 0, 0, 1, 1 | no | \(ABC\) | 0, 2, 2, 2 | yes |
| \(AB\) | 0, 1, 2, 1 | no | \(ABD\) | 0, 1, 3, 2 | no |
| \(AC\) | 0, 2, 1, 1 | no | \(ACD\) | 0, 2, 2, 2 | yes |
| \(AD\) | 0, 1, 2, 1 | no | \(BCD\) | 0, 1, 2, 3 | no |
| \(ABCD\) | 0, 2, 3, 3 | no |
Check: \(ABC \cdot ACD = A^{2}BC^{2}D = BD\), the generalised interaction. \(\checkmark\)
(c) For \((0,0), (1,1), (2,2)\): \(x_1 + x_2 \equiv 0, 2, 1\) (not constant); \(x_1 + 2x_2 \equiv 0, 0, 0\) (constant).
(a) \(AB\) is confounded — a bad design, since a two-factor interaction has been sacrificed when \(ABC\) was available. (b) \(ABC\), \(ACD\) and their generalised interaction \(BD\) are confounded: in blocks of four a \(2^{4}\) must give up at least one two-factor interaction. (c) The \(AB^{2}\) component is confounded; both main effects remain orthogonal to blocks.
(a) Arrange a \(2^{4}\) in two blocks of eight with \(ABCD\) confounded. (b) Arrange it in four blocks of four, and compare the choices \(\{ABC, BCD\}\), \(\{ABC, ABD\}\) and \(\{ABCD, ABC\}\) as generators. (c) Arrange a \(3^{2}\) in three blocks of three with \(AB\) confounded.
To construct a blocked factorial in which a chosen effect is confounded with blocks, checking the generalised interactions before adopting it.
Sacrifice the highest-order interaction available, and compute every generalised interaction before the design is adopted (Unit 3, section 1).
Applying it:
(a) Even number of letters:
\[ \text{Block 1: } (1),\ ab,\ ac,\ ad,\ bc,\ bd,\ cd,\ abcd; \qquad \text{Block 2: } a,\ b,\ c,\ d,\ abc,\ abd,\ acd,\ bcd. \]Eight in each; block 1 contains \((1)\) and is closed under multiplication.
(b)
\[ ABC \cdot BCD = AD, \qquad ABC \cdot ABD = CD, \qquad ABCD \cdot ABC = D. \]The first two each confound a two-factor interaction; the third confounds a main effect. A better choice is \(\{ABC, ACD\}\), whose interaction is \(BD\) (Practical 7(b)): every four-block arrangement of a \(2^{4}\) loses one two-factor interaction, so the choice is which one.
(c)
| \(x_1+x_2 \bmod 3\) | Block |
|---|---|
| 0 | (0,0), (1,2), (2,1) |
| 1 | (0,1), (1,0), (2,2) |
| 2 | (0,2), (1,1), (2,0) |
(a) The two blocks above confound \(ABCD\) only. (b) None of the three proposed pairs is acceptable — two lose a two-factor interaction they need not and the third a main effect; compute the generalised interaction first. (c) Each block holds every level of A once and every level of B once, so both main effects are orthogonal to blocks and the two block degrees of freedom are exactly the \(AB\) component.
The \(2^{3}\) data of Practical 5 were run as two replicates of two blocks of four, with \(ABC\) confounded in both:
| Replicate | Block 1: (1), ab, ac, bc | Block 2: a, b, c, abc | Replicate total |
|---|---|---|---|
| I | 9 + 21 + 18 + 15 = 63 | 16 + 12 + 11 + 25 = 64 | 127 |
| II | 11 + 23 + 20 + 15 = 69 | 16 + 14 + 13 + 27 = 70 | 139 |
Complete the analysis of variance, and give the degrees of freedom for a \(2^{4}\) in \(r\) replicates of two blocks with \(ABCD\) confounded.
To analyse a factorial with one effect totally confounded with blocks, and to see that the block sum of squares is that effect's.
Applying it:
and \(SS_{ABC} = 2^{2}/16 = 0.25\) (Practical 5): equal.
| Source | SS | df | MS | \(F\) |
|---|---|---|---|---|
| Replicates | 9.00 | 1 | ||
| Blocks within replicates (= \(ABC\)) | 0.25 | 2 | ||
| A | 272.25 | 1 | 272.25 | 544.5 |
| B | 90.25 | 1 | 90.25 | 180.5 |
| C | 30.25 | 1 | 30.25 | 60.5 |
| AB | 12.25 | 1 | 12.25 | 24.5 |
| AC | 2.25 | 1 | 2.25 | 4.5 |
| BC | 0.25 | 1 | 0.25 | 0.5 |
| Error | 3.00 | 6 | 0.50 | |
| Total | 419.75 | 15 |
\(F_{0.05;\,1,\,6} = 5.987\). For a \(2^{4}\) in \(r\) replicates of two blocks with \(ABCD\) confounded: replicates \(r - 1\), blocks within replicates \(r\) (= \(ABCD\)), 14 unconfounded effects, error \(14(r-1)\), total \(16r - 1\).
A, B, C and AB are significant; AC, BC are not; \(ABC\) cannot be tested at all — its sum of squares is the block sum of squares. Confounding cost one degree of freedom of error and the whole of \(ABC\), and bought blocks of four instead of eight, which in a real experiment would lower the error mean square (worked also in Unit 3, Example 3.1).
The \(2^{3}\) data of Practical 5 were run in two replicates of two blocks of four, with \(ABC\) confounded in replicate I and \(AB\) in replicate II:
| Replicate | Block 1 | Block 2 |
|---|---|---|
| I (\(ABC\) confounded) | (1) 9, ab 21, ac 18, bc 15 = 63 | a 16, b 12, c 11, abc 25 = 64 |
| II (\(AB\) confounded) | (1) 11, ab 23, c 13, abc 27 = 74 | a 16, b 14, ac 20, bc 15 = 65 |
Complete the analysis of variance.
To analyse a factorial in which different effects are confounded in different replicates, each being estimated from the replicates where it is free.
Unconfounded effects are estimated from all replicates; a partially confounded one only from the replicates in which it is free (Unit 3, section 2).
Applying it:
From both replicates (Practical 5): \(SS_A = 272.25\), \(SS_B = 90.25\), \(SS_C = 30.25\), \(SS_{AC} = 2.25\), \(SS_{BC} = 0.25\).
\[ [AB]_{\text{I}} = (9 + 21 + 11 + 25) - (16 + 12 + 18 + 15) = 66 - 61 = 5, \qquad SS_{AB} = \frac{5^{2}}{1 \times 8} = 3.125, \] \[ [ABC]_{\text{II}} = (16 + 14 + 13 + 27) - (11 + 23 + 20 + 15) = 70 - 69 = 1, \qquad SS_{ABC} = \frac{1^{2}}{8} = 0.125. \] \[ SS_E = 419.75 - 9 - 10.25 - 395.25 - 3.125 - 0.125 = 2.0 \ \text{ on } 15 - 1 - 2 - 7 = 5 \text{ d.f.}, \ MS_E = 0.4. \]| Source | SS | df | MS | \(F\) |
|---|---|---|---|---|
| Replicates | 9.000 | 1 | ||
| Blocks within replicates | 10.250 | 2 | ||
| A | 272.250 | 1 | 272.250 | 680.6 |
| B | 90.250 | 1 | 90.250 | 225.6 |
| C | 30.250 | 1 | 30.250 | 75.6 |
| AB (from replicate I) | 3.125 | 1 | 3.125 | 7.81 |
| AC | 2.250 | 1 | 2.250 | 5.63 |
| BC | 0.250 | 1 | 0.250 | 0.63 |
| ABC (from replicate II) | 0.125 | 1 | 0.125 | 0.31 |
| Error | 2.000 | 5 | 0.400 | |
| Total | 419.750 | 15 |
\(F_{0.05;\,1,\,5} = 6.608\).
A, B, C and AB are significant; AC, BC and ABC are not. Unlike Practical 9, every effect is tested — AB and ABC on half the information each (\(r'/r = 1/2\)), so their estimates have twice the variance of the others. AB's sum of squares, 3.125 from one replicate, differs from Practical 5's 12.25 from two: an effect estimated from fewer replicates is estimated less precisely, which is the price of keeping it.
A one-half fraction of a \(2^{4}\) with defining contrast \(I = ABCD\) was run in two replicates:
| Run | A | B | C | D = ABC | Rep I | Rep II | Total |
|---|---|---|---|---|---|---|---|
| (1) | − | − | − | − | 14 | 16 | 30 |
| ad | + | − | − | + | 22 | 22 | 44 |
| bd | − | + | − | + | 19 | 17 | 36 |
| ab | + | + | − | − | 25 | 27 | 52 |
| cd | − | − | + | + | 17 | 15 | 32 |
| ac | + | − | + | − | 23 | 25 | 48 |
| bc | − | + | + | − | 20 | 22 | 42 |
| abcd | + | + | + | + | 30 | 28 | 58 |
| Total | 170 | 172 | 342 | ||||
Write the alias structure, estimate the effects and complete the analysis of variance.
To analyse a half-replicate of a two-level factorial, reading each estimate together with its alias.
The fraction is the principal block of \(ABCD\): set \(D = ABC\) on a full \(2^{3}\) in A, B, C. Its resolution is IV: main effects are clear of two-factor interactions, which are aliased in pairs (Unit 3, section 3).
Applying it:
| Source (alias) | Contrast | Effect | SS | \(F\) (1, 7) |
|---|---|---|---|---|
| A (= BCD) | 62 | 7.75 | 240.25 | 122.31 |
| B (= ACD) | 34 | 4.25 | 72.25 | 36.78 |
| C (= ABD) | 18 | 2.25 | 20.25 | 10.31 |
| D (= ABC) | −2 | −0.25 | 0.25 | 0.13 |
| AB (= CD) | 2 | 0.25 | 0.25 | 0.13 |
| AC (= BD) | 2 | 0.25 | 0.25 | 0.13 |
| AD (= BC) | 6 | 0.75 | 2.25 | 1.15 |
| Replicates | 0.25 | |||
| Error (7 d.f.) | 13.75 | MS 1.964 | ||
| Total (15 d.f.) | 349.75 |
\(SS_E = 349.75 - 0.25 - 335.75 = 13.75\). \(F_{0.05;\,1,\,7} = 5.591\).
A, B and C are significant; D and the three pairs of two-factor interactions are not. Each reading rests on the assumption that the three-factor alias is negligible — the estimate labelled A is really \(A + BCD\). Eight runs per replicate instead of sixteen have bought clear main effects; the two-factor interactions could not have been separated from each other if any had been large.
A one-quarter fraction of a \(2^{5}\), with generators \(D = AB\) and \(E = AC\), was run in two replicates:
| Run | A | B | C | D = AB | E = AC | Rep I | Rep II | Total |
|---|---|---|---|---|---|---|---|---|
| de | − | − | − | + | + | 18 | 20 | 38 |
| a | + | − | − | − | − | 25 | 25 | 50 |
| be | − | + | − | − | + | 20 | 22 | 42 |
| abd | + | + | − | + | − | 29 | 27 | 56 |
| cd | − | − | + | + | − | 22 | 20 | 42 |
| ace | + | − | + | − | + | 27 | 29 | 56 |
| bc | − | + | + | − | − | 21 | 23 | 44 |
| abcde | + | + | + | + | + | 33 | 31 | 64 |
| Total | 195 | 197 | 392 | |||||
Write the defining relation and aliases, estimate the effects and complete the analysis of variance.
To analyse a quarter-replicate of a two-level factorial in five factors, a resolution III screening design.
The third word is the product of the first two, \(ABD \cdot ACE = BCDE\). The shortest word has three letters, so the design is of resolution III: main effects are aliased with two-factor interactions.
Applying it:
| Estimates | Aliased with |
|---|---|
| \(A\) | \(BD\), \(CE\), \(ABCDE\) |
| \(B\) | \(AD\), \(CDE\), \(ABCE\) |
| \(C\) | \(AE\), \(BDE\), \(ABCD\) |
| \(D\) | \(AB\), \(BCE\), \(ACDE\) |
| \(E\) | \(AC\), \(BCD\), \(ABDE\) |
| \(BC\) | \(DE\), \(ACD\), \(ABE\) |
| \(BE\) | \(CD\), \(ADE\), \(ABC\) |
| Source (main alias) | Contrast | SS | \(F\) (1, 7) |
|---|---|---|---|
| A (= BD = CE) | 60 | 225 | 114.55 |
| B (= AD) | 20 | 25 | 12.73 |
| C (= AE) | 20 | 25 | 12.73 |
| D (= AB) | 8 | 4 | 2.04 |
| E (= AC) | 8 | 4 | 2.04 |
| BC (= DE) | 0 | 0 | 0 |
| BE (= CD) | 4 | 1 | 0.51 |
| Replicates | 0.25 | ||
| Error (7 d.f.) | 13.75 | MS 1.964 | |
| Total (15 d.f.) | 298 |
\(SS_E = 298 - 0.25 - 284 = 13.75\). \(F_{0.05;\,1,\,7} = 5.591\).
A, B and C are significant; D, E and the interaction strings are not. Because the design is resolution III, each significant estimate is really a main effect plus two-factor interactions — "B" is \(B + AD\). Five factors in eight runs is a screening bargain: it says A, B and C are worth a follow-up experiment that separates them from their aliases.
Two irrigation methods (A, whole plots) and three varieties (B, sub-plots) are compared in three replicates. Each whole plot was split into three sub-plots, to which the varieties were randomised:
| Replicate | A₁: B₁, B₂, B₃ | A₂: B₁, B₂, B₃ | Total |
|---|---|---|---|
| 1 | 20, 24, 22 (66) | 26, 30, 29 (85) | 151 |
| 2 | 18, 23, 21 (62) | 27, 31, 29 (87) | 149 |
| 3 | 21, 26, 23 (70) | 27, 30, 30 (87) | 157 |
Analyse the experiment.
To analyse a split-plot design, testing the whole-plot factor against the whole-plot error and the sub-plot factor and interaction against the sub-plot error.
A is compared between whole plots, so it is tested against the whole-plot error; B and \(A \times B\) within whole plots, against the sub-plot error (Unit 3, section 4).
Applying it:
\(G = 457\), \(N = 18\), \(\text{CF} = 457^{2}/18 = 11602.722222\), raw sum of squares 11877, \(SS_{\text{total}} = 274.277778\).
\[ SS_R = \frac{151^{2}+149^{2}+157^{2}}{6} - \text{CF} = 5.777778, \qquad SS_A = \frac{198^{2}+259^{2}}{9} - \text{CF} = 206.722222, \] \[ \sum\frac{(RA)^{2}}{3} - \text{CF} = \frac{66^{2}+85^{2}+62^{2}+87^{2}+70^{2}+87^{2}}{3} - \text{CF} = 218.277778, \qquad SS_{\text{WP error}} = 218.277778 - 5.777778 - 206.722222 = 5.777778. \]The variety totals are 139, 164, 154 and the A × B totals 59, 73, 66 (A₁) and 80, 91, 88 (A₂):
\[ SS_B = \frac{139^{2}+164^{2}+154^{2}}{6} - \text{CF} = 52.777778, \qquad SS_{AB} = 1.444444, \qquad SS_{\text{SP error}} = 274.277778 - 218.277778 - 52.777778 - 1.444444 = 1.777778. \]| Source | SS | df | MS | \(F\) | \(F_{0.05}\) |
|---|---|---|---|---|---|
| Replicates | 5.777778 | 2 | 2.888889 | ||
| A (irrigation) | 206.722222 | 1 | 206.722222 | 71.56 | 18.51 |
| Whole-plot error | 5.777778 | 2 | 2.888889 | ||
| B (variety) | 52.777778 | 2 | 26.388889 | 118.75 | 4.459 |
| A × B | 1.444444 | 2 | 0.722222 | 3.25 | 4.459 |
| Sub-plot error | 1.777778 | 8 | 0.222222 | ||
| Total | 274.277778 | 17 |
Irrigation (\(F = 71.6\) on 1 and 2 d.f.) and variety (\(F = 118.8\)) are both significant; their interaction is not (\(F = 3.25\)). The whole-plot error mean square, 2.89, is thirteen times the sub-plot one, 0.22: the design has bought precision on the varieties and the interaction at the cost of precision on irrigation, which is tested on only 2 error degrees of freedom. Analysing this as an ordinary factorial, with one pooled error, would overstate the evidence on irrigation.
Four treatments are compared in six blocks of two, every pair of treatments together exactly once:
| Block | Contents | Block total \(B_j\) |
|---|---|---|
| 1 | A = 10, B = 12 | 22 |
| 2 | A = 11, C = 15 | 26 |
| 3 | A = 9, D = 18 | 27 |
| 4 | B = 13, C = 16 | 29 |
| 5 | B = 11, D = 17 | 28 |
| 6 | C = 14, D = 20 | 34 |
Carry out the intra-block analysis.
To analyse a balanced incomplete block design, comparing treatments after adjusting for the blocks each one appeared in.
Applying it:
The treatments differ significantly (\(F = 59.7\)). The adjusted means are 10.08, 12.08, 14.08, 19.08 for A, B, C, D, against unadjusted 10, 12, 15, 18.33: C has been pulled down and D pushed up, because D happened to appear in the heaviest blocks. The design recovers two thirds of the information of a complete block design, in exchange for blocks of two (worked also in Unit 3, Example 3.2).
Four treatments are compared in a \(3 \times 4\) Youden square: three rows (positions) and four columns (days), each column a block of three:
| Col 1 | Col 2 | Col 3 | Col 4 | Row total | |
|---|---|---|---|---|---|
| Row 1 | A 12 | B 15 | C 18 | D 21 | 66 |
| Row 2 | B 14 | C 17 | D 20 | A 13 | 64 |
| Row 3 | C 16 | D 22 | A 14 | B 16 | 68 |
| Column total | 42 | 54 | 52 | 50 | 198 |
Analyse the experiment.
To analyse a Youden square: a Latin square with rows missing, whose columns form a symmetric BIBD.
Every treatment appears once in each row, so rows are orthogonal to treatments and are taken out first; the columns are the blocks of a BIBD (Unit 4, section 3).
Applying it:
\(T_A = 39\), \(T_B = 45\), \(T_C = 51\), \(T_D = 63\); \(G = 198\), \(\text{CF} = 198^{2}/12 = 3267\), raw sum of squares 3380, \(SS_{\text{total}} = 113\). A appears in columns 1, 3, 4 (totals 42, 52, 50):
\[ Q_A = 39 - \frac{144}{3} = -9, \quad Q_B = 45 - \frac{146}{3} = -\frac{11}{3}, \quad Q_C = 51 - \frac{148}{3} = \frac53, \quad Q_D = 63 - \frac{156}{3} = 11; \quad \sum Q = 0. \checkmark \] \[ SS_{\text{tr(adj)}} = \frac{3}{8}\left(81 + \frac{121}{9} + \frac{25}{9} + 121\right) = 81.833333, \quad SS_{\text{rows}} = \frac{66^{2}+64^{2}+68^{2}}{4} - 3267 = 2, \quad SS_{\text{cols}} = \frac{42^{2}+54^{2}+52^{2}+50^{2}}{3} - 3267 = 27.666667, \] \[ SS_E = 113 - 2 - 27.666667 - 81.833333 = 1.5 \ \text{ on } 11 - 2 - 3 - 3 = 3 \text{ d.f.}, \quad MS_E = 0.5. \]| Source | SS | df | MS | \(F\) |
|---|---|---|---|---|
| Rows | 2.000000 | 2 | 1.000000 | |
| Columns (unadjusted) | 27.666667 | 3 | 9.222222 | |
| Treatments (adjusted) | 81.833333 | 3 | 27.277778 | 54.56 |
| Error | 1.500000 | 3 | 0.500000 | |
| Total | 113.000000 | 11 |
\(F_{0.05;\,3,\,3} = 9.277\).
The treatments differ significantly (\(F = 54.6\)): adjusted means 13.13, 15.13, 17.13, 20.63 for A, B, C, D, each difference with standard error 0.61. Both nuisance factors have been removed — rows exactly, as in a Latin square, and columns as the blocks of a BIBD — with only three positions for four treatments.
Six treatments, the pairs \(12, 13, 14, 23, 24, 34\) from \(\{1, 2, 3, 4\}\), are compared in the triangular PBIBD with four blocks of three (block \(x\) holds the three pairs not containing \(x\)):
| Block | Treatments and yields | Block total |
|---|---|---|
| 1 | 23 = 15, 24 = 17, 34 = 21 | 53 |
| 2 | 13 = 12, 14 = 14, 34 = 22 | 48 |
| 3 | 12 = 9, 14 = 13, 24 = 16 | 38 |
| 4 | 12 = 10, 13 = 12, 23 = 16 | 38 |
Carry out the intra-block analysis.
To analyse a partially balanced incomplete block design with two associate classes, and to see that its treatment comparisons have two standard errors.
Here \(v = 6\), \(b = 4\), \(k = 3\), \(r = 2\); first associates share a symbol (\(n_1 = 4\), \(\lambda_1 = 1\)), second associates do not (\(n_2 = 1\), \(\lambda_2 = 0\)) (Unit 4, Example 4.1).
Applying it:
| Treatment | 12 | 13 | 14 | 23 | 24 | 34 |
|---|---|---|---|---|---|---|
| \(T_i\) | 19 | 24 | 27 | 31 | 33 | 43 |
| blocks containing it | 3, 4 | 2, 4 | 2, 3 | 1, 4 | 1, 3 | 1, 2 |
| \(Q_i\) | −19/3 | −14/3 | −5/3 | 2/3 | 8/3 | 28/3 |
| \(\hat\tau_i = (5Q_i - Q_{i''})/8\) | −5.125 | −3.250 | −1.125 | 0.625 | 2.250 | 6.625 |
For 12, whose second associate is 34: \(\hat\tau = (5(-19/3) - 28/3)/8 = -41/8 = -5.125\). \(\sum Q = 0\) and \(\sum\hat\tau = 0\). \(\checkmark\)
\[ SS_{\text{tr(adj)}} = \sum\hat\tau_iQ_i = 117.75, \qquad G = 177, \ \text{CF} = 2610.75, \qquad SS_{\text{total}} = 2785 - 2610.75 = 174.25, \] \[ SS_{\text{blocks(unadj)}} = \frac{53^{2}+48^{2}+38^{2}+38^{2}}{3} - \text{CF} = 56.25, \qquad SS_E = 174.25 - 56.25 - 117.75 = 0.25 \ \text{ on } 12 - 4 - 6 + 1 = 3 \text{ d.f.} \]| Source | SS | df | MS | \(F\) |
|---|---|---|---|---|
| Blocks (unadjusted) | 56.25 | 3 | 18.75 | |
| Treatments (adjusted) | 117.75 | 5 | 23.55 | 282.6 |
| Error | 0.25 | 3 | 0.083333 | |
| Total | 174.25 | 11 |
Adjusted means \(14.75 + \hat\tau_i\): 9.625, 11.500, 13.625, 15.375, 17.000, 21.375. The standard error of a difference is 0.3227 for first associates and 0.3536 for second associates. \(F_{0.05;\,5,\,3} = 9.013\).
The treatments differ significantly (\(F = 282.6\)); 34 is best (adjusted mean 21.38). Two treatments that share a symbol — first associates, which met in a block once — are compared with standard error 0.32; two that never met, second associates, with 0.35. That is the practical meaning of "partially" balanced, and why a BIBD is preferred when one exists; here none does, since \(\lambda = r(k-1)/(v-1) = 4/5\) is not a whole number.