A Bio-fertiliser Trial with a Pre-planned Contrast and a Tempting Outlier
A plant breeder compares four formulations of a bio-fertiliser (A, B, C, D) on paddy yield. Twenty-four plots of equal size in one field are allotted six plots each, completely at random, and the yield in kg per plot is recorded. The data file is corrupted, but the following survive: treatment totals A = 132, B = 150, C = 144, D = 174; grand total = 600; and Σy² over all 24 plots = 15,600. The breeder remembers that his project plan had named one primary question: whether the new patented formulation D beats the average of the other three. After running a one-way analysis he is disappointed by the F ratio and proposes to drop one unusually low-yielding plot in treatment A as an 'outlier', expecting the result then to become significant. The field has a slight fertility gradient running north to south, and the plots, numbered in that order, were allotted to treatments by drawing lots.
1. Complete the analysis-of-variance table and carry out the F test at the 5% level, interpreting the result.
Read the worked answer
N = 24, k = 4, n = 6 per treatment.
Correction factor CF = G²/N = 600²/24 = 360,000/24 = 15,000. SST = Σy² − CF = 15,600 − 15,000 = 600, df = 23. SSTr = (132² + 150² + 144² + 174²)/6 − CF = (17,424 + 22,500 + 20,736 + 30,276)/6 − 15,000 = 90,936/6 − 15,000 = 15,156 − 15,000 = 156, df = 3, MSTr = 52.0. SSE = 600 − 156 = 444, df = 20, MSE = 22.2.
ANOVA table:
| Source | SS | df | MS | F |
|---|---|---|---|---|
| Treatments | 156 | 3 | 52.0 | 2.34 |
| Error | 444 | 20 | 22.2 | |
| Total | 600 | 23 |
F = 52.0/22.2 = 2.34 against F(0.05; 3, 20) = 3.10. Fail to reject: p ≈ 0.10. The data are consistent with equal population mean yields; there is no evidence of an overall formulation effect. Useful by-products: within-treatment SD = √22.2 = 4.71 kg, and the standard error of a treatment mean = √(22.2/6) = 1.92 kg, so a single treatment mean is estimated rather imprecisely - which is why the trial cannot resolve small differences.
2. Test the breeder's pre-planned question: does D exceed the average of A, B and C? Is it legitimate to do this after the overall F was not significant?
Read the worked answer
Use the planned contrast c = (−1, −1, −1, 3), with Σc_i T_i = −132 − 150 − 144 + 522 = 96 and Σ(c_i²/n_i) = (1 + 1 + 1 + 9)/6 = 12/6 = 2.
SS(contrast) = 96²/(6 × 12) = 9,216/72 = 128, df = 1, F = 128/22.2 = 5.77.
Equivalently, in t-form: D mean = 174/6 = 29.0; the other three = 426/18 = 23.67; difference = 5.33 kg; SE = √(MSE × (1/6 + 1/18)) = √(22.2 × 0.2222) = 2.221; t = 5.33/2.221 = 2.40 against t(0.025; 20) = 2.086, p ≈ 0.026 (and 2.40² = 5.77, as it must). 95% CI for the difference: 5.33 ± 2.086 × 2.221 = 5.33 ± 4.63, i.e. (0.70, 9.96) kg.
So D yields significantly more than the average of the other three, by about 0.7 to 10 kg per plot.
Legitimacy: yes, provided the question really was written down before the trial. A pre-planned contrast is a single comparison and does not need the omnibus F as a gatekeeper - the omnibus test spreads 3 df over questions the breeder never asked. Note that this contrast accounts for 128 of the 156 treatment SS, leaving only 28 for the remaining 2 df: the omnibus F is diluted by contrasts of no interest. If, however, the 'planned' question was constructed after inspecting the treatment means, the nominal 5% level is meaningless and the finding needs confirmation in a new trial.
3. Advise on the proposal to drop the low-yielding plot in treatment A, and on the fertility gradient in the field.
Read the worked answer
Dropping a plot because it reduces a treatment mean is not outlier screening; it is choosing results. A low yield may be genuine plot-to-plot variation, which is exactly what MSE measures. A value may be discarded only if an objective, recorded reason exists - a weighing error, a torn irrigation channel, pest damage confined to that plot - identified independently of the analysis, applied by the same rule to all four treatments, and reported together with the analysis of the full data. Otherwise the reported p-value is not a p-value at all. Note also that the design is no longer balanced once a plot goes: the neat formulas above fail, unequal replication changes every SS, and the breeder would need a proper missing-plot (non-orthogonal) analysis rather than a re-run of the same arithmetic.
The gradient is the more serious defect. Complete randomisation protects the test against bias only in expectation, and with only six plots per treatment a random draw can still put D disproportionately on the fertile south end; the resulting treatment effect is then confounded with soil fertility, and precision is lost because the gradient inflates the error mean square. The right design is a randomised complete block design - four blocks of four plots running along the gradient, with each formulation appearing once in each block - which removes the north–south variation from MSE. Failing that, a uniformity trial before the experiment, or a covariate in an analysis of covariance, would have helped. The breeder should also confirm that 'drawing lots' was actually honoured: if plots were assigned in numbered order, the gradient is confounded with the treatment order and the data cannot be rescued by analysis.