Question
A gardener tries a new fertiliser on three of six labelled plots, chosen uniformly from the \(\binom63\) possible groups of three. All plots are harvested once. The six yields, in kilograms, are 2, 4, 7, 9, 12 and 15; the fertilised plots happened to be those yielding 4, 12 and 15.
The gardener proposes a one-sided randomisation test: use treated mean yield minus untreated mean yield, and test the sharp null that the fertiliser changes no plot's yield at all. What is the exact p-value for a difference at least as large as the one observed? Is the result significant at a 5% level? Do not assume the yields are normally distributed.
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Hint 1
Under the sharp null, the six numerical yields remain fixed even if the assignment labels are switched.
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Solution
Under the sharp null, each plot would have produced its observed yield under either treatment. Randomness comes solely from which three plots were labelled treated. Each of the \(\binom63=20\) assignments therefore has probability \(1/20\), without any population normality assumption.
The treated sum is \(4+12+15=31\); the untreated sum is \(2+7+9=18\). Thus the observed difference in means is
\[T_{\mathrm{obs}}=\frac{31}{3}-\frac{18}{3}=\frac{13}{3}\text{ kg}.\]
The sum of all yields is 49. If an alternative treated group has sum \(s\), the corresponding test statistic is \(T(s)=s/3-(49-s)/3=(2s-49)/3\). Consequently \(T(s)\ge T_{\mathrm{obs}}\) exactly when \(s\ge31\).
Enumerating triples from \(\{2,4,7,9,12,15\}\), the qualifying treated sets are \(\{4,12,15\}\) (sum 31), \(\{7,9,15\}\) (31), \(\{7,12,15\}\) (34) and \(\{9,12,15\}\) (36). No other triple reaches 31. Hence the exact one-sided randomisation p-value is
\[p=\frac4{20}=0.20.\]
Since \(0.20>0.05\), this test does not reject the sharp null at the 5% level. This is a probability over the assignment mechanism assuming the sharp null, not a 20% probability that the null itself is true.
Assumptions
Exactly three of six distinctly labelled plots were assigned uniformly at random to treatment.
The sharp null specifies each plot's potential treated and untreated outcomes are identical; there is no interference between plots.
The prespecified statistic is the treated-minus-control mean, and the p-value is an inclusive upper-tail probability.
Final Answer
Observed mean difference = \(13/3\) kg; exact one-sided p-value = \(4/20=0.20\). Not significant at 5%.