5.3 Checking the model

The residuals \(e_{ij} = y_{ij} - \bar y_{i\cdot } - \bar y_{\cdot j} + \bar y_{\cdot \cdot }\) are plotted as in Section 3: against the fitted values, against treatments and blocks, and on a normal probability plot. In an RCBD the assumption that most needs checking is additivity. A treatment that does well in good blocks and badly in poor ones produces residuals that are positive in the corners of the table and negative in the middle, or the reverse.

Tukey’s one-degree-of-freedom test for non-additivity checks for the commonest form of interaction, one proportional to the product of the main effects, \[ y_{ij} = \mu + \tau _i + \beta _j + \gamma \tau _i\beta _j + \varepsilon _{ij} . \] Writing \(a_j = \bar y_{\cdot j} - \bar y_{\cdot \cdot }\) and \(c_i = \bar y_{i\cdot } - \bar y_{\cdot \cdot }\), the sum of squares for non-additivity is \[ SS_N = \frac {\bigl (\sum _i\sum _jc_ia_jy_{ij}\bigr )^2}{\sum _ic_i^2\,\sum _ja_j^2} , \] on one degree of freedom, and it is tested against the remainder of the error: \[ F_0 = \frac {SS_N}{\bigl (SSE - SS_N\bigr )/\bigl ((t - 1)(b - 1) - 1\bigr )} , \] which has the \(F_{1,\,(t-1)(b-1)-1}\) distribution under additivity. \(SS_N\) is the regression sum of squares of the residuals on the products \(c_ia_j\); if additivity holds, \(\gamma = 0\) and the products explain nothing beyond chance. When the test is significant, a transformation of the response, often the logarithm (which turns a multiplicative model into an additive one), will frequently remove the interaction.

Example 5.3.1. For the fertiliser data, the residuals are

field \(A\) \(B\) \(C\) \(D\)
\(1\) \(0.60\) \(0.60\) \(0.00\) \(-1.20\)
\(2\) \(-0.65\) \(-0.65\) \(-0.25\) \(1.55\)
\(3\) \(-0.40\) \(-0.40\) \(1.00\) \(-0.20\)
\(4\) \(0.35\) \(-0.65\) \(-0.25\) \(0.55\)
\(5\) \(0.10\) \(1.10\) \(-0.50\) \(-0.70\)

with no pattern and none larger than \(1.55/\sqrt {0.808} = 1.7\) standard deviations. For Tukey’s test, \(\sum \sum c_ia_jy_{ij} = 17.835\), \(\sum c_i^2 = 16.51\) and \(\sum a_j^2 = 20.675\), so \[ SS_N = \frac {17.835^2}{16.51 \times 20.675} = 0.932 , \qquad F_0 = \frac {0.932}{(9.70 - 0.932)/11} = 1.17 , \] with \(p = 0.30\). There is no evidence of non-additivity.

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