Solution. Here the rows are the \(r = 4\) fertilisers and the columns the \(c = 3\) varieties, so \(N = rc = 12\) and \(g = 154\). (Note that the first row totals \(17 + 25 + 25 = 67\).)

1.

Correction factor: \(cf = \frac {g^2}{rc} = \frac {154^2}{12} = 1976.33\).

2.

Total: \(SS_T = \sum \sum x_{ij}^2 - cf = 2586 - 1976.33 = 609.67\).

3.

Columns (varieties): \(SS_C = \frac {1}{r}\sum C_j^2 - cf = \frac {48^2 + 64^2 + 42^2}{4} - 1976.33 = 64.67\).

4.

Rows (fertilisers): \(SS_R = \frac {1}{c}\sum R_i^2 - cf = \frac {67^2 + 18^2 + 42^2 + 27^2}{3} - 1976.33 = 459.00\).

5.

Error: \(SSE = 609.67 - 459.00 - 64.67 = 86.00\).

At \(\alpha = 1\%\) the critical values are \(F_{0.01,\,3,\,6} = 9.78\) for rows and \(F_{0.01,\,2,\,6} = 10.92\) for columns.

(a)

Varieties: \(F_C = 2.26<10.92\), so there is no evidence of a difference between the varieties.

(b)

Fertilisers: \(F_R = 10.67>9.78\), so we reject \(H_0\): the fertilisers differ, \(f_1\) giving by far the highest yields and \(f_2\) the lowest.

Solution. From tables \(t_{0.025,\,6}=2.447\), so \begin {align*} LSD &= 2.447\times \sqrt {1.57\left (\frac {1}{4}+\frac {1}{4}\right )}\\[1ex] &= 2.447\times \sqrt {0.785}\\[1ex] &= 2.447\times 0.886 = 2.168 \end {align*}

Comparing each difference against \(2.168\):

\(v_2-v_1=2.25>2.168\), so \(v_1\) and \(v_2\) differ significantly.

\(v_3-v_1=3.00>2.168\), so \(v_1\) and \(v_3\) differ significantly.

\(v_4-v_1=4.00>2.168\), so \(v_1\) and \(v_4\) differ significantly.

All three exceed the \(LSD\), so \(v_1\) differs from each of the others. Note that \(v_2-v_1\) only just clears the bar, at \(2.25\) against \(2.168\); a conclusion resting on that small a margin should be reported as such rather than stated flatly.