A contrast among the corn varieties
For the corn data, with variety means 48.93, 55.07, 59.80, 48.30 and 51.93, three plots each, and \mathrm{MS}_{\mathrm{error}}=10.57 on 10 degrees of freedom, test whether variety 2 differs from the average of varieties 1 and 4, using a contrast. Use F_{1,10;0.95}=4.96.
Worked answer (click to show)
The contrast is C=\mu_2-(\mu_1+\mu_4)/2, with coefficients (-\tfrac12,1,0,-\tfrac12,0), which sum to zero. Its estimate is
\widehat C=55.07-\frac{48.93+48.30}{2}=6.45, \qquad \sum_j\frac{c_j^2}{n_j}=\frac{\tfrac14+1+\tfrac14}{3}=0.5.
So \mathrm{SS}_C=6.45^2/0.5=83.2 and F=83.2/10.57=7.87 on (1,10) degrees of freedom. Since 7.87>4.96, reject H_0:C=0 at level 0.05; the p-value is 0.019. This contrast is valid only if the comparison was planned before the yields were seen.