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A mean square in ANOVA divides a sum of squares in ANOVA by its positive number $d$ of statistical degrees of freedom. If the corresponding residual subspace has scalar error covariance $\lambda I$, its expectation is $\lambda$. Its scale depends on whether raw observations or group means were projected.

Both factors are assigned at orchard level. Therefore the experimental units are the twelve orchards; the trees are observational units within them. The six combinations form a balanced factorial design, replicated twice. The orchard ANOVA stratum has $12-1=11$ statistical degrees of freedom. Spray uses $2-1=1$, pruning uses $3-1=2$, and their interaction term uses $(2-1)(3-1)=2$, leaving six for error.

Dividing each treatment sum of squares in ANOVA by its statistical degrees of freedom and using $1440/6=240$ as the denominator gives all missing entries:

The unrounded F-test statistics are $998/240=4.15833\dots$, $560/240=2.33333\dots$ and $202/240=0.841667\dots$. The within-orchard tree mean square in ANOVA, 180, is not the treatment error denominator: using it would confuse subsampling with independent replication. The tree ANOVA stratum has $12(30-1)=348$ statistical degrees of freedom; $1+11+348=360$ is the uncorrected total, and the corrected total is 359.

The orchard residual mean square in ANOVA is on the original tree-response scale: it equals 30 times the corresponding residual mean square in ANOVA for orchard means. Thus the estimated variance of one orchard's sample mean is $240/30=8$.

Each pruning marginal sample mean averages four independent orchard means, giving variance $8/4=2$. Two different pruning marginals use disjoint orchards, so the variance of their estimated treatment contrast isEach spray marginal sample mean uses six orchards. Similarly,The corresponding standard errors are $2$ and $\sqrt{8/3}$ in the units of weight per tree. These compare per-tree marginal sample means, averaging equally over the other factor, even when an interaction term is fitted. Comparing orchard totals instead would multiply these variances by $30^2$.