= Solution
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 \b[all missing entries]:
|| Orchard source
|| Degrees of freedom
|| Mean square
|| Variance ratio, one significant figure
| Spray
| 1
| 998
| 4
| Pruning
| 2
| 560
| 2
| Spray by pruning
| 2
| 202
| 0.8
| Residual
| 6
| 240
| Not applicable
The unrounded <F-test> statistics are $998/240=4.15833\ldots$, $560/240=2.33333\ldots$ and $202/240=0.841667\ldots$. 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.
Back to article page
