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Analysis of variance decomposes variation into contributions associated with model terms and residual error. For nested normal linear models, differences of residual sums of squares give sequential or partial F-tests.

For nested normal linear models, the decrease in residual sum of squares obtained by adding coefficients is their extra sum of squares. Dividing by the number of added coefficients and by the full-model residual mean square gives the partial F-test. A sequential analysis of variance attributes this decrease according to the specified order of entry; it need not equal the partial comparison when other terms are retained.

Test $H_0:{\gamma }_d=0$ for both nonreference days, against at least one nonzero day fixed effect, conditional on chocolate. There are two added regression coefficients and $24-6=18$ residual statistical degrees of freedom. The nested-model F-test statistic isUnder $H_0$ and the normal linear model assumptions, $F\sim F_{2,18}$. Its 5% upper critical value is $3.555$, so do not reject the absence of day effects; the p-value is approximately $0.456$. Thus the missing row has two statistical degrees of freedom, mean square $4.875$, and the F-test value above. These observations do not provide evidence that including day improves the chocolate-adjusted mean model. They do not prove that every possible day effect or chocolate–day interaction term is absent.

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.

Under the usual normal linear model for independent orchard means with common error variance, the treatment ratios have null F-distributions with denominator six statistical degrees of freedom. Their upper-tail P-values are approximatelyUse the unrounded ratios when evaluating these P-values. Thus none of spray, pruning or their interaction term is significant at a 5% significance level. Spray gives modest evidence if a 10% significance level was chosen in advance, but this is not strong evidence.

Failure to reject does not establish absence of treatment effects. Only two orchards per combination and six residual statistical degrees of freedom leave considerable uncertainty and potentially low statistical power. The table alone cannot give the direction or magnitude of individual effects: that requires treatment means. These model-based F-tests also rely on appropriate orchard allocation and comparable residual variation; they should not be interpreted as unconditional conclusions from the table alone.