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The additive model gives useful evidence for both predictors. To test for no adjusted sex effect, use $H_0:{\beta }_M=0$ against $H_1:{\beta }_M\ne 0$. The statistic $T={\hat{\beta }}_M/\mathrm{se}({\hat{\beta }}_M)$ has a Student t-distribution with 97 statistical degrees of freedom under the normal linear model and the null hypothesis. Here $T=3.016$ and $p=0.00327$. Similarly, $H_0:{\beta }_h=0$ against a nonzero common slope gives $T=4.397\sim t_{97}$ under the null hypothesis, with $p=2.81\times 10^{-5}$. Thus, at a fixed height males have a higher fitted mean weight, and within sex greater height is associated with greater weight.

The interaction model adds $\delta M(x-165)$. Its female height slope is $0.6696$ and its male height slope is $0.6696-0.0977=0.5719$. Test $H_0:\delta =0$ against $H_1:\delta \ne 0$ using $T=-0.344\sim t_{96}$ under the null hypothesis, or equivalently $T^2\sim F_{1,96}$. The $p$-value is 0.73139. The additive model is a reasonable parsimonious choice; the data give no evidence for different height slopes between sexes. This does not establish exact equality of population slopes. The additive model also has the higher adjusted coefficient of determination, 0.405 versus 0.3996, and a slightly smaller residual standard error.

Ignoring sex gives a height slope of 0.8769, appreciably above the adjusted slope 0.6211. This is omitted-variable bias in the descriptive regression: a taller group with a larger intercept makes the pooled height association steeper. Algebraically, with $z=x-165$,The printed estimates imply a positive height-sex covariance. Pooled and adjusted slopes therefore describe different comparisons; the difference is not evidence that adding sex changed a physical effect of height. For predictions, retain both sex and height, check regression diagnostics, and assess performance on new students. The residual standard error near 5.7 remains substantial and the coefficient of determination near 0.42 does not promise precise individual prediction.

A regression factor is a categorical predictor represented by level indicators and contrasts. Treating year as a regression factor permits arbitrary differences between the three yearly means, reflecting conditions such as weather, rather than imposing a linear time trend.

Let $Y_{ijk}$ be the yield for seed variety $i\in \{1,2\}$, fertiliser level $j\in \{L,H\}$, and year $k\in \{2004,2005,2006\}$. A treatment-coded two-factor normal linear model with additive year effects iswith independent $N(0,{\sigma }^2)$ errors. The unrestricted coefficients are ${\beta }_0,a,b,c_{2005},c_{2006},d$: six parameters for twelve observations. Equivalently use baseline-zero variety, fertiliser, and interaction arrays. There are no year-treatment interactions in this fit.

Dividing each sum of squares by its statistical degrees of freedom gives the completed analysis of variance:In particular, the missing interaction statistic is $16.0083/(2.8414/6)\simeq 33.803$. Its null hypothesis is $H_0:d=0$, against $H_1:d\ne 0$: the change in mean yield from high fertiliser is the same for both varieties. Under the null hypothesis and the independent homoscedastic normal linear model, this nested-model F-test has distribution $F_{1,6}$. The printed $p$-value 0.001137 strongly rejects no interaction.

Do not drop fertiliser just because its main-effect ANOVA row has $p=0.7011$. In this balanced factorial design that row measures a fertiliser effect averaged over varieties; large opposite effects can cancel. The strongly supported interaction requires retaining its associated main effects by regression model hierarchy. Seed and year also have evidence of effects in their ANOVA rows, so this table alone gives no compelling simplification of the fitted terms.

The interaction contrast is $-4.6200$, with standard error 0.7946. The fitted high-minus-low fertiliser effect is $+2.4700$ for variety 1 but $2.4700-4.6200=-2.1500$ for variety 2. The fitted variety-2-minus-variety-1 contrast is $+3.6500$ under low fertiliser and $3.6500-4.6200=-0.9700$ under high fertiliser. Thus variety 2 performs best under low fertiliser, while high fertiliser benefits variety 1 and reduces variety 2's fitted yield. The marginal fertiliser effect is only $(2.4700-2.1500)/2=0.1600$.

The four fitted means in the baseline year are $5.5050,7.9750,9.1550,7.0050$ for $(1,L),(1,H),(2,L),(2,H)$ respectively. Add $1.5775$ in 2005 and $0.0875$ in 2006 to every fitted mean. The 2005 coefficient has $p=0.017646$ under a $t_6$ test of a zero contrast with 2004; the analogous 2006 contrast has $p=0.863216$. The latter is lack of evidence for a difference, not proof that the yearly means coincide. Standard errors for the combined contrasts require the coefficient covariance matrix. The experiment is small: residual standard error is 0.6882 on only six statistical degrees of freedom. If the same fields were reused, possible within-field dependence and field allocation would need checking before treating the fitted error independence as established.

Compare the species-only model against the additive day-and-species model. The null hypothesis is $H_0:d_2=d_3=0$, with the alternative that at least one is nonzero. The likelihood-ratio statistic is the reduction in residual deviance,Under the null hypothesis, regular binomial likelihood asymptotics give $G^2\stackrel{\mathrm{approx}}{\sim }{\chi }_2^2$, because the larger model has two additional parameters. For this chi-squared distribution the upper-tail probability is $\exp (-0.341/2)\simeq 0.843$. There is no evidence of a day effect after controlling for species, and the species-only model is a reasonable simpler choice. The sequential day-only statistic 0.273 from the preceding table answers a different question and must not replace this comparison.

The saturated model estimates each Poisson mean separately as $Y_i$, permitting the boundary value zero when $Y_i=0$. Subtracting fitted from saturated log-likelihood gives the Poisson deviancewith $0\log (0/{\hat{\mu }}_i)=0$. A zero observation contributes $2{\hat{\mu }}_i$. If the model includes an intercept, the score equation gives ${\sum }_i{Y}_i={\sum }_i{\hat{\mu }}_i$, so the summed linear terms cancel; the unsimplified expression applies generally.

For goodness of fit, the null hypothesis is that the specified independent Poisson mean model is correct; the alternative is an unrestricted collection of means. When a quadratic likelihood approximation is valid, for example with fixed $n$ and large expected counts, $D$ has approximately a chi-squared distribution with $n-p$ statistical degrees of freedom. Reject for a large upper-tail deviance. Merely increasing the number of sparse observations does not make this saturated-model reference reliable: the alternative dimension also increases. With small fitted means, replication-based checks or a parametric bootstrap are more defensible. This qualification concerns the absolute goodness-of-fit deviance; a fixed-dimensional nested likelihood-ratio comparison can have a better chi-squared approximation.