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The commands read the 52 cell values into a vector and build the matching two-factor layout. In expand.grid(country, goods) the first factor varies fastest, so the four countries are listed successively within each goods category. This matches row-by-row reading of the displayed table. Converting Goods to a factor is essential: its labels identify 13 categories and are not a quantitative trend. Country supplies the other categorical factor. The summaries suppress coefficient correlations, not coefficient standard errors.

The first fit is the additive two-factor normal linear modelThere are $1+12+3=16$ free coefficients and $52-16=36$ residual statistical degrees of freedom. This is two-way analysis of variance without an interaction term. For this complete balanced layout, ordinary least squares givesThese formulas follow from the row- and column-sum normal equations: fitted row and column totals equal their observed totals. The balanced-design coefficient standard errors are $s\sqrt{2/4}$ for a goods contrast, $s\sqrt{2/13}$ for a country contrast, and $s\sqrt{1/4+1/13-1/52}$ for the reference-cell intercept. These yield the common contrast standard errors and the different intercept standard error in each summary. In particular the intercept is the fitted Software–UK value, not the grand mean or the observed Software–UK entry.

Thus the first intercept predicts $149.8269$ million euros for Software in the UK. The Books coefficient adds $254.7$ million euros to the fitted value in any country, relative to Software; every other goods coefficient has this same arithmetic-difference interpretation. The country coefficients subtract $193.7$, $57.0308$ and $97.3769$ million euros from the UK prediction for France, Germany and the rest of Europe respectively, regardless of goods. The model therefore assumes common absolute country differences across goods categories. For Software–France its prediction iswhich is impossible for positive spending and indicates a limitation of this additive original-scale specification.

The residual standard error is $80.98$ million euros, obtained from $s^2=\mathrm{RSS}/36$; direct calculation gives $\mathrm{RSS}=236088.7435$. Each coefficient's printed statistic is $\hat{\theta }/\mathrm{se}(\hat{\theta })$, with a $t_{36}$ reference distribution under this normal linear model. The two-sided p-values compare each goods category with Software, or each other country with the UK, after adjustment for the other factor. For example France and the rest of Europe differ significantly from the UK at 5%, but Germany does not. The raw-scale significant goods differences are Books, Clothing, Electronic Goods, Groceries and Leisure Travel; these are individual reference-level comparisons, not an overall test of the goods factor.

The coefficient of determination is $R^2=1-\mathrm{RSS}/\mathrm{TSS}=0.7743$. The omnibus F-test compares all 15 nonintercept coefficients with zero:with reference $F_{15,36}$ and $p=1.2544\times 10^{-7}$. This says at least one factor contrast is nonzero under the assumed error model. There is one observation per cell, so an unrestricted goods-country interaction would saturate the model, leaving zero residual degrees of freedom. Without replication or a specified interaction structure, the residual variation cannot be separated into interaction and independent error.

The second fit applies a logarithmic transformation and fits the same factor design to $Z_{gc}=\log Y_{gc}$. The Gaussian log-response model isIts intercept exponentiates to a fitted Software–UK conditional median of $e^{4.5267}\approx 92.45$ million euros. Goods coefficients now give multiplicative ratios to Software, and country coefficients give multiplicative ratios to the UK. The estimated country ratios areFor example the Books ratio is $e^{1.6415}\approx 5.16$ and the Sports Equipment ratio is $e^{-1.8364}\approx 0.1594$, consistently across countries. The model assumes common relative, rather than common absolute, country effects. It also guarantees positive retransformed fitted values. Under normal log errors, the original-scale conditional mean is $e^{\eta +{\sigma }^2/2}$, whereas $e^{\eta }$ is the conditional median and geometric mean. Correcting the retransformation bias here multiplies the naive predictions by approximately $e^{0.127406/2}=1.0658$.

The log-scale residual standard error is $0.3569$, with the same 36 residual degrees of freedom, corresponding to a typical multiplicative error multiplier $e^{0.3569}\approx 1.429$. France and the rest remain significant reference-country contrasts; Germany remains nonsignificant at 5%. The log-scale significant reference-goods contrasts are Books, Music, Videos/DVDs, Clothing, Sports Equipment, Electronic Goods, Groceries and Leisure Travel. Event Tickets, Toys, Video Games and Housewares do not individually differ significantly from Software at that level. These p-values remain subject to the usual multiple testing qualification.

The log-scale $R^2=0.9381$ is computed using variation in the logarithms, so it cannot by itself establish better fit on the original response scale. The change is plausible because spending varies widely in magnitude and relative discrepancies are more meaningful than constant absolute discrepancies, but it should be assessed with regression diagnostics, especially a residual-versus-fitted plot and a normal Q-Q plot. The minimum residual near $-1.01$ merits inspection; the command summaries alone do not verify independent normal errors. Moreover these responses are forecasts, so the inferential p-values have a literal sampling interpretation only if an appropriate stochastic model for their uncertainty is justified.

The log-scale analysis of variance adds Goods and then Country. In this balanced design their effect spaces are orthogonal after removing the intercept, so the sequential sums of squares equal the corresponding adjusted factor sums and are unchanged by reversing the two factors:The residual sum of squares is $4.58662$, giving $M{S}_E=4.58662/36=0.127406$. ThusTheir tiny upper-tail p-values show strong overall goods and country effects under the log-error model; a globally significant country factor is compatible with the nonsignificant individual Germany contrast. The omnibus fit statistic is $36.3433$ on $(15,36)$ degrees of freedom, whose p-value is approximately $3.14\times 10^{-17}$. The printed zero reflects numerical formatting, not an exactly zero probability.

If two cell entries were missing, retain their correct factor labels and mark those responses unavailable, rather than shifting the remaining scan values into different cells. Fit the additive models to the 50 observed cells, assuming the missingness mechanism permits those observations to be analyzed without selection bias. All factor levels remain represented, and the row-column incidence graph remains connected: any two goods rows still share an observed country, since only two edges of the complete layout have been removed, and every country still has many observed goods. By identifiability of an incomplete two-factor additive design, the design therefore still has rank $13+4-1=16$. HenceThe coefficients and their covariance must be recomputed by the observed-cell ordinary least squares equations. The balanced row-mean-plus-column-mean formulas and common within-factor standard errors no longer apply. Goods and Country effects are generally no longer orthogonal, so sequential ANOVA sums can depend on order; factor tests adjusted for the other factor should use nested model comparisons. Missing cells do not restore replication or make an unrestricted interaction test possible: that larger model still fits every observed cell exactly. If missingness depends on the unavailable spending values, additional modelling of that mechanism would be needed.

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.

Index chocolate by $a\in \{A,B,C,D\}$, day by $d$ in the three observed categories, and replicate by $r\in \{1,2\}$. The additive two-factor normal linear model isHere $Y_{adr}$ is the board count, $\mu$ is the mean for the reference chocolate and reference day, and ${\alpha }_a,{\gamma }_d$ are chocolate and day fixed effects. With corner-point constraints, set ${\alpha }_A=0$ and ${\gamma }_{d_0}=0$, where $d_0$ is the first level in the day factor. The printed coefficient-free output does not determine that factor ordering; the model is unchanged by a different reference category. There is no chocolate–day interaction term in this fit. The six free mean statistical parameters consist of one baseline, three chocolate contrasts and two day contrasts; the common error variance supplies a further statistical parameter.

Let $W_{aj\ell }$ be recall for age level $a$, processing level $j$, and replicate $\ell =1,\dots ,10$. Write $u_a=1$ for Younger and $0$ for Older. With Older and A as the reference levels in a regression factor, the additive two-factor normal linear model under treatment coding iswhere the errors are independent $N(0,{\sigma }^2)$, with common positive variance. The design matrix is fixed; subjects supply independent observations; the specified conditional mean is correct; and the absence of an interaction term means the age difference is constant across processing levels. These are model assumptions, not consequences of random assignment. The treatment allocation supports independence between processing assignment and background characteristics, but age itself was not randomized.

An older subject in A has $u_a=0$ and ${\gamma }_A=0$, so the estimated mean is $11.35$ words. Its standard error is the intercept's $0.7632$. There are $100-6=94$ residual degrees of freedom, giving the 95% confidence interval for this mean:This is a mean-response confidence interval, not a prediction interval for a new individual's recall; the latter also includes the new observation's error variance.

The second two-factor normal linear model adds age-by-processing interaction terms:Its ten free mean parameters are equivalent to one mean for each of the ten cells. To compare the fits, the null hypothesis is ${\delta }_B={\delta }_C={\delta }_D={\delta }_E=0$. The full residual sum of squares is $722.30$ on $90$ degrees of freedom. Adding interaction reduces the additive model's residual sum of squares by $190.30$, so the reduced value is $912.60$. The nested-model F-test statistic isIts p-value is $0.0002793$. Reject additivity and retain the interaction model. The significant interaction means that a single age effect averaged over all processing methods does not adequately summarize the data.

The regression intercept $11.0$ is the older-A mean. The age coefficient $3.8$ is the younger-minus-older contrast specifically in A. The processing coefficients $(-4.0,2.4,1.0,-4.1)$ compare B, C, D and E with A specifically among older subjects. The interaction coefficients $(-4.3,0.4,3.5,-3.1)$ are differences between the age contrast in each of those methods and its value in A. Adding the relevant coefficients givesThus A, C and especially D favour younger subjects in their estimated means, whereas B and E have little estimated age difference. In the younger group, D has the largest estimated recall, followed by C and A; B and E are much lower. Among older subjects C is largest, D and A are intermediate, and B and E are lowest. Comparing these orders is a description of estimates, not a claim that every pairwise difference is significant.

Each coefficient's printed standard error estimates its uncertainty; its $t$ statistic divides the estimate by that error, and its two-sided p-value uses $t_{90}$ under the relevant zero-contrast null hypothesis. For example, the older B-versus-A and E-versus-A contrasts have $p=0.00217$ and $0.00170$. The C-versus-A and D-versus-A older contrasts have $p=0.06139$ and $0.43201$. The B interaction has $p=0.01846$, while the D interaction is borderline at $0.05387$ and the other two individual interaction tests are not significant at $5\%$. These individual tests do not override the joint interaction F-test, and multiple comparisons require care. A simple younger-versus-older contrast outside A combines coefficients and must use their estimated covariance, rather than adding their marginal errors.

The residual standard error $2.833$ estimates $\sigma$. The coefficient of determination $0.7293$ is the fraction of centered sample recall variation explained by the ten-cell model. The overall $F=26.93$ with null law $F_{9,90}$ tests all nine non-intercept coefficients jointly against zero, giving extremely strong evidence that the cell means are not all equal. The sequential analysis of variance separates age, process, and their interaction; the balanced design makes the main-effect sums orthogonal, whereas the coefficient table uses the specified reference-level contrasts.

To assess pooling into three processing types, retain age-by-type interaction because the previous test rejects additivity. The reduced model $W\sim \mathrm{Age}\ast \mathrm{Type}$ has six cell means. Its null hypothesis says A and C have equal means within each age, and B and E have equal means within each age: four restrictions. Compare it with the ten-cell model usingIn fact the supplied fitted cell means and equal cell sizes determine the increase without raw observations. Pooling two ten-person cells with means $m_1,m_2$ adds $5(m_1-m_2{)}^2$ to the residual sum of squares. Here the four differences are $2.4,2.8,-0.1,1.1$, soTherefore $F_{\mathrm{pool}}=2.3083$ and $p\simeq 0.0640$. At $5\%$, the data do not reject the three-type reduction, though the result is borderline and is not proof that the pooled means are identical. This factor-level pooling test keeps the previously supported interaction structure; reducing the categories and removing all interaction simultaneously would test a different set of restrictions.