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In a complete two-factor design with equal cell replication, centred factor indicator columns for different factors have zero cross-products. Consequently the two main effects in an additive normal linear model are orthogonal, and their extra sums of squares do not depend on the order in which these main effects are entered. An omitted interaction term can still require a separate adequacy check.

A logarithmic count-component contrast $b$ multiplies the susceptible mean by $e^b$, while a zero-logit contrast $g$ multiplies the structural-zero odds by $e^g$. If the baseline zero predictor is $\eta$, the marginal mean ratio isThus the count ratio alone is not a population-average effect when a covariate changes both components. With interaction terms, the relevant count contrast must first include the interactions for the stated reference group.

A factorial design includes every combination of specified factor levels. Balanced replication permits separate estimation of marginal effects and interaction terms. The factor allocation, not the number of subsamples, determines the appropriate error term for each effect.

To pool factor levels in a normal linear model, impose equality of their fitted means while retaining any needed interaction terms. Compare the resulting nested model with the full model using a nested-model F-test. For two cells each of size $m$, pooling their fitted means adds $(m/2)({\overline{Y}}_1-{\overline{Y}}_2{)}^2$ to the residual sum of squares. If pooling two pairs separately within two other-factor levels, there are four restrictions.

An interaction contrast compares a treatment difference across levels of another factor. A nonzero difference of differences shows that an additive representation of those cell means is inadequate. Its standard error must use the error stratum in which that interaction term is randomized.

A statistical interaction means that the effect of one predictor depends on another on the chosen response scale. In a linear regression, an interaction term ${\beta }_{3}XZ$ changes the slope in $X$ to ${\beta }_1+{\beta }_{3}Z$. Whether interaction is present depends on the comparison scale; additive effects and multiplicative effects use different null contrasts. Comparing subgroup effects requires an interaction contrast, rather than contrasting their separate significance labels.

At second order in the interaction term, attach the two external fermion fields to different vertices and contract the remaining two fermion fields with each other and the two vector fields with each other. The factor $1/2!$ in the expansion is cancelled by the two ways to assign the external ends. There is no closed fermion loop, and hence no extra closed-loop fermionic sign.

After removing the two external Dirac propagators, the internal momentum assignment $l$ for the photon and $p+l$ for the fermion gives the one-loop insertionThus the insertion is proportional to exactly the propagator product specified in the source. Define $\mathcal{I}=-i\Sigma$. With Fourier convention $\psi (x)\sim e^{ip\cdot x}$, the reduced free inverse kernel is $K_0(p)=-/p-m$, since the printed Dirac propagator equals $i{K}_0^{-1}$. Repeated insertions giveThe factors of $i$ and the sign of the named self-energy are convention dependent; the cancelling counterterm is fixed once this insertion convention is specified. The internal photon joins two vertices on one directed fermion line, as in the first panel of the figure in 1b.

Choose the covariant derivative $D_{\mu }={\partial }_{\mu }+i{\mu }^{ϵ/2}e{A}_{\mu }$, so the interaction term is $-{\mu }^{ϵ/2}e\bar{\psi }{\gamma }^{\mu }{A}_{\mu }\psi$. The gauge transformation is $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha$ and $\psi \mapsto e^{-i{\mu }^{ϵ/2}e\alpha }\psi$. The derivative of this local phase cancels the added vector-potential term in $D_{\mu }\psi$, making $i\bar{\psi }{\gamma }^{\mu }{D}_{\mu }\psi$ gauge-invariant.

For renormalized fields and parameters, a complete local QED counterterm parametrization isWith ${\psi }_0=\sqrt{Z_2}\psi$, $A_0=\sqrt{Z_3}A$, the definitions are $\delta Z_i=Z_i-1$, $\delta m_L=Z_2{m}_0-m$, and $e_0={\mu }^{ϵ/2}e{Z}_1/(Z_2\sqrt{Z_3})$. In particular $\delta m_L$ is the coefficient in the Lagrangian density, not just $m_0-m$.

The Ward identity requires $Z_1=Z_2$ in a gauge-preserving renormalization scheme. The kinetic and vertex counterterms then combine into $i\delta Z_2\bar{\psi }/D\psi$, so gauge invariance is explicit and $e_0={\mu }^{ϵ/2}e/\sqrt{Z_3}$. A photon mass term is not an allowed counterterm in unbroken QED. For perturbative inversion one separately adds gauge fixing, ${\mathcal{L}}_{\mathrm{gf}}=-({\partial }_{\mu }{A}^{\mu }{)}^2/(2\xi )$, taking $\xi =1$ in Feynman gauge. That quantization term is not itself part of the gauge-invariant density. The Abelian Faddeev-Popov determinant is field independent for this linear gauge and introduces no interacting ghost contribution.

Gauge invariance and power counting permit the same local operators as in the original renormalizable action. With renormalized fields, a convenient counterterm parametrization isHere $D_{\mu }={\partial }_{\mu }-i{\mu }^{ϵ/2}e{A}_{\mu }$. Expanding the first term supplies kinetic, scalar-scalar-photon, and seagull counterterms with the relative coefficients required by the Ward identity. Equivalently introduce separate vertex constants first and then impose that identity. For ${\varphi }_0=Z_{\varphi }^{1/2}\varphi$, $A_0=Z_A^{1/2}A$, and $e_0={\mu }^{ϵ/2}{Z}_{e}e$, it gives $Z_e{Z}_A^{1/2}=1$. Thus charge renormalization can be represented through the photon field factor, rather than adding an independent gauge-noninvariant interaction.

The mass coefficient defined here is $\delta M^2=Z_{\varphi }{m}_0^2-m^2$; it must be distinguished from $m_0^2-m^2$. Gauge fixing may require the corresponding gauge-parameter redefinition, and vacuum diagrams can require an additive vacuum-energy counterterm. No photon mass counterterm is allowed by the unbroken gauge symmetry. The quartic scalar counterterm remains necessary for the full theory even at a scale where the renormalized $\lambda$ is zero, because gauge interactions generate scalar four-point divergences.

The one-particle-irreducible scalar two-point function has three one-loop topologies: a scalar-photon exchange loop built from two cubic vertices; a photon tadpole at one seagull vertex; and a scalar tadpole at one quartic vertex. The last is absent when $\lambda =0$. Reducible chains of self-energy insertions belong to the Dyson expansion and are not additional one-loop self-energy graphs. The scalar-QED photon-tadpole symmetry factor is $1/2$ because its two identical photon legs are joined; the oriented complex-scalar tadpole has no such identical-real-line factor.

Route the photon momentum as $l$ and the internal scalar momentum as $l+k$. The transverse projector obeys $P_{\alpha \beta }{l}^{\beta }=0$, so the two cubic vertices giveMultiplying the two vertices and propagators therefore givesFor the seagull, $g^{\alpha \beta }{P}_{\alpha \beta }=d-1$. Including the photon-loop symmetry factor gives insteadThe printed seagull coefficient is twice this result. The vertex's factor two differentiates the two photon fields; closing them into one line requires the compensating symmetry factor. Direct expansion of the action gives the same check: the interaction density $-e^2{A}^2{\varphi }^{†}\varphi$ has one photon contraction, yielding coefficient $-e^2$, not $-2e^2$. The identical-photon origin of the vertex factor is also explicit in the standard scalar-electrodynamics rules.^[ref] This discrepancy does not affect the massless-photon limit in part d or the ultraviolet pole in part f.

At first order the Dyson expansion inserts $i{S}_{\mathrm{int}}=\lambda \int d^{4}x{\bar{\psi }}_{\alpha }(x)({\gamma }_5{)}_{\alpha \beta }{\psi }_{\beta }(x)\varphi (x)$. Contract the scalar field at the vertex with the external scalar. The four remaining fermion fields have two admissible pairings: both external fermions attach to the vertex, or the two vertex fermions contract with each other while the external pair contracts freely. Wick's theorem givesThe minus sign in the second term is the fermion sign of the closed contraction: $⟨T{\bar{\psi }}_{\alpha }(x){\psi }_{\beta }(x)⟩=-i{S}_F(0{)}_{\beta \alpha }$. The first term is connected; the second is a scalar-attached fermion tadpole multiplied by a disconnected free external fermion propagator. The full time-ordered expectation includes both topologies, whereas the connected generating functional retains only the first. There is no order-${\lambda }^2$ contribution in this pure Yukawa expansion because the total number of scalar fields would be odd.

For a parity-preserving free propagator, the tadpole coefficient is $\mathrm{tr}[{\gamma }_5{S}_F(0)]$, whose integrand contains only $\mathrm{tr}{\gamma }_5$ and $\mathrm{tr}({\gamma }_5{\gamma }^{\mu })$. Both vanish. Thus the second term is a legitimate formal Wick contraction but is zero with a compatible regulator; normal ordering the interaction removes it from the outset.

Conditioning on marital status produces five contingency tables. In each, compare the conditional row percentages. The following differences are the sample proportion with some bad experiences minus the sample proportion without them, in percentage points; negative very-safe differences mean fewer respondents feeling very safe in the former group.

The p-values test independence within the respective strata; the magnitudes and directions come from the conditional percentages and association measures. Differences in the table are calculated from the counts before rounding.

Among single respondents, the very-safe sample proportion changes from $33.95\%$ to $29.52\%$, while the unsafe sample proportion changes from $36.28\%$ to $38.10\%$. The direction resembles the weak pooled pattern, but $p=0.6114$ supplies little evidence against independence in this stratum. This is not evidence that the population association is exactly zero.

Among married respondents, the very-safe sample proportion changes from $41.18\%$ to $32.75\%$ and the fairly-safe sample proportion rises from $27.68\%$ to $35.39\%$. Unsafe remains almost unchanged, $31.14\%$ versus $31.87\%$. Thus the within-stratum association again chiefly shifts very safe toward fairly safe. Its $p=0.0257$ is below the $5\%$ significance level, while $V=0.092$ still indicates a small association.

For respondents living with a partner, very safe changes from $41.46\%$ to $28.32\%$, fairly safe from $30.08\%$ to $40.71\%$, and unsafe from $28.46\%$ to $30.97\%$. The very-safe difference is larger than in the married stratum. The printed p-value $0.04885$ would be just below the conventional $5\%$ significance level. However, this particular output is internally inconsistent: recalculating Pearson's chi-squared test from its six displayed counts gives $X^2=4.910389$, $p=0.08585$, and $V=0.14425$. The reported $V=0.144$ agrees with the counts, whereas the printed $X^2=6.900389$ would itself imply $p=0.03174$ for two statistical degrees of freedom, rather than the printed $0.04885$. Using the displayed counts, this stratum does not reject independence at $5\%$. The source PDF itself contains these incompatible figures; this is not resolved by rounding.

Among divorced respondents, very safe changes from $41.18\%$ to $37.17\%$, but unsafe also decreases, from $36.76\%$ to $30.97\%$, as fairly safe increases from $22.06\%$ to $31.86\%$. This is a redistribution toward the middle category, not a simple shift toward greater unsafety. The near-zero gamma, $-0.013$, reflects the lack of a consistent ordinal direction, while $V=0.106$ can still register distributional differences without a direction. The p-value $0.3583$ does not reject independence.

Among widowed respondents, very safe changes from $51.43\%$ to $42.47\%$, and unsafe from $14.29\%$ to $27.40\%$. The ordinal association is positive on the chosen ordering, with gamma $0.220$. Its lack of conventional statistical significance, $p=0.3158$, does not imply a smaller estimated association than in the married group: the widowed stratum has only $108$ complete observations, versus $857$ in the married stratum, and therefore less precision. All five tables have sufficiently large expected counts for the reported approximations; even the smallest expected count is about $8.10$.

The plot uses large-sample Wald confidence intervals for the difference of independent sample proportions within each stratum. They concern the individual category contrasts, whereas the tabulated Pearson's chi-squared tests compare the full three-category distributions. The intervals are unadjusted for multiple hypothesis testing; the pooled row reuses the stratum observations, so it is not a sixth independent sample.

In the elaboration model, the persistence of a generally weak positive ordinal pattern within four strata is closer to partial replication than to a complete explanation of the pooled association by marital status. The differences in estimated patterns suggest possible specification: the divorced group's shift toward the middle category differs from the widowed group's shift toward unsafe. They do not by themselves establish population effect modification. In particular, significant and nonsignificant results need not differ significantly; compare the effects using an interaction contrast or a fitted model with the appropriate interaction terms, rather than comparing their separate p-values.^[ref]

One additional omnibus check can be derived from the supplied tables without assuming a common effect. Recalculating the five statistics from the counts, including the corrected living-with-partner statistic, the joint conditional independence test for stratified contingency tables givesUnder independent-stratum sampling this tests independence within every stratum jointly; it does not test whether the association is homogeneous across strata. It does not reject at $5\%$, but this alone does not establish that stratification removed the population association: its ten-degree-of-freedom alternative is less focused than the pooled two-degree-of-freedom test, and nonrejection is not proof of equality.

The elaboration model also distinguishes an antecedent confounder from an intervening causal mediator. These cross-sectional tables do not establish the relevant time ordering or identify either role for marital status. Five separate stratum tests further require caution about multiple hypothesis testing: for example a Bonferroni correction at familywise $5\%$ would use $0.01$ per test, below all five individual p-values. That caution changes the interpretation of selected significance labels, not the observed conditional percentages.

For a group with sample size $n_s$, let $Y_s$ be its overweight count. The natural sampling model is a binomial distribution, $Y_s\sim \mathrm{Bin}(n_s,{\pi }_s)$, with observed proportion $p_s=Y_s/n_s$. A grouped-binomial logistic regression specifiesFitting event/nonevent counts, or fitting the proportions with weights $n_s$, gives the same binomial likelihood. The weights represent trial counts; unweighted regression of the proportions would give small and large groups the same influence. If printed percentages are rounded, exact original event counts are preferable to manufacturing fractional counts from $n_s{p}_s$.

Year should ordinarily be a factor with three levels unless a linear trend over time is specifically intended. Country, sex and age group are also possible factors. The actual subset, reference levels and formula are absent from this PDF. For a year-by-age interaction term, writeWith treatment contrasts, $\alpha$ is the baseline logit, $u_a$ and $v_t$ are effects at the reference level of the interacting factor, and $w_{at}$ is an additional difference of log-odds differences. Exponentiating a main-effect contrast gives an odds ratio; exponentiating an interaction contrast gives a ratio of odds ratios. It is incorrect to interpret a main effect as a constant effect across the other factor when the interaction term is present.

A coefficient table should be read as estimate, standard error, a Wald estimate/error ratio and its reference-distribution probability. A whole multi-level interaction term is better tested by a nested likelihood-ratio test, using the decrease in deviance and the number of added independent coefficients. Null and residual deviances, residual degrees of freedom, convergence and fitted probabilities should all be checked. Large residual dispersion suggests lack of fit or dependence within sampled schools; a quasibinomial regression or an appropriate clustered analysis may then be preferable. Residual deviance is not automatically a valid chi-squared goodness-of-fit statistic for sparse groups or after estimating a free dispersion.

An interaction plot should display the group means or fitted probabilities against year, with separate lines for the interacting factor, and should also inspect the link scale. Nonparallel probability curves alone do not establish a logit interaction: even an additive logit model transforms nonlinearly to probabilities. As an independent descriptive illustration, the original study's Table 2 reports English boys' age-specific percentages. The following original plot shows their widening age differences after 1984 on both scales. It uses published descriptive values, not a reconstruction of the absent examination fit. The data are from pmc.ncbi.nlm.nih.gov/articles/PMC26603/ .

For example, on the probability scale the 1984–1994 increase for the oldest group exceeds that for the youngest by $6.9-0.8=6.1$ percentage points using the rounded table values. On the logit scale the corresponding difference of differences isThis describes the interaction pattern, but its sampling uncertainty requires the group counts. The absent output prevents identifying the exam fit's coefficient estimates and significance tests.

There are two different uses of Poisson regression. A rare-event approximation models $Y_s\sim \mathrm{Pois}(n_s{\pi }_s)$ withThe fixed offset $\log n_s$ accounts for exposure, and exponentiated slopes are risk ratios, not odds ratios. The approximation is most appropriate when probabilities are small; it is not exact for appreciable prevalence because binomial variance is $n_s{\pi }_s(1-{\pi }_s)$ rather than $n_s{\pi }_s$.

An exact alternative stacks event and nonevent counts in each group and fits a log-linear model with a separate group intercept and event-by-predictor effects:Conditioning on their total $n_s$ gives the binomial event probability ${\mu }_{s1}/({\mu }_{s0}+{\mu }_{s1})$, whose logit is exactly $x_s^T\beta$. Profiling the group intercepts enforces the observed totals and reproduces the grouped-binomial logistic regression. This Poisson surrogate for a conditional multinomial model is exact and should not be confused with the rare-event approximation.

The missing output is needed to identify the actual regression formula, response column, transformations and coefficient estimates. A useful possible analysis regresses the maintained-school percentage among acceptances on the corresponding percentage among applications, allowing for year; it must be described as a model of college-level composition, not a direct estimate of individual acceptance probabilities.

For instance, if $a_{jt}$ and $p_{jt}$ are these two percentages, a normal linear model might specify $\mathbb{E}(a_{jt})=\alpha +\beta p_{jt}+\delta 1_{\{t=2000\}}$, with a slope-by-year interaction term if needed. The intercept is the expected response at zero predictor, often an extrapolation; the slope is a percentage-point change in acceptance composition per percentage-point change in application composition. A slope of one is not by itself evidence of equal acceptance chances: equal chances within every college would give the entire relation $a_{jt}=p_{jt}$, including zero intercept and no year shift. A fit with positive slope mainly confirms that applicant composition helps predict acceptance composition.

Regression diagnostics should include residuals versus fitted values and predictors, a residual normal Q-Q plot, checks of heteroscedasticity, regression leverage, studentized residuals and Cook's distance. The aggregate mature-college category merits an influence check and a sensitivity analysis: combining several colleges can conceal different within-college patterns. Repeated observations from the same college across years may have correlated errors, so treating them as independent needs justification. Check nonlinearity, the year interaction and the effects of excluding no observation except with a substantive reason.

Percentages are bounded, and their precision depends on their denominators. A simple weighted least squares or binomial analysis requires the underlying numbers accepted, not merely percentages. For a proportion based on $m$ people, the working variance is $\pi (1-\pi )/m$ under independent sampling, rather than constant across colleges. If the predictor percentage is itself estimated, an ordinary regression also ignores its uncertainty. The supplied summaries cannot recover all four cells of each college contingency table or the actual fitted numerical output.

The published Cambridge percentage table nevertheless allows a descriptive comparison of acceptance rates. Write $p$ for the maintained-school proportion among applicants and $a$ for its proportion among acceptances. The relative risk from group compositions givesThe overall acceptance probability cancels by Bayes' theorem. This risk ratio compares acceptance probabilities; it is not the acceptance odds ratio, which also needs the overall acceptance probability.

For Christ's, the two published application/acceptance pairs give risk ratios $37/77\approx 0.481$ and $147/187\approx 0.786$. If the displayed percentages were rounded to the nearest whole percentage point, the relative risk bounds from rounded group compositions give ranges approximately $[0.461,0.501]$ and $[0.755,0.818]$. Thus both comparisons remain below one under that rounding assumption. These are descriptive comparisons of observed groups, without adjustment for applicant qualifications or a causal effect interpretation. The published compositions determine approximate relative acceptance rates, but not their standard errors or the original regression results.

These are separate data, so results from the randomized trial must not be carried over. First inspect the construction of the survival response: a command creating a survival time object must pair follow-up in months with an event indicator, whose death/censoring coding should be checked explicitly. A censoring time is not an observed death time. Stratified product-limit fits would then summarize Kaplan–Meier estimators for the dose groups before adjusted regression.

If the fitted command is a Cox proportional-hazards model with dose, sex and age, its specification isThere is no separate intercept estimate because an intercept is absorbed into the unspecified baseline hazard $h_0$. With the stated coding, $e^{{\beta }_G}$ compares high-dose with low-dose death hazards at equal sex and age; $e^{{\beta }_F}$ compares female with male hazards at equal dose and age; $e^{{\beta }_A}$ is the hazard multiplier for a one-year age increase. A negative treatment coefficient would correspond to a lower fitted hazard under high dose. A hazard ratio is not a ratio of survival probabilities, nor directly a ratio of survival times. Under this model the survivor function is $S(t\mid x)=S_0(t{)}^{\exp (x^T\beta )}$.

The coefficient standard errors, Wald ratios and exponentiated confidence intervals should be read together. The common large-sample interval for a coefficient's hazard ratio isGlobal likelihood-ratio, score and Wald tests assess the included slopes jointly; for exactly these three one-coefficient terms their reference degrees of freedom are three. They need not coincide in a small sample. The Cox partial likelihood uses the event risk sets; the fitted log-likelihood is not a full likelihood for an estimated baseline survival curve. The method for tied event times should be recorded.

Check the proportional hazards assumption using scaled Schoenfeld residuals and time trends; check the functional form of age, influential subjects and independent censoring. Dose-by-age or dose-by-sex interaction terms are scientifically possible but require enough information and a stated model comparison. Adjustment can alter a crude dose association, and the allocation mechanism here is not specified as randomized. No coefficients, confidence limits, significance results, exact fitted calls or diagnostics for this dataset are recoverable without the output sheet. The displayed model is an interpretation template, not a claim that it was the exact missing command.

Both fits are the same additive two-way analysis of variance model: a common intercept, five country effects represented by four contrasts, and eight item effects represented by seven contrasts. The three missing cells are omitted, leaving $37$ observations and model rank $1+4+7=12$, hence $25$ residual degrees of freedom. The residual sum of squares is $659.44$ in either ordering, giving

These are sequential sums of squares, so each row measures the improvement when that factor is added after the preceding factors. The country-first sum $1115.56$ describes its unadjusted contribution before accounting for item; its displayed $F=10.573$ and $p=3.73\times 10^{-5}$ are not the country-adjusted-for-item test. When country is added after item, its extra sum is $1616.74$, givingThus the additive normal linear model gives strong evidence of country differences after accounting for the different products. Conversely, item added after country has extra sum $16910.20$, soThe printed zero for this probability is numerical formatting, not a probability literally equal to zero. Product differences are much larger than residual variation, as expected from their different scales.

The explanation for the order dependence is that missing cells can destroy factor orthogonality in two-way ANOVA. A complete equally replicated crossed design would make centered country and item indicator columns orthogonal. Here the missing observations change the item mix across countries and the country mix across items, so the two factor subspaces are not orthogonal. The two sequential partitions still have the same total explained sum: $1115.56+16910.20=16409.02+1616.74=18025.76$ to the displayed precision. For inference about either factor conditional on the other, compare the full model with the reduced model that omits that factor, rather than interpreting its first-entry sum as adjusted evidence.

The omnibus country test does not identify the particular country contrasts. As an extension, fitting the supplied price table gives the item-adjusted UK-minus-US contrast about $21.33$ pounds, with standard error $2.824$ and an ordinary Gaussian-model $95\%$ confidence interval approximately $[15.51,27.14]$ pounds. This is a derived model contrast, with Student t-distribution reference on $25$ degrees of freedom; simultaneous exploration of many country pairs requires control for multiple hypothesis testing.

Check residuals against fitted values, product and country, together with a normal Q-Q plot and influential observations. The homoscedasticity assumption on absolute prices may be doubtful because expensive products can have larger absolute variation. A model for logarithmic prices gives multiplicative country effects and can make the scale assumption more plausible. Country-by-item interaction terms would allow discounts to differ by product, but there is at most one observed value per cell: an unrestricted interaction saturates the $37$ observed means and leaves no pure-error degrees of freedom, so it cannot be tested against independent measurement noise from these data alone. Additional products or genuine cell replication would support that extension. The selected products and non-random missing entries also limit claims about a whole country's consumer prices; the conditional inference is for the stated additive model and observed price collection.

The S-Plus input and the data table differ in one entry: the former uses $33$ for the first year in the second male age group, whereas the table has $22$. The displayed coefficients are reproduced by the software vector with $33$, so that is the input analysed here. Using $22$ instead would give a different fit, including an intercept of $36.175$ rather than $37.275$.

The factor grid has $5\times 4\times 2=40$ rows. Its first factor varies fastest, so the five years run within each age group, and the age groups run within each sex. All three predictors are regression factors: in particular the year coefficients are category contrasts, not a fitted linear slope in calendar time. Under reference-level coding, take the first year, youngest age group and men as the reference levels in a regression factor. Write $s=1$ for women and $0$ for men, and $I_{yk},I_{aj}$ for the other year and age indicators. The normal linear model isThe star in the formula includes both main effects and their statistical interaction. There are twelve coefficients: an intercept, four year contrasts, one sex contrast, three age contrasts and three sex-by-age contrasts. Year effects are assumed common across all sex-age combinations; there is no year interaction term. Independent errors with common variance are additional assumptions of the reported exact tests.

The least-squares estimator is $\hat{b}=(X^{T}X{)}^{-1}{X}^{T}P$, where $X$ has rank twelve. Under the normal linear model,and these quantities are independent. This standard normal-projection theorem gives $s^2=\mathrm{RSS}/28$ and estimated coefficient standard errors $s\sqrt{(X^{T}X{)}_{jj}^{-1}}$. A coefficient estimate divided by its standard error has a Student t-distribution with 28 degrees of freedom under its zero-coefficient null. These are the printed t statistics and two-sided p-values; a printed zero is rounding, not a probability of exactly zero. The option suppresses the coefficient-correlation display.

The fitted intercept $37.275$ is the expected percentage for the reference sex, age and year. Relative to that year, the year changes are $0.375,0.500,1.750,3.000$ percentage points. The last two are individually significant at the conventional five-percent level; the first two are not. The sex coefficient $-18.8$ is the female-minus-male difference in the youngest group. The male age changes are $-7,-12.8,-23$ points. Adding the three interaction term coefficients gives the female age changes $-5.2,-8.2,-13.8$. Thus the sex gap shrinks with age. An individual insignificant interaction term coefficient does not justify deleting the whole interaction: the joint partial F-test for the three interaction terms gives $F=18.6822$ on $(3,28)$ degrees of freedom, with $p\simeq 7.46\times 10^{-7}$.

The regression residuals are observed minus fitted percentages; their printed five-number summary describes their spread. Direct calculation gives $\mathrm{RSS}=60.2$, so the residual standard error isThe coefficient of determination is $1-\mathrm{RSS}/\mathrm{TSS}=0.985827$. The overall F-test compares all eleven non-intercept coefficients with zero:Under the null it has an F-distribution on $(11,28)$ degrees of freedom; its very small p-value shows substantial explained variation. This does not independently validate the Gaussian error assumptions. Survey percentages can have different sampling variances and correlations, and the survey denominators and design are not supplied. The output's exact inferential interpretation is conditional on the stated normal linear model.

The final command makes an interaction plot, not another regression fit. Its points are arithmetic averages over the five years for each sex-age combination. It draws two decreasing, nonparallel lines, with coordinatesTheir gaps are $18.8,17.0,14.2,9.6$ percentage points, illustrating the sex-by-age statistical interaction. The average uses the software vector's disputed entry, consistently with the displayed fit. These are means of percentages, not denominator-weighted pooled prevalence estimates.

For a population question use a population-averaged logistic model for repeated binary outcomes. Code $z_i=1$ for the new treatment and $0$ for the control, with visit times $t_j=2j$ months for $j=1,2,3$. Let $w_{ij}=(1,z_i,x_i^T,t_j{)}^T$, ${\gamma }_M=({\alpha }_M,{\varphi }_M,{\beta }_M^T,{\delta }_M{)}^T$, andDifferent subjects are independent. Each response has a Bernoulli distribution with this marginal mean, but no independence of a subject's three responses is assumed. The linear time effect and absence of treatment-by-time interaction are substantive mean assumptions; if inappropriate they can be replaced by time indicators and interaction terms.

For complete data, set $m_i=(m_{i1},m_{i2},m_{i3}{)}^T$, $A_i=\mathrm{diag}(m_{ij}(1-m_{ij}))$, and $D_i=\partial m_i/\partial {\gamma }_M^T=A_i{W}_i$, where $W_i$ has rows $w_{ij}^T$. A working exchangeable correlation matrix has unit diagonal and off-diagonal value $\rho \in (-1/2,1)$; set $V_i=A_i^{1/2}C(\rho )A_i^{1/2}$. The generalized estimating equation isWith correct means and independent subjects, a positive-definite working covariance matrix and regularity give consistency even if the working correlation is wrong. Write $H={\sum }_i{D}_i^T{V}_i^{-1}{D}_i$ and $u_i=D_i^T{V}_i^{-1}(Y_i-m_i)$. The fitted subject-level sandwich covariance matrix isevaluated at the estimates. This inference is asymptotic in the number of independent subjects, not the number of visits.

For the actual incomplete data, define $R_{ij}=1$ if visit $j$ is observed, with $R_{i0}=1$ and monotone dropout. Let ${\mathcal{H}}_{i,j-1}$ contain treatment, baseline covariates and responses observed before visit $j$. Under sequential missing at random, specifyThe assumption is that the retention decision, conditional on this history, does not additionally depend on unseen current or future outcomes. Fit the $q_{ij}$, for example by visit-specific logistic regressions. A simple valid choice uses working independence and the inverse-observation-weighted estimating equations for longitudinal dropout:Terms with $R_{ij}=0$ are zero and require no missing response. Given the full response vector, sequential missing at random makes $\mathbb{E}(R_{ij}/{\rho }_{ij}\mid Y_i,z_i,x_i)=1$. Consequently the weighted score has the same expectation as the complete-data score, proving its mean-zero property. Correct retention probabilities, positivity and a correctly specified marginal mean are required. Use cluster-robust uncertainty, accounting for fitted retention-model coefficients, or resample entire subjects and refit both models. Under missing completely at random, the simpler unweighted observed-response GEE can suffice; ordinary unweighted GEE is not valid under arbitrary history-dependent missing at random.

The treatment coefficient gives a population conditional-on-baseline odds ratio $e^{{\varphi }_M}$ at each visit. To communicate an absolute population treatment benefit, standardize predicted probabilities over the full baseline population, rather than only completers:A causal reading additionally uses the randomized controlled trial assignment, consistency of the treatment definition, no interference and the dropout assumptions; the statistical model alone does not supply those conditions.

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.