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scan reads the numerical prices, including the three missing-value markers, into a vector. The data must be read row by row for the subsequent factor construction to match them. The sample size available for analysis is therefore $37$, rather than $40$. The numerical summary describes the observed prices: the sample median is £43.54 and the sample mean is £46.94. The upper tail, particularly the boots, raises the sample mean above the sample median. These summaries mix very different products and are not adjusted country comparisons.

The two character scans create labels. gl(8,5,length=40,labels=item) makes a regression factor with each of eight item labels repeated five times. gl(5,1,length=40,labels=country) cycles through the five country labels once for each item. For a regression factor and numerical response, the two plot commands produce grouped box plots. The country box plots suggest lower US prices, while the item box plots show particularly expensive boots and inexpensive cardigans. They are descriptive displays of this small basket, with different item compositions where prices are missing; they cannot separate country effects from item effects.

The first lm fits an additive two-way analysis of variance:The constraints are R's usual treatment coding. The normal linear model used for the printed tests assumes independent errors with common variance and approximately a normal distribution, and no item-by-country interaction in the mean. Missing prices are omitted, leaving $37$ responses and $1+4+7=12$ independent mean coefficients. Thus the residual degrees of freedom are $37-12=25$. This deletion is defensible for the conditional price model when missingness does not depend on the unobserved error after conditioning on the two factors; the output itself does not establish such a missing-data mechanism.

The anova tables report sequential sums of squares, rather than order-independent adjusted effects. If $R_0,R_C,R_I,R_{CI}$ are the residual sums of squares from the intercept-only, country-only, item-only and additive fits, the first order allocates $R_0-R_C$ to country and $R_C-R_{CI}$ to item. The second order allocates $R_0-R_I$ to item and $R_I-R_{CI}$ to country. Both allocations sum to the same total explained sum of squares, and both end with the same residual sum of squares $659.4$. The three missing combinations destroy balanced factorial orthogonality, so the allocation changes with order. To test country after allowing for items, use the country row in the second table; to test item after allowing for countries, use the item row in the first table. The first-entered rows do not supply those adjusted tests.

Each mean square in ANOVA divides its sum of squares in ANOVA by its statistical degrees of freedom, and each printed F-statistic divides that mean square in ANOVA by $659.4/25=26.38$. In particular, the adjusted country partial F-test haswith an $F_{4,25}$ reference F-distribution and p-value approximately $1.86\times 10^{-6}$. The adjusted item partial F-test gives $F=91.58$ on $7,25$ statistical degrees of freedom. Both factors contribute strongly under the additive normal linear model. The larger item sum of squares in ANOVA measures the much greater differences between these products, but is not evidence that every pair of items differs significantly.

The intercept $50.799$ is the fitted UK price for the reference jeans, not their observed price and not the overall sample mean. Each country regression coefficient is its adjusted price difference from the UK, assumed common across items: Sweden is £5.224 lower, France £10.708 lower, Germany £7.676 lower and the US £21.328 lower. Each item regression coefficient is its adjusted difference from the reference jeans, assumed common across countries. Thus the fitted UK boots price is $50.799+52.854=103.653$, and the fitted US boots price is $103.653-21.328=82.325$.

The coefficient standard errors come from $s^2(X^{T}X{)}^{-1}$, where $s^2=659.4/25$ and $X$ is the design matrix. Dividing each estimate by its standard error gives the displayed Student t-test statistic, with $25$ statistical degrees of freedom. The stars summarize the corresponding two-sided p-values. The individual tests compare each indicated country or item with its reference level, conditional on all other additive terms; they are neither tests of every pairwise comparison nor a simultaneous confidence interval procedure. In particular, Sweden's p-value $0.0626$ provides weaker evidence than the other country contrasts, and does not establish equality with the UK. The intercept test has little substantive relevance, and interpreting many individual stars also raises multiple hypothesis testing issues.

The residual standard error $s=5.136$ measures unexplained prices in pounds. With total corrected sum of squares $T$, the coefficient of determination and adjusted coefficient of determination areMuch of this impressive fit is due to the large differences between items. The overall F-test compares the additive fit with a constant mean: $F=((T-659.4)/11)/(659.4/25)=62.12$ on $11,25$ statistical degrees of freedom. It tests that all eleven non-intercept coefficients vanish, not that the additive model is adequate. Regression diagnostics should examine regression residuals, changing price variability across items and possible interaction. Percentage differences may be more plausible than common pound differences, motivating a natural logarithm of price. No random sample of countries or products, and no independent replication of a price within a cell, is supplied; broad claims about all designer goods or the causes of country differences require more than these conditional tests.

Finally, Item*Country expands to both main effects and their full interaction. A cell-means parameterization has one separate fitted mean for every observed cell. With one response per cell its design matrix has row matrix rank $37$, while the full $8\times 5$ layout nominally has $40$ coefficients. Three coefficient combinations are aliased because their cells are missing. This is saturation of an unreplicated two-factor regression: every observed price is fitted exactly and the residual degrees of freedom are zero. The final fit has zero residuals, $R^2=1$ and no residual-based estimate of the error variance or ordinary interaction significance test. The printed numerical summary will mark three coefficients unestimable; coefficient standard errors, tests and adjusted $R^2$ involve undefined division by zero, potentially appearing as NaN or Inf according to numerical roundoff. Perfect interpolation here is a consequence of a saturated statistical model, not evidence for a superior predictive model.

The Poisson deviance compares the fitted Poisson regression with a saturated statistical model:using $0\log 0=0$. If the usual deviance goodness-of-fit test approximation is adequate, compare $27.2$ with a chi-squared distribution on $29$ statistical degrees of freedom. Its upper-tail p-value is about $0.561$, and $D/29=0.938$. There is no detected lack of fit or evidence of substantial overdispersion from this deviance. This neither proves the Poisson regression correct nor establishes that the slope is nonzero. Small expected counts, dependence or inadequate covariate specification can undermine the reference approximation, so deviance residuals and other regression diagnostics still matter.

read.table(...,header=T) reads four grouped rows with column names. The displayed outcome counts give the observed risks $0.1519,0.2145,0.2611,0.5111$ in the indicated factor order. rbind in the first test pools over predisposition: the two rows represent no cannabis use and some use, with event/non-event counts $(341,1775)$ and $(82,238)$. Thus the pooled risks are $341/2116=0.1612$ and $82/320=0.2563$. This test concerns the marginal statistical association between use and outcome, not an association adjusted for predisposition.

Under independence, expected counts in the two-by-two contingency table are $E_{ij}=O_{i+}{O}_{+j}/N$. The ordinary Pearson chi-squared statistic is $17.5183$. The printed $16.8618$ is reproduced by the default two-by-two Yates continuity correction:All absolute differences here exceed one half. Comparison with a chi-squared distribution on one statistical degree of freedom gives $p=4.02\times 10^{-5}$. There is strong evidence of marginal association. The simplified output omits the usual continuity-correction wording, but its numerical value shows which statistic was used.

attach makes the data columns available by name, and tot=with+without obtains group sizes $1936,275,180,45$. The grouped binomial regression fits $Y_g\sim \mathrm{Bin}(n_g,{\pi }_g)$, with response $Y_g/n_g$ and prior weights $n_g$. These weights tell R the binomial denominator; treating the four proportions as four equally precise individual observations would be wrong. The default logit link isLet $A$ indicate some use and $B$ indicate predisposition. The additive logistic regression specifies $\eta =a+bA+cB$. Its log-likelihood, including the binomial constants, isDifferentiation gives score equations $X^T(Y-n\pi )=0$ and Fisher information matrix $X^{T}WX$, where $W_{gg}=n_g{\pi }_g(1-{\pi }_g)$. Fisher scoring solves these equations, and the inverse fitted Fisher information matrix gives the reported approximate coefficient variances. The number of scoring iterations records numerical convergence, not degrees of freedom or statistical evidence.

The intercept corresponds to no use and no predisposition and gives baseline fitted risk ${\mathrm{logit}}^{-1}(-1.73881)=0.1495$. The additive coefficients imply common adjusted odds ratios:They multiply the odds, not the risks. The Wald tests divide the estimates by their standard errors and use a standard normal distribution; both predictors have small two-sided p-values. Approximate $95\%$ confidence intervals for these odds ratios are $(1.296,2.266)$ and $(1.685,3.110)$, obtained by exponentiating the log-scale intervals. The intercept test against zero asks whether the reference risk equals one half, and is not a test of either exposure association. The binomial dispersion is assumed to be one because $\mathrm{Var}(Y_g)=n_g{\pi }_g(1-{\pi }_g)$; the output does not demonstrate independence or rule out extra-binomial variation.

There are four grouped probabilities and three fitted coefficients, leaving one residual degree of freedom. The fitted risks are approximately $(0.1495,0.2314,0.2869,0.4080)$. The binomial deviance compares this fit with the four unrestricted observed proportions:The deviance goodness-of-fit test against ${\chi }_1^2$ has p-value $0.0796$. The common-odds-ratio fit is not rejected at $5\%$, although the doubly exposed group's observed risk exceeds its fitted risk appreciably and suggests possible interaction.

The second formula adds $\delta AB$. Its four coefficients fit all four probabilities, making a saturated statistical model with zero grouped residual degrees of freedom. The near-zero deviance is roundoff from exact interpolation; it supplies no remaining grouped lack-of-fit test. Unlike Q1's unreplicated Gaussian fit, however, these grouped responses represent many independent Bernoulli trials, and their binomial variances are specified by their sizes and probabilities. Coefficient standard errors remain available from binomial information without estimating dispersion from the zero residual degrees of freedom.

Indeed, let ${\eta }_{ab}=\log [Y_{ab}/(n_{ab}-Y_{ab})]$ be the fitted cell logit link value. The saturated coefficients areThese calculations yield the printed estimates. The delta method gives $\mathrm{Var}({\hat{\eta }}_{ab})\approx 1/Y_{ab}+1/(n_{ab}-Y_{ab})$, and independent groups make the variance of $\hat{\delta }$ the sum of all eight reciprocal counts. Its square root is $0.37857$, reproducing the printed standard error.

The saturated logistic regression describes the conditional odds ratios as follows:Thus the main coefficients now concern effects at the other factor's reference level. Logistic interaction as a ratio of odds ratios gives $e^{\delta }=1.939$, with approximate $95\%$ confidence interval $(0.923,4.073)$. The interaction Wald test has $p=0.0802$, while the nested likelihood-ratio test uses deviance difference $3.0733$ on one statistical degree of freedom and has $p=0.0796$. The data suggest a larger odds association in predisposed subjects, but do not establish this interaction at the conventional $5\%$ level. This conclusion is specific to the odds scale; absolute-risk differences can vary even with no logit-scale interaction.

The Akaike information criterion is $-2\hat{\ell }+2k$. The additive fit has $k=3$ and AIC $31.766$; the interaction fit has $k=4$ and AIC $30.692$. The extra coefficient improves twice the log-likelihood by $3.0733$ but incurs penalty $2$, so AIC falls by only $1.0733$. It weakly favors the larger fit rather than decisively establishing an interaction, and AIC is not a p-value.

The marginal odds ratio $1.793$ and the fitted conditional odds ratios are different quantities. Predisposition is associated with both use frequency and outcome, so confounding is a substantive possibility; noncollapsibility of the odds ratio also means marginal and conditional odds ratios need not agree. This is an observational association analysis with only one adjustment factor. It does not by itself establish a causal effect, exclude other confounding, or validate the assumed independent binomial sampling.

For independent binomial distributions, the log-likelihood, up to known binomial coefficients, isThe saturated statistical model maximizes each term at ${\tilde{p}}_i=y_i/n_i$. Under the logit link, the fitted probabilities are ${\hat{p}}_i=(1+e^{-x_i^T\hat{\beta }}{)}^{-1}$. Subtracting the fitted log-likelihood from the saturated log-likelihood gives the binomial devianceA zero-count summand uses the continuous convention $0\log 0=0$. For the restricted model, replace ${\hat{p}}_i$ by ${\hat{p}}_i^{(0)}=(1+e^{-{\tilde{x}}_i^T\hat{\tilde{\beta }}}{)}^{-1}$ to obtain $D_0$. Since the saturated likelihood cancels, $D_0-D_1=2({\hat{\ell }}_1-{\hat{\ell }}_0)$ is the likelihood-ratio test statistic. For the nested restriction setting $p-k$ coefficients to zero, Wilks theorem gives the asymptotic chi-squared distribution ${\chi }_{p-k}^2$ under regularity and an interior true parameter; reject for a large difference.

For the six anther cells, ${\omega }_0$ gives a common success probability, ${\omega }_1$ gives a separate probability for each storage condition, ${\omega }_2$ gives a common intercept and a log-force slope, and ${\omega }_3$ gives storage-specific intercepts with a common log-force slope. Their parameter counts are $1,2,2,3$, so their residual statistical degrees of freedom are $5,4,4,3$. In ${\omega }_3$, storage has a constant effect on log odds; multiplying force by $r$ multiplies the odds by $r^{\beta }$ in either storage group.

The relevant nested likelihood-ratio tests at the $5\%$ level are:

The two two-parameter models ${\omega }_1$ and ${\omega }_2$ are not nested, so their deviance difference has no ordinary nested chi-squared distribution calibration. The preferred parsimonious model is ${\omega }_1$, with a storage effect and no log-force effect. Its deviance goodness-of-fit test is acceptable: $D_1=5.173<9.49$ on four statistical degrees of freedom. The common-probability model also passes its separate goodness-of-fit comparison, $10.452<11.07$, but this does not contradict the relative evidence for adding storage: the two tests assess different null hypotheses. These are large-sample approximations; sufficiently large expected successes and failures, not merely the number of anthers, underpin their calibration.

For an exponential dispersion family, differentiating the normalization integral with respect to $\theta$ gives an expected score of zero. The score is $(Y-b^{\prime }(\theta ))/\varphi$, hence $E(Y)=b^{\prime }(\theta )$. Differentiating again, the expected squared score is $b^{\prime \prime }(\theta )/\varphi$; it is also $\mathrm{Var}(Y)/{\varphi }^2$. ThusThe canonical link function sets the linear predictor equal to the natural parameter: $g(\mu )=\theta =(b^{\prime }{)}^{-1}(\mu )$. The variance function is $V(\mu )=b^{\prime \prime }((b^{\prime }{)}^{-1}(\mu ))$, so $\mathrm{Var}(Y_i)=\varphi V({\mu }_i)$; the dispersion parameter $\varphi$ is separated from the mean-dependent factor.

The saturated statistical model has a separate mean for each response and fits ${\tilde{\mu }}_i=y_i$, with boundary parameters interpreted by limits. At the same fixed dispersion, its log likelihood and the fitted model's log-likelihood define the scaled devianceThe unscaled deviance isKeeping dispersion fixed in this definition is essential; it is not a comparison with a separately estimated zero variance in a Gaussian saturated fit.

For the normal distribution, expanding its density exponent givesConsequentlyHere $V(\mu )=1$ and the canonical link is the identity, so ${\mu }_i={\beta }^T{x}_i$. The saturated natural parameter is ${\tilde{\theta }}_i=y_i$. Substitution into the deviance formula yields

For the cholesterol fits, all formulae include an intercept. Model 1 has age and BMI with three coefficients; models 2 and 3 have one of these covariates with two coefficients; model 4 has only the intercept. Their residual statistical degrees of freedom are respectively $27,28,28,29$. Their reported deviances are residual sums of squares, because these are Gaussian linear models.

The relevant hypothesis is $H_0:{\beta }_{\mathrm{BMI}}=0$ in the full model, against $H_1:{\beta }_{\mathrm{BMI}}\ne 0$, with age already present. Thus compare model 2 with model 1 using the partial F-test. Its extra sum of squares is $31.64-27.10=4.54$, and the common variance is estimated from model 1 by ${\hat{\sigma }}^2=27.10/27$. ThereforeSince $4.52325>4.210$, reject at $5\%$; the two-sided coefficient-test equivalent has $p\simeq 0.0427$. BMI is associated with cholesterol after adjustment for age, under the linear Gaussian model. The RSS values alone do not identify the sign of its coefficient or the size of the association.

For comparison, the marginal BMI test would compare model 4 with model 3, giving $F=13.88/(35.82/28)=10.8498$ on $(1,28)$ statistical degrees of freedom, with $p\simeq 0.00268$. This does not answer the adjusted question and makes the BMI evidence appear stronger than the conditional test. Age also contributes after BMI: comparing model 3 with model 1 gives $F=8.72/(27.10/27)=8.6878$ on $(1,27)$ statistical degrees of freedom, with $p\simeq 0.00653$. Thus retaining both predictors is supported at the stated level.

A raw deviance difference must not simply be compared with the supplied ${\chi }_1^2$ cutoff: Gaussian dispersion is unknown and the differences here are unscaled. If ${\sigma }^2$ were known, the scaled difference $4.54/{\sigma }^2$ would have a ${\chi }_1^2$ law. Estimating it from the full fit gives the exact finite-sample $F$ comparison above, with the appropriate denominator statistical degrees of freedom.

factor(day) treats the nine days as categorical rather than imposing a numerical time trend. rep(400,27) supplies the number of trials for every dish, and prop is the observed surviving proportion. The contrast setting uses treatment coding with day 1 as reference; the ordered-factor polynomial option is unused. Supplying these proportions to glm with the binomial family and weights $400$ gives precisely the grouped-binomial logistic regression likelihood for the counts. The weights represent trial counts, not additional independent dish observations. The binomial family's default link is the logit link and its dispersion is fixed at $1$.

Let $Y_{ij}$ be survivors in dish $j=1,2,3$ on day $i=1,\dots ,9$. Both fitted models assume independent $Y_{ij}\sim \mathrm{Bin}(400,p_i)$. Model 1 specifiesso it has one common probability and one fitted coefficient. Its estimated probability is the total surviving count divided by the total trial count, $\hat{p}=3539/10800\simeq 0.327685$.

Model 2 specifiesIt has nine coefficients and allows a separate probability for each day, while retaining a common probability for the three dishes within that day. The MLEs are the day totals divided by $1200$. In day order they areThe intercept is the first day's logit, and each other coefficient is its day's logit minus that reference. These are additive effects on log odds, not additive changes in survival probability.

The analysis of deviance adds the eight day indicators after an intercept-only model. Its residual deviance compares a fitted model with the saturated statistical model assigning one separate probability to each of the $27$ dishes. For fitted dish probabilities ${\hat{p}}_{ij}$, the binomial deviance isThe binomial scale is one, so this is also scaled deviance. The reported decrease measures the extra fit provided by day. There are $27$ observational units for these residual statistical degrees of freedom, not $10800$ separately fitted Bernoulli responses.

For independent random variables $Y_i\sim \mathrm{Bernoulli}(p_i)$, the full logistic regression has linear predictorLayout A is the reference level in a regression factor. The smaller model sets ${\beta }_B={\beta }_C=0$, leaving the same logit link and the price predictor. In either caseThe saturated statistical model allows a separate probability per observation; its maximizing probabilities are $p_i=y_i$, with log-likelihood zero under $0\log 0=0$. Hence the binomial deviance for either fitted model isEquivalently, write the summands as $2[y_i\log (y_i/{\hat{p}}_i)+(1-y_i)\log ((1-y_i)/(1-{\hat{p}}_i))]$. The fitted probabilities differ between the two models, but the definition of binomial deviance is the same. A residual deviance for individual binary data need not itself have an accurate chi-squared distribution approximation; the nested-model difference used next has a different asymptotic justification.

The unscaled residual deviance of an exponential dispersion family compares its fitted mean with the saturated statistical model at a common dispersion parameter. Its normalization removes the factor $1/\varphi$ in the log-likelihood. In Poisson regression and binomial regression, $\varphi =1$, so it is twice the saturated-minus-fitted log-likelihood. Under suitable regularity, $D/\varphi$ has approximately a chi-squared distribution with the residual degrees of freedom; this is not an automatic accurate approximation for sparse responses. Nested deviance reductions with estimated dispersion lead to approximate F-tests after scaling by the dispersion estimate.

A two-way analysis of variance with the full interaction and one observation in each observed factor cell has one independently varying fitted cell mean per observation. A cell-indicator design matrix has full row matrix rank, even when some combinations are missing, so the model is a saturated statistical model. Its residual sum of squares is zero and its residual degrees of freedom are zero. Missing combinations make some coefficients nonidentifiable; they do not restore a pure-error estimate. Standard errors and an ordinary residual-based F-test for interactions therefore cannot be recovered merely by fitting this larger model.

For an exponential dispersion family with a common dispersion parameter $\varphi$, scaled deviance compares a fitted model with the saturated statistical model at the same dispersion:For a normal distribution with variance ${\sigma }^2$, unscaled deviance is the residual sum of squares and scaled deviance is that sum divided by ${\sigma }^2$. For the binomial distribution, $\varphi =1$ and the two coincide. Unknown Gaussian dispersion requires variance estimation; raw deviance differences should not be treated as chi-squared statistics without the scale.