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The first fit is a Bernoulli logistic-regression model for the probability of a positive response. Its fitted linear predictor isand its fitted probability is $\hat{\pi }=(1+e^{-\hat{\eta }}{)}^{-1}$. There are eight coefficients. The null degrees of freedom identify $n=200$, and the residual degrees of freedom are $200-8=192$.

Each slope is a change in the log odds per unit of its covariate, holding the others fixed. Exponentiating gives conditional odds ratios; for example a ten-unit glucose increase multiplies fitted odds by $e^{10(0.032115958)}=1.379$, and a ten-year age increase by $e^{10(0.041182353)}=1.510$. These are changes in odds, not multiplicative changes in probability, and the observational associations are not automatically causal effects. The intercept corresponds to all numerical covariates being zero, generally an extrapolation beyond meaningful subjects.

The standard errors come from the inverse Fisher information matrix of the logistic regression. With binomial dispersion parameter fixed at one, the displayed t ratios are asymptotic normal Wald statistics. In the full fit, glucose and the pedigree measure have clear individual evidence of nonzero slopes, with two-sided $p\simeq 2.09\times 10^{-6}$ and $0.00609$. The body-mass and age terms are borderline, with $p\simeq 0.0504$ and $0.0618$; pregnancy count is weaker, $p\simeq 0.110$. Blood pressure and skinfold have little individual evidence after adjustment. These labels should not be converted into a claim that the weaker predictors have no effect, or used as automatic model-selection rules.

A deviance residual here isIts square is the observation's contribution to binomial deviance, so the squared residuals sum to the printed residual deviance. The five-number summaries describe the distribution of these signed residuals. The null deviance $256.4142$ belongs to an intercept-only logistic regression; the full fitted deviance is $178.3907$. Four Fisher scoring iterations record convergence of the numerical likelihood fit.

The second model uses only glucose:Its two coefficients leave $198$ residual degrees of freedom. A ten-unit glucose increase multiplies its fitted odds by $e^{0.3778371}\simeq 1.459$. This differs from the adjusted multiplier $1.379$, because the models condition on different covariates; both adjustment and noncollapsibility of the odds ratio can change logistic coefficients. The glucose-only model has deviance $207.3727$, so its maximized log-likelihood is lower than the full model's.

For these fixed-dimensional nested models, analysis of deviance and Wilks theorem giveThus the additional six predictors collectively improve the glucose-only model, even though several individual Wald tests are not significant. These comparisons assume independent subjects, an appropriate logistic mean structure, full rank, and regular finite parameter estimates without separation.

It would be incorrect to infer that absolute fit is good merely because $178.39<192$. Individual Bernoulli deviance need not have a chi-squared calibration: there is just one binary observation per saturated probability, and the saturated dimension increases with sample size. A simple counterexample is an intercept-only model with true probability $1/2$, for which $D/n\to 2\log 2$, not one. For absolute adequacy, inspect residual patterns and predicted-risk calibration, use meaningful grouped comparisons when appropriate, or calibrate a chosen fit statistic by parametric bootstrap. The regular nested-model tests above are not subject to that saturated-model dimensionality problem. Only the displayed summaries and a few example rows are supplied, so patient-level diagnostics or complete numerical refitting cannot be reconstructed from this page alone.

Use the logit link $g(\pi )=\log \{\pi /(1-\pi )\}$ and independent counts $R_{ij}\sim \mathrm{Bin}(n_{ij},{\pi }_{ij})$. This gives a grouped-binomial logistic regression with 24 cells and 13 free coefficients: an intercept, 11 centre contrasts, and one period contrast. Reference treatment coding makes the intercept the log odds for Centre 1 in the earlier of the two included periods. The two factors are assumed to have been coded as specified in the question.

The binomial log-likelihood, apart from known constants, isMaximize by Fisher scoring or iteratively reweighted least squares. Using response proportions $r_{ij}/n_{ij}$ with binomial trial weights $n_{ij}$ is equivalent to using the count response $(r_{ij},n_{ij}-r_{ij})$; omitting those weights would give the wrong likelihood and standard errors. The coefficient covariance matrix is approximately the inverse of $X^T\mathrm{diag}\{n_{ij}{\hat{\pi }}_{ij}(1-{\hat{\pi }}_{ij})\}X$.

The fitted reference log odds are $-0.809585$, and the period change is $-0.426758$, givingThese are fitted probabilities under a common period effect, so they need not equal the separate empirical proportions at Centre 1. The estimated period odds ratio isThus the later-period death odds are estimated to be about 35% lower at each centre, with strong evidence of a period effect: the normal Wald statistic is $-5.415$. The displayed “t value” is a coefficient divided by its standard error; with fixed binomial dispersion its usual large-sample calibration is standard normal, not a Student distribution with an estimated Gaussian-error variance. The odds ratio is not a mortality probability ratio.

Each centre coefficient is a log odds ratio relative to Centre 1, adjusted for period. All its point estimates are negative, so all other centres have lower fitted mortality probabilities in either period. For example, Centre 11's odds ratio is $e^{-1.079474}=0.3398$, or about $2.94$ times greater odds at Centre 1 than at Centre 11. Individual unadjusted two-sided Wald tests find significant differences at the 5% level for centres 2, 4, 6, 7, 8, 9, 11 and 12; the coefficients for 3, 5 and 10 do not individually reach that threshold. These separate tests are correlated comparisons against a common reference and should not be presented as 11 independent discoveries without considering multiple testing. They adjust only for time period, not for differing patient case mix.

For fit assessment, the residual binomial deviance is $15.73891$ on $24-13=11$ statistical degrees of freedom. Its ${\chi }_{11}^2$ upper tail is approximately $0.151$, so there is no appreciable evidence of lack of fit to these binomial cell counts. In particular, comparing with a saturated centre-by-period model gives this same deviance and 11 additional parameters; the data do not strongly require a centre-period interaction. The moderate deviance residuals are consistent with that assessment, although a residual plot by centre and period would be more informative than the five-number summary alone.

The reduction from null to fitted deviance isIt tests the 12 centre-and-period coefficients jointly against an intercept-only model, with reference ${\chi }_{12}^2$ and $p\approx 1.24\times 10^{-16}$. It is not a centre-only test adjusted for period. For that test, fit the period-only model and compare its deviance with the full fit: the pooled binomial probabilities in each period give deviance $90.34578$, soThus centre heterogeneity persists after allowing for period. Dispersion one is the stipulated binomial variance, and the five scoring iterations describe numerical convergence rather than statistical evidence. The fitted associations alone do not establish a causal difference in surgical quality.

Treat the counts as independent binomial distributions, $R_i\sim \mathrm{Bin}(n_i,{\pi }_i)$, conditional on the known numbers of operations. The null hypothesis is that the probabilities ${\pi }_i$ are all equal, with no assumed numerical value. Centre 4 has no operations in this period and hence contributes likelihood one, supplying no information about its mortality probability. The empty groups in a binomial homogeneity test principle leaves 11 contributing centres and 10 test degrees of freedom.

Under the common-probability model the maximum-likelihood estimator isUse the two-column contingency table of deaths and survivors, deleting its all-zero row. The Pearson chi-squared test of homogeneity has statisticUnder the null, the large-sample reference distribution is ${\chi }_{10}^2$. The expected counts are adequate for this approximation here, including at the smallest nonempty centre. Calculating givesAn alternative is the likelihood-ratio test against separate centre probabilities, with binomial devianceThe unrestricted probabilities are $r_i/n_i$; subtracting one fitted common probability leaves the same 10 degrees of freedom. Here $D=39.5033$ and $p\approx 2.07\times 10^{-5}$. Thus there is strong evidence against a common mortality probability. This omnibus test does not identify Centre 1 as the sole source of the heterogeneity. If its comparison with the remaining centres is the primary scientific question, specify a separate planned one-versus-rest comparison; inspection of many pairwise comparisons calls for consideration of multiple testing.

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.

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.

Index city by $g\in \{0,1\}$ and age group by $j$, with the youngest group the reference level. For each of the 15 observed cells, the grouped-binomial logistic regression assumeswith independent cell counts. The response supplied to R is $C_{gj}/N_{gj}$, and the weights specify the binomial numbers of trials $N_{gj}$. They are not 15 arbitrary precision weights: the conditional variance of the proportion is ${\pi }_{gj}(1-{\pi }_{gj})/N_{gj}$. The logit link models log odds, and the two sets of factor coefficients use corner-point constraints.

There are nine parameters: one intercept, seven age contrasts and one city contrast. Hence the residual degrees of freedom are $15-9=6$, not the total population minus nine. The null model has only an intercept and $14$ residual degrees of freedom. Adding age uses seven more parameters and reduces the binomial deviance from $2330.46$ to $232.28$, a drop of $2098.19$. Adding city then uses one parameter and reduces it by $227.12$ to $5.15$. These are sequential likelihood-ratio tests, asymptotically compared with chi-squared distributions on seven and one degrees of freedom. Both provide extremely strong evidence of an effect; the printed age p-value $0.00$ is rounding, not an exactly zero probability. Because the table is incomplete, sequential main-effect sums need not be invariant to the order of fitting.

The intercept estimate $-11.69364$ is the log odds for the youngest Minneapolis–Saint Paul group. It implies baseline odds $e^{-11.69364}$ and a fitted probability about $8.35\times 10^{-6}$. Its standard error is $0.44923$; its reported Wald test tests log odds zero, or probability $1/2$, rather than a scientifically interesting comparison of groups.

The age coefficients are log odds ratios relative to the youngest group within the same city. Their exponentials are approximately $13.86$, $46.82$, $99.03$, $162.23$, $284.38$, $497.07$ and $484.57$ as age increases through the seven other groups. All these comparisons have large positive Wald statistics and small p-values. Thus the fitted probabilities are much higher in the older groups than in the youngest group. The slightly smaller coefficient in the oldest group than in the preceding group does not itself prove a real decline: that claim needs a contrast and its standard error, including the covariance of the two estimates.

The city coefficient $0.85492$ is the log odds ratio for Fort Worth relative to Minneapolis–Saint Paul, adjusted for age. Its standard error $0.05969$ gives $z=14.322$, so the null city effect is strongly rejected by the Wald test, consistent with the sequential likelihood-ratio test. The fitted constant-over-age city odds ratio and an approximate 95% confidence interval areThis is an odds ratio, not generally a probability ratio. The probabilities here are sufficiently small that the two ratios are close. The absence of an age-by-city interaction is an assumption of the fitted additive log-odds model.

The final binomial deviance is $5.1509$ on six degrees of freedom; its approximate upper-tail p-value is $0.525$. It gives no indication of substantial lack of fit against the saturated observed-cell model. The deviance residuals range from $-1.2830$ to $1.0820$, with median zero and no conspicuously large value in the five-number summary. Neither statement proves that every model assumption holds. In particular, the small case counts motivate checking the accuracy of the chi-squared asymptotic approximation, for example by a fitted-model parametric bootstrap.

The dispersion is fixed at one by the binomial distribution; it was not estimated as a free scale parameter. Four Fisher scoring iterations describe the numerical optimization, not four additional parameters. Residual and cellwise observed-versus-fitted comparisons would help assess adequacy and potential overdispersion.

The inverse logit link is ${\mu }_i(\beta )=e^{{\beta }^T{x}_i}/(1+e^{{\beta }^T{x}_i})$. independence therefore gives the Likelihood function and log-likelihoodDifferentiation yields the score ${\nabla }_{\beta }\ell ={\sum }_i{x}_i(y_i-{\mu }_i)$. The matrix whose rows are $x_i^T$ has dimension $n\times p$, not the PDF's printed $p\times n$. With this necessary dimension correction, any finite maximum Likelihood function estimate satisfiesThe Hessian matrix is $-X^{T}WX$, with $W_{ii}={\mu }_i(1-{\mu }_i)$, so the log-likelihood is concave. These score equations apply to a finite interior estimate; complete or quasi-complete separation can instead make its supremum occur only as coefficients diverge.

The deviance is twice the difference between maximized saturated and fitted log-likelihoods. A saturated Bernoulli distribution has fitted means $y_i\in \{0,1\}$ and log-likelihood zero, interpreted by limits. Hence the binomial deviance here isZero-multiplier logarithms are interpreted as zero.

Let $C_i$ be the disease count among $m_i$ miners in exposure group $i$, with representative exposure $t_i$, for $i=1,\dots ,8$. The grouped-binomial logistic regression isThe model assumes independent miners, a common disease probability within each group at its representative exposure, and binomial variation without an extra dispersion parameter. The use of average exposure is a grouping approximation, not a claim that every miner in a group has identical exposure. The input response is $C_i/m_i$ and weights=miners supplies the binomial denominators $m_i$, not inverse variances; together they yield this count likelihood function. The default binomial link is the logit link.

For $H_0:\beta =0$, the likelihood-ratio test compares the intercept-only model with the linear-exposure model. The reduction in binomial deviance isUnder the usual regular large-sample approximation it has a chi-squared distribution with one degree of freedom. There is very strong evidence of exposure dependence, with a positive estimated coefficient $\hat{\beta }=0.09346$.

For goodness of fit against separate probabilities in all eight groups, the residual binomial deviance $6.0508$ has approximately $8-2=6$ degrees of freedom. Its upper-tail probability is about $0.418$, giving no evidence of lack of fit. This is an approximate check, particularly because the earliest exposure groups have small expected disease counts. It supports adequacy of this mean structure, not proof that all its assumptions or possible confounding variables have been addressed.

The predict call asks for a prediction at exposure $40$, sets type="response" to return the fitted probability rather than its log odds, and requests the standard error of that fitted probability. It givesFor $z=(1,40{)}^T$, the linear predictor is $z^T\hat{\beta }$, and the delta method givesThe intercept-slope covariance is included in this expression. This measures uncertainty in the fitted mean probability; it is not the standard deviation of a new individual's binary disease outcome or of an unspecified new sample proportion.

The model's odds at exposure $t$ are $O(t)=e^{\alpha +\beta t}$. Hence $O(t+h)/O(t)=e^{h\beta }$, independently of $t$. By invariance of maximum likelihood estimation, estimate this odds ratio by $e^{h\hat{\beta }}$. A Wald confidence interval for $\beta$ is $\hat{\beta }\pm 1.96\mathrm{se}(\hat{\beta })$; exponentiating its endpoints after multiplication by $h>0$ gives the interval for the multiplier. Using the printed coefficient and standard error $0.01543$ yieldsThus the estimated odds rise by about $9.8\%$ per year, or by a factor about $2.55$ per decade. These are changes in odds, not fixed multiplicative changes in probabilities.

The second fit keeps the same independent binomial distributions but permits a quadratic linear predictor,In the sequential analysis, each exposure term adds one coefficient. With the first model's deviances, all nine missing entries are recovered in this completed table:The final reduction is $6.0508-3.282\approx 2.769$, with rounding to the precision of the printed output. Testing $\gamma =0$ therefore gives $p\approx 0.096$ on one degree of freedom. At the $5\%$ significance level I would retain the simpler linear-exposure model: it already fits adequately and the extra curvature is not clearly established. The quadratic fit does improve the likelihood somewhat; its Akaike information criterion is about $0.77$ smaller, since $\Delta \mathrm{AIC}\approx -2.769+2$. Thus different reasonable selection criteria may weakly prefer different models, but these data do not give strong evidence that the additional term is necessary.

The first fit is grouped-binomial logistic regression: independently, $Y_i\sim \mathrm{Bin}(N_i,{\pi }_i)$, with the observed group sizes $N_i$ andThe reference group is under 50 with a nonmalignant initial tumour. The weights specify binomial numbers of trials, not six equally weighted individual observations. There is no age-by-malignancy interaction. Each age coefficient compares that age group with the reference age at fixed malignancy; $\exp {\beta }_3$ is the common malignancy odds ratio.

The maximum-likelihood estimates are obtained by maximizing the binomial likelihood. The reported standard errors come from the inverse fitted Fisher information. Each displayed Wald test tests a coefficient equal to zero, using $z=\hat{\beta }/\mathrm{se}(\hat{\beta })$ and the approximate standard normal distribution; its two-sided probability is $2\Phi (-|z|)$. At the 5% level the middle age coefficient and the malignancy coefficient are significant, with negative signs; the oldest age coefficient is not, although its 9.17% probability is weaker evidence in the same direction. The oldest group has only 19 observations and its larger standard error does not establish absence of an age effect. The intercept tests reference log-odds zero, equivalently reference survival probability $1/2$, rather than an age contrast. The reference fitted probability is about $0.888$; the fitted malignancy odds ratio is about $0.481$.

The binomial deviance is twice the saturated-minus-fitted log-likelihood. The null fit has one parameter, hence $6-1=5$ residual degrees of freedom; the first fit has four parameters, hence $6-4=2$. The improvement $12.65585-0.49409=12.16176$ is an approximate likelihood-ratio test on three degrees of freedom, with probability about $0.0069$. The residual deviance $0.49409$, compared with ${\chi }_2^2$, gives about $0.781$, so there is no evidence of lack of fit. These are asymptotic comparisons, especially approximate for the small oldest strata. The Akaike information criterion is $-2\ell +2p$, here with $p=4$.

Pooling the two older age groups gives a grouped-binomial logistic regression with a common coefficient for age at least 50 and three parameters in total. The similar older-age estimates and the sparse oldest group motivate examining this restriction, but the significance of either coefficient separately is not a test of their equality. The actual analysis of deviance for pooling compares the nested fits: the deviance increases by only $0.784-0.49409\simeq 0.28991$ for one lost parameter, giving a ${\chi }_1^2$ probability about $0.590$. The pooled model also has smaller Akaike information criterion, $28.723$ rather than $30.433$. Prefer the pooled model for its simpler adequate fit. Its common age and malignancy coefficients are negative and significant at 5% in the reported Wald tests.

There is a literal parenthesis error in the printed final R fragment: the first sum is not closed before subtracting the second sum. As printed, the expression is incomplete. Closing that sum gives twice the saturated-minus-fitted log-likelihood, including the binomial constants which cancel. Its intended final value is therefore the second fit's binomial deviance, $\boxed{0.784\text{ approximately}}$.

The cumulative-response logistic model uses $C_{it}={\sum }_{s<t}{Y}_{is}$ instead of the two individual lag predictors. It and the two-lag history-dependent logistic regression are nonnested: the entire accumulated history generally cannot be represented using only the two recent outcomes. Their binomial deviance difference therefore has no ordinary nested chi-squared distribution calibration.

Use the Akaike information criterion, which up to the same saturated-model constant equals $D+2k$ for these Bernoulli distribution conditional likelihoods. The two-lag fit has six coefficients, while the cumulative fit has five:Thus prefer the cumulative-history model, whose Akaike information criterion is lower by 35 despite its smaller number of statistical parameters. This is a model-selection comparison of conditional histories, not a nested likelihood-ratio test.

The second logistic regression adds ${\gamma }_2{Y}_{i,t-2}$ while retaining the first lag and baseline covariates. The models are nested under $H_0:{\gamma }_2=0$. Their likelihood-ratio test statistic is the reduction in binomial deviance,Under the null and regular large-sample conditions for the correctly specified conditional likelihood, this has an approximate chi-squared distribution with one statistical degree of freedom. Since $9.4>3.841$, reject at 5%; the approximate p-value is $0.0022$. Prefer the two-lag model to the one-lag model. Its additional lag captures statistically useful information in the history.

The preferred cumulative-response logistic model estimatesIts logit link describes conditional smoking-free probability given baseline predictors and prior successful weeks. Holding the other predictors fixed, males have $e^{-0.364092}=0.695$ times the female success odds; the reported p-value is $0.0152$, and the approximate 95% odds ratio confidence interval is $(0.518,0.932)$. Age has estimated odds ratio $e^{0.001224}=1.0012$ per additional year, with p-value $0.918$, giving little evidence for an age association in this fit.

The combined treatment has conditional success odds ratio $e^{0.327118}=1.387$ versus the reference treatment, with approximate 95% confidence interval $(1.035,1.859)$ and p-value $0.0285$. The fitted conditional odds are about 39% higher for the combined treatment. The trial randomization supports treatment comparisons, but conditioning on accumulated post-treatment outcomes means this coefficient is not directly the marginal total treatment effect.

Each previous successful week multiplies current success odds by $e^{0.399296}=1.491$, with approximate 95% confidence interval $(1.329,1.673)$ and very small p-value $1.06\times 10^{-11}$. This is strong fitted persistence. It can reflect true state dependence, unobserved heterogeneity, or an omitted calendar-time trend; this fit alone cannot distinguish them. The baseline intercept implies success probability ${\mathrm{logit}}^{-1}(-1.417630)\simeq 0.195$ for a reference-treatment female aged zero with no previous success. That age is outside the study's useful interpretation range, so the intercept chiefly anchors the regression. The Bernoulli distribution dispersion parameter is fixed at one, and the residual binomial deviance is 1122 on 995 statistical degrees of freedom. With individual binary outcomes, comparing that binomial deviance mechanically to a chi-squared distribution is not a reliable general goodness-of-fit test.

For independent patients, the Bernoulli logistic-regression model iswhere $d_i=1_{\{{\mathrm{frail}}_i=1\}}$ with nonfrail patients as the reference level in a regression factor. The fitted linear predictor isThe null model fits one intercept, so its 99 residual degrees of freedom imply 100 patients; the full model's $100-3=97$ confirms this. Unit weights correspond to individual binary responses, not grouped binomial counts.

The Akaike information criterion estimates expected relative out-of-sample discrepancy, using the maximized log-likelihood plus a complexity penalty:Here $k=3$, since the Bernoulli dispersion is fixed at one. The saturated model puts probability one on each observed binary outcome and has log likelihood zero, so the binomial deviance is $D=-2\ell (\hat{\beta })$. ConsequentlyThere is no estimated error-variance parameter to count. The printed name “Aikake” should be read as Akaike.

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.

There are $120$ observations, since the intercept-only model has $119$ residual degrees of freedom. Adding price introduces one coefficient and decreases the binomial deviance by $148.263-101.920=46.343$. Adding a three-level factor introduces two coefficients and decreases it by $101.920-97.571=4.349$. Thus the missing entries areThe likelihood-ratio test comparing the smaller and full logistic regression tests $H_0:{\beta }_B={\beta }_C=0$ against at least one nonzero layout contrast. Its statistic is $2({\hat{\ell }}_1-{\hat{\ell }}_2)=4.349$. Under ordinary interior-parameter, full-rank and finite-estimate regularity conditions, Wilks theorem gives an asymptotic chi-squared distribution with two degrees of freedom. The reported p-value $0.1136$ exceeds $0.05$, so at that level retain the simpler price-only model.

The table is an analysis of deviance for nested generalized linear models: price is entered before layout. Thus its price row tests the price-only model against an intercept, whereas the layout row tests layout after allowing for price. The latter is the relevant comparison of model1 and model2. Failing to reject a layout effect is not proof that layouts have identical effects; it means these data do not justify the extra two parameters under this test.

For observed proportions $y_i$ and fitted proportions ${\hat{p}}_i$, the binomial deviance isA second-order Taylor expansion around $y_i={\hat{p}}_i$ givesthe generalized Pearson chi-squared statistic. Under the fitted binomial model this is approximately ${\chi }_{4-3}^2={\chi }_1^2$. The observed deviance is $1.6102$, whose upper-tail probability is about $0.20$. Since this exceeds $0.05$, there is no significant evidence of overdispersion.

A saturated model has enough independently varying mean parameters to match every observed response or response cell. In individual Bernoulli distribution data its fitted probabilities are $p_i=y_i$, and its maximum log-likelihood is zero under $0\log 0=0$. It defines the reference in a binomial deviance.