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For a continuous predictor $x$ and a categorical factor, an additive Poisson regression with log link has $\log {\mu }_j(x)=a_j+bx$. The log-mean curves are parallel and ${\mu }_j(x)/{\mu }_k(x)=e^{a_j-a_k}$ is constant in $x$. Adding a predictor-by-factor interaction gives $\log {\mu }_j(x)=a_j+b_{j}x$, so this ratio becomes $e^{a_j-a_k+(b_j-b_k)x}$. With $J$ factor levels, the common-slope restriction has $J-1$ independent constraints. A likelihood-ratio test compares the two deviances with an asymptotic chi-squared distribution on $J-1$ degrees of freedom. Evidence against common slopes establishes varying effects on the log scale; it does not by itself determine their signs or the ordering of the mean curves.

A deviance residual is the signed square root of an observation's contribution $d_i$ to fitted-model deviance. For a Poisson regression, $d_i=2[y_i\log (y_i/{\hat{\mu }}_i)-(y_i-{\hat{\mu }}_i)]$, with $0\log 0=0$. Its squared sum is the Poisson deviance. Leverage and estimated dispersion can further standardize these residuals; a normal quantile-quantile plot is a diagnostic approximation, not a requirement that count errors have a normal distribution.

Maximum likelihood estimation is a method of statistical estimation: within a specified statistical model, choose the statistical parameter values that maximize the likelihood function of the observed data. For a parameter set $\Theta$ and observations $x$, its exact requirement isThe equality requires an attained, finite maximum; a supremum approached only at an excluded boundary is not enough. The resulting rule is a maximum-likelihood estimator. The likelihood function uses a probability mass function for discrete data and a probability density function for continuous data. See MIT 18.05, Reading 10b, sections 3 and 3.2.

The direction of the comparison is the interesting part: hold the observations fixed and vary the statistical parameter. Ten independent Bernoulli trials with eight successes and two failures give the maximum-likelihood estimate $\hat{p}=0.8$, as derived in log-likelihood. This chooses the member of the statistical model that assigns the greatest mass to that outcome. It does not assert an $80\%$ posterior probability that this parameter value is true. Bayesian statistics combines the same likelihood function with a prior distribution to obtain a posterior distribution; maximum a posteriori estimation maximizes that posterior mass or density instead.

The method turns many fitting tasks into one maximization problem. In the normal distribution model with known variance, it recovers the sample mean and ordinary least squares, while binomial regression and Poisson regression use different sampling assumptions to build their own likelihood functions. The log-likelihood turns independent contributions into a sum, exposing score equations and observed information. A peak's location gives an estimate; its shape records how much the observed data distinguish nearby parameter values.

Existence, uniqueness, bias, and statistical consistency are separate questions. Normal mean and variance maximum-likelihood estimators include a variance estimate with bias, likelihood supremum at an excluded boundary gives nonexistence, and the Neyman-Scott incidental parameter problem gives inconsistency. The usual asymptotic normality of a maximum likelihood estimator requires assumptions: a consistent estimator at an interior true parameter in a fixed finite-dimensional model with identifiability, sufficient smoothness and control of log-likelihood derivatives, and finite nonsingular Fisher information. A common parameter-independent support of a probability distribution is part of a standard sufficient framework; see Rigollet, MIT 18.650, Maximum Likelihood Estimation, slides 22-23. This is a description of the regular setting, not a complete theorem's hypotheses. The exact endpoint limit for a shifted exponential distribution shows why changing support can produce a different rate and a nonnormal limit. These properties do not follow just from maximizing a function.

In his 1922 paper in the Royal Society's Philosophical Transactions A, R. A. Fisher explicitly distinguished likelihood from probability and described choosing the parameter values with greatest likelihood. Section 6 also discusses comparing likelihoods after giving the maximum value one, the convention used in the figures here. See Fisher, On the Mathematical Foundations of Theoretical Statistics, section 6, pp. 326-328, DOI 10.1098/rsta.1922.0009.

The response is a device count, while the predictor is a skill score. A simple linear regression models $\mathbb{E}(Y\mid x)={\beta }_0+{\beta }_{1}x$, where $Y$ is the device count and $x$ the skill score. Ordinary least squares chooses the intercept and regression slope to minimize the residual sum of squares. The regression slope describes a conditional association, not the causal effect of training or acquiring a device.

The descriptive statistics put the device count between $0$ and $8$, with sample mean about $4.494$ and sample standard deviation about $2.128$ among $2053$ records. Skill scores range from $0$ to $24$, with sample mean about $13.546$ and sample standard deviation about $5.989$ among $2051$ records. Two missing skill values account for the stated two deleted records; because device counts have no missing values, the regression complete-case sample has $2051$ observations. Estimating two regression coefficients leaves $2051-2=2049$ residual statistical degrees of freedom, matching that part of the output. The units of the two standard deviations differ, so comparing $5.989$ with $2.128$ does not establish that skill is more variable in a substantive sense.

If the coefficient estimates are trusted, the fitted values follow $\hat{y}=0.392122+0.209240x$. The intercept predicts about $0.392$ devices at a skill score of zero; zero is within the stated score range, but precision and the number of observations near zero are not shown. A five-point increase in score corresponds to about $1.0462$ additional fitted devices and a ten-point increase to about $2.0924$. At a score of ten the fitted value is about $2.4845$, and at 24 it is about $5.4139$. Expected counts may be fractional even though individual counts are integers. These are conditional interpretations of the coefficient fields, not reconstructed raw observations or validated individual predictions.

A coefficient's standard error estimates its sampling uncertainty, whereas the residual standard error describes unexplained response variation. For the intercept, $0.392122/0.114570\approx 3.42255$, agreeing with the printed $t=3.423$ and rounded $p=0.000633$. Conditional on the model assumptions, this would reject a zero intercept at conventional significance levels. The intercept test is not a test of whether skills predict devices. For the slope, the printed $p=0.013672$ would reject $H_0:{\beta }_1=0$ at 5% but not 1%; it does not assign a probability to the null hypothesis or measure the usefulness of prediction. The significance stars simply encode thresholds and add no independent evidence.^[ref]

The slope row fails its own arithmetic check:At $2049$ statistical degrees of freedom the ratio gives a two-sided p-value about $0.00845$, while the printed $t=2.882$ gives about $0.003993$; neither is $0.013672$. Thus the choice of which source field is trusted changes the apparent strength of evidence, including the 1% decision. A 95% confidence interval built conditionally from the printed estimate and standard error would be about $[0.0536,0.3649]$, but it is not a verified interval from a coherent fitted model. In fact, using the printed residual standard error $1.793$, the complete predictor sample size and predictor standard deviation gives a slope standard error near $0.006612$, not $0.079377$.

The printed coefficient of determination $R^2=0.2893$ would mean that the fitted line explains $28.93\%$ of the complete-case sample's centred device-count sum of squares relative to an intercept-only fit; $71.07\%$ remains unexplained by that comparison. It is not the percentage of devices caused by skill, the fraction of correctly predicted households, or a measure of future predictive accuracy. The adjusted coefficient of determination $0.289$ is consistent with that $R^2$ and the given sample size, allowing rounding. The small adjustment reflects fitting one slope among many cases; it does not validate the model.

For a one-predictor ordinary least squares fit with an intercept and the usual common-variance standard errors, the simple-regression output identities requireHere the printed slope t gives $F\approx 8.305924$, the printed F field says $83.2$, and the printed $R^2$ instead implies $F\approx 834.073$. These cannot all hold. Moreover, $F=83.2$ on $1$ and $2049$ statistical degrees of freedom has p-value about $1.72\times 10^{-19}$, not $0.001978$. Reading the printed F p-value alone would give a 1% significant test of the zero-slope model, but its reported calibration is incompatible with its statistic. No significance label resolves these identities.

There is also a contradiction that does not depend on tail probabilities. Ordinary least squares with an intercept makes the arithmetic mean of fitted values equal the arithmetic mean of observed responses on the analysed records. The printed coefficients and predictor mean give $0.392122+0.209240(13.546)\approx 3.22649$. Deleting two device counts anywhere in the printed range $[0,8]$, and allowing the reported overall mean to be rounded, would leave the complete-case device mean between approximately $4.49008$ and $4.49888$. The mean bounds after deleting bounded observations therefore rule out reconciling $3.22649$ with the response summary merely by citing those two missing predictor values. This is not the mistaken assumption that full-sample and complete-case sample means must always be identical.

If its own field were trusted, residual standard error $1.793$ estimates the unexplained response standard deviation, in device-count units. It is neither the standard error of the slope nor a guarantee that every prediction is accurate to $1.793$ devices. The residual five-number summary has extremes near $-5.206$ and $5.086$ and median near $-0.060$, with central quartiles near $-1.372$ and $1.252$. The roughly central median does not establish normally distributed errors, homoscedasticity, independence or a linear conditional expectation. A residual-versus-fitted plot could check curvature and spread; a normal probability plot could assess an error-shape approximation. Neither diagnostic can be recovered from five residual quantiles alone. Individual prediction intervals include response variation as well as coefficient uncertainty and are wider than confidence intervals for the conditional expectation.

For conventional finite-sample t and F calibration, a normal linear model assumes independent mean-zero normal errors of common variance with an appropriate predictor specification. Mean-zero errors conditional on skill and adequate sampling remain essential even when large-sample approximations relax normality. A bounded count response can have nonconstant conditional variance, so residual diagnostics and a scientifically suitable response model matter; count data do not become normally distributed just because the sample is large. Poisson regression is one possible alternative to investigate, but its own conditional mean–variance assumptions and the bounded device range require assessment, not automatic substitution. Complete-case selection can also bias the fitted association if it depends on the response in a relevant way. The only safe summary is conditional: the printed slope is positive, but the numerical effect, its standard error, t/F evidence and explanatory fit require a corrected, coherent output before a single definitive regression claim is made.

The Poisson generalized linear mixed model assumes conditionally Poisson distributed counts with a log link. The fitted variance components are both zero, so this fit lies on a boundary and reduces to ordinary Poisson regression. It estimatesThus the fitted annual increase is about 6.6%, with approximate 95% confidence interval for the rate ratio $(e^{0.001},e^{0.128})\simeq (1.001,1.137)$. The age evidence is much weaker than in the Gaussian fit: the quoted p-value is $0.047$. Zero estimated group variances do not prove that real bird and year differences are absent.

The Poisson distribution forces conditional variance to equal the conditional expected value. Here fitted means are around two to three hatchlings, whereas the Gaussian model estimated a residual variance of only $0.188$. This suggests checking underdispersion and the bounded nature of hatching counts, rather than assuming a Poisson response solves the diagnostic problems. Inspect simulated count residuals and predictive distributions; if the number of eggs is observed, fit a binomial mixed model and check that model's dispersion too.

The displayed Gaussian and Poisson Akaike information criterion values should not simply be ranked as if they used the same likelihood convention: one evaluates a continuous density and the other a probability mass. Compare properly specified count models on the same observed response scale, with compatible likelihoods and predictive checks. A boundary fit, doubtful dispersion assumptions, and possible nonlinear age effects all warrant further work before a strong biological conclusion.

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 printed formula expands to the main effects of school type, acceptance and college, together with school-type-by-college and acceptance-by-college interactions. It contains neither a school-type-by-acceptance interaction nor the corresponding three-way interaction. Its Poisson regression model is thereforeFor every college the mean table has odds ratioThus it tests conditional independence of school type and acceptance given college, while permitting both margins to vary freely across colleges. It does not merely assert a common, possibly nonzero association.

The maximum likelihood estimation score equations match every school-type-by-college and acceptance-by-college margin. Solving them gives the stratified two-by-two conditional independence model's fitted countsIndeed these counts factor by row and column and have exactly the observed margins. For nondegenerate tables there are three free mean parameters per college: one total and two independent margin proportions. There are $100$ cells and $75$ parameters, hence 25 residual degrees of freedom. The residual deviance iswith zero-count contributions interpreted as zero. The Poisson linear terms cancel because the fitted totals equal the observed totals. Under the null and suitable large-count regularity, compare $G^2$ with ${\chi }_{25}^2$. Sparse cells, structural zeros or degenerate margins require a different calibration, such as an appropriate conditional test or simulated reference distribution.

A large residual deviance is evidence of school-type/acceptance association in at least some colleges. A small one indicates compatibility with conditional independence, not proof of identical admission processes or absence of selection on qualifications. Residuals and individual college contrasts locate departures. Adding the school-type-by-acceptance term gives a common odds ratio, one extra parameter and 24 residual degrees of freedom. Adding the full three-way term permits each college's association and saturates the tables. The first comparison tests a common association; the second checks whether a common association is adequate. No frequencies are supplied here to evaluate these tests numerically.

Use $e_{ij}$ for the expected accident count, not for the accident rate. With a design row $z_{ij}$ consisting of an intercept, seven non-reference site indicators and the after-treatment indicator,Here $\log p_{ij}$ is a known generalized linear model offset, with coefficient fixed at one. Differentiating the Poisson regression log-likelihood gives the score function equationsThese nine equations estimate the intercept, seven site contrasts and one treatment contrast, with the stated reference-level constraints.

The Fisher information is $Z^{\mathsf{T}}WZ$, where $W=\mathrm{diag}(e_{ij})$. A Fisher scoring update adds $(Z^{\mathsf{T}}WZ{)}^{-1}{Z}^{\mathsf{T}}(y-e)$ to the current coefficient vector. In iteratively reweighted least squares, the working response is $z_{ij}^{\ast }=\log e_{ij}+(y_{ij}-e_{ij})/e_{ij}$; regress $z_{ij}^{\ast }-\log p_{ij}$ on the design with weights $e_{ij}$. Thus the reported four iterations solve the likelihood equations while accounting for unequal exposure periods.

The common adjusted rate ratio is obtained by exponentiating the after coefficient:This estimates a $54.2\%$ lower accident rate after the intervention, conditional on the additive site model. The standard error $0.2751810$ gives normal Wald statistic $-2.8369$, two-sided $p\simeq 0.0046$, and an approximate $95\%$ confidence interval for the rate ratioThe label t value is again a normal-reference Wald ratio under fixed Poisson dispersion, rather than an exact Student statistic.

The intercept gives the before rate at reference site one: $e^{0.2707792}\simeq 1.311$ accidents per year. The seven site coefficients are log baseline-rate ratios relative to that site. In site order two through eight, their exponentials are about $0.615,2.767,1.711,0.769,1.797,0.615,1.221$. The third site has the largest estimated rate and a nominal Wald ratio about $3.118$ against the reference. These are adjusted baseline comparisons, not differences in absolute accident counts. The overall unadjusted before/after rate ratio is $(15/18)/(114/68)\simeq 0.497$, and it differs from $0.458$ because site risks and their relative exposure weights differ.

An independent way to solve the fit is the profile likelihood for a common Poisson rate ratio with unequal exposures. Put $t_i=y_{i1}+y_{i2}$ and ${\lambda }_i=\exp (\mu +{\alpha }_i)$. For fixed $r$, the site score gives ${\hat{\lambda }}_i=t_i/(p_{i1}+r{p}_{i2})$. The remaining equation isIts positive root reproduces the reported treatment estimate. Comparing the same site model with $r=1$ to the fitted model gives a deviance improvement about $9.750$ on one degree of freedom, with nominal likelihood-ratio $p\simeq 0.0018$.

There are $16$ count observations and $9$ fitted mean parameters, leaving $7$ residual degrees of freedom. The null deviance $132.9485$ on $15$ degrees of freedom is for the intercept-only rate model, still including the exposure offset. The residual Poisson deviance is $16.27524$ on $7$ degrees of freedom. A chi-squared comparison gives approximately $p=0.023$, suggesting potential lack of fit of a single common treatment ratio; the extreme deviance residuals near $-2.03$ and $2.14$ also deserve inspection. Several after-cell expected counts are below one or near two, so this absolute-fit calibration is only approximate. Site-specific treatment effects, overdispersion or temporal dependence are possible explanations to investigate, with a parametric bootstrap available to calibrate the sparse-count fit statistic. A before/after association alone does not isolate a causal effect from secular changes or regression to the mean at selected sites.

Use $\alpha$ for the intercept, to avoid confusing it with the observation means. Let $e_i=\exp (\hat{\alpha }+{\hat{\beta }}^T{x}_i)$ be the fitted Poisson regression mean. Apart from constants, its log-likelihood isAt a finite interior maximum-likelihood estimate, the intercept score equation gives the Poisson regression margin-matching score equations:The statistical saturated model fits each mean to $y_i$, taking the boundary mean zero if $y_i=0$. Subtracting fitted from saturated log-likelihood gives the Poisson devianceThe intercept balance cancels the sum of linear terms. ConsequentlyThe unsimplified formula makes nonnegativity transparent; individual terms in the simplified formula need not each be positive.

For the deviance goodness-of-fit test, if the fitted design has rank $r$, compare a suitably calibrated $D$ with ${\chi }_{n-r}^2$, rejecting for large values. This approximation applies in a regular large-count regime, such as a fixed set of cells with increasing exposure and positive fitted proportions. It is not automatically accurate for many sparse Poisson observations. Check residual patterns and overdispersion; where the approximation is doubtful, simulate independent Poisson random variables with fitted means, refit each sample, and use a parametric bootstrap distribution of the same statistic. A large p-value means no detected lack of fit, not proof of the model. If a finite interior estimate fails to exist, the score proof needs a boundary-limit interpretation; for example all-zero data give a supremum with every fitted mean tending to zero and limiting deviance zero.

Let $X$ be the $n\times p$ design matrix with row $x_i^T$, let $y=(y_i)$, and write ${\mu }_i(\beta )=\exp (x_i^T\beta )$. Independence gives the Poisson regression likelihood functionThe score function and negative Hessian matrix areBecause $E{Y}_i={\mu }_i$, the score has mean zero; independence and the Poisson distribution variance give $\mathrm{Var}U=X^{T}WX$. Thus the observed and expected Fisher information matrices are the same here.

Assume $X$ has full column matrix rank $p$. For a nonzero vector $a$, $a^{T}Ia={\sum }_i{\mu }_i(x_i^{T}a{)}^2>0$. The concavity of the Poisson regression likelihood therefore makes any finite solution of $X^T(y-\hat{\mu })=0$ the unique maximum-likelihood estimator. If the design is rank deficient, parameters differing by a vector in $\ker X$ give the same distribution, so only identifiable combinations can be estimated. Full rank alone does not ensure a finite maximizer: for an intercept-only model with all observations zero, $\ell =-n{e}^{\beta }$ approaches its supremum as $\beta \to -\infty$. More generally, directions that decrease only fitted means attached to zero counts can produce boundary fits. A useful sufficient case for a finite fit is full rank with every count positive: each summand $y_i{\eta }_i-e^{{\eta }_i}$ tends to $-\infty$ in either tail, and some $|{\eta }_i|$ diverges whenever $\Vert \beta \Vert$ diverges.

For an interior fit, Newton's method and Fisher scoring give the same update:This is also iteratively reweighted least squares. Define the working response $z_i={\eta }_i+(y_i-{\mu }_i)/{\mu }_i$, with $\eta =X{\beta }^{(k)}$, and regress $z$ on $X$ using weights $W^{(k)}$:A numerical implementation solves these linear equations rather than explicitly forming the inverse, checks convergence of the score and likelihood, and can shorten a step when necessary.

Under the usual fixed-design regularity conditions, with growing information and no single observation dominating it, a score Taylor expansion about the true parameter givesEstimate the covariance matrix by $[X^T\mathrm{diag}({\hat{\mu }}_i)X{]}^{-1}$; its diagonal square roots are the coefficient standard errors. This justifies large-sample Wald confidence intervals and Wald tests for specified coefficients or contrasts. Finite-sample unbiasedness is not being asserted. A likelihood-ratio test compares nested fits through twice their maximized log-likelihood difference. The Poisson deviance against the saturated fit iswith $0\log 0=0$. Residual diagnostics and a check of overdispersion assess whether the assumed Poisson variance is adequate; the variance correction required when it is not is derived in the next part.

The Wald test of $H_0:\beta =0$ useswith an approximate standard normal distribution under the null. Its two-sided p-value is $0.403$, and the approximate $95\%$ Wald confidence interval isThere is no convincing evidence of a covariate effect, and the slope is very imprecisely estimated. The model can fit acceptably while its slope is statistically indistinguishable from zero. In the Poisson regression, a unit covariate increase multiplies the mean by $e^{\beta }$; exponentiating the two interval limits shows how poorly this multiplier is determined on the stated scale. One might compare with an intercept-only model using a likelihood-ratio test, but its numerical deviance difference is not supplied. Failure to reject does not establish no effect.

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.

The Poisson distribution mass function isIt is an exponential dispersion family of order one withIndeed, $b^{\prime }(\theta )=e^{\theta }=E(Y)$ and $\varphi b^{\prime \prime }(\theta )=e^{\theta }=\mathrm{Var}(Y)$. A generalized linear model specifies independent random variables from a response exponential family, together with a linear predictor and a link function relating it to the response mean. The Poisson canonical link is $g(\mu )=\log \mu$.

For the insurance counts, let class $1$ and merit $0$ be the reference levels in a regression factor. The fitted Poisson regression assumesThe insured exposure $n_{ij}$ is known; its logarithm is an offset, not a coefficient to estimate. The model has eight free mean parameters and fixes the dispersion parameter at one. It assumes additive class and merit effects on the log-rate scale, hence multiplicative effects on the rate scale without a class-by-merit interaction.

Apart from constants independent of the parameters, the log-likelihood isDifferentiating gives the Poisson regression margin-matching score equationsThus the fitted totals equal the observed totals in each class and merit category, including the reference categories by subtraction. In design matrix notation these equations are $Z^T(Y-\hat{\mu })=0$. Fisher scoring solves them iteratively; the reported three iterations describe the numerical fit, not an additional statistical result.

The regression intercept gives the baseline fitted claim rate $e^{-2.0357359}=0.130584$ per insured car year. The merit rate ratios, relative to merit $0$ at the same class, areThe class rate ratios, relative to class $1$ at the same merit, areEach predicted cell rate is the baseline rate times its class and merit multipliers. The output's coefficient-to-standard error ratios are asymptotic normal Wald test statistics under the fitted Poisson regression, despite the software label “t value”; there is no estimated Gaussian residual scale in this model. All nonreference effects have very large absolute ratios.

The null Poisson deviance is $33854.16$ on nineteen statistical degrees of freedom; adding the seven factor coefficients reduces it by $33274.6437$. But the final Poisson deviance is still $579.5163$ on twelve statistical degrees of freedom, and the deviance residuals include values near $-10.79$ and $11.63$. These are far beyond what a well-fitting unit-dispersion model would normally produce. The additive Poisson model fits substantially better than a common rate, but remains grossly inadequate. A class-by-merit interaction, overdispersion, or statistical dependence between claims could contribute; the summary alone does not distinguish these explanations. Its very small nominal standard errors and formal Wald tests rely on the rejected model, so improved mean structure and error assumptions are needed before treating them as reliable uncertainty assessments.

The R command fits a Poisson regression with the default log link:with independent annual counts conditional on the supplied temperature index. The fitted mean is $\hat{\lambda }(T)=\exp (2.26061+0.48870T)$. The intercept gives about $9.59$ expected named storms at index zero. A one-unit increase in the index multiplies the expected count by $e^{0.48870}\simeq 1.63$; a $0.1$ increase multiplies it by about $1.050$.

The reported standard errors estimate the sampling uncertainty of the coefficients. Each displayed $z$ is the estimate divided by its standard error, compared with a standard normal distribution under a zero-coefficient null hypothesis. The slope has $z=2.879$ and two-sided $p=0.00399$, evidence of positive association within this model. The intercept test is against mean one at index zero and is not itself a test of a climatic mechanism. Association does not establish causation.

There are $61$ annual observations and two fitted coefficients, giving $59$ residual degrees of freedom. The residual deviance iswith $0\log 0=0$. It is not large relative to $59$ and gives no obvious overdispersion signal, although temporal dependence and omitted covariates still require substantive assessment.

For 2005 insert the index into the fitted mean:This is an expected count, approximately fourteen storms, rather than a guarantee of an integer outcome. In R one would use predict(Mod, newdata=data.frame(Temp=0.743), type="response"). A predictive distribution must include Poisson variation and, if desired, coefficient-estimation uncertainty; the latter needs the coefficient covariance matrix, not just the two printed standard errors.

This is a family-level cluster-randomized trial, but there are only about 1.1 contributing children per family. Most families therefore contribute a single eligible child, leaving little opportunity for within-family intraclass correlation to add much uncertainty. The equal-size exchangeable approximation has design effect $1+(m-1)\rho$, with cluster size $m$ and correlation $\rho$; when $m$ is close to one, this is close to one. A multilevel Poisson regression can consequently give nearly the same standard error as the independent-child working analysis.

Agreement is compatible with limited within-family dependence and little change from covariate adjustment; it does not prove all dependence absent. Mean family size alone does not determine the exact design effect for highly unequal cluster sizes.

Larger differences would be expected with larger families and stronger within-family correlation, especially with unequal family sizes. Pronounced between-family heterogeneity, recurrent attendances concentrated in particular children, or important imbalanced covariates can also create overdispersion or change the adjusted comparison.

The first command fits independent Poisson random variables with means ${\mu }_j$ and a log link. For concentration $x$ and strain indicator $s$, its fitted linear predictor isThe interaction allows different concentration slopes in the two strains. At $x=0$ the second strain's mean is multiplied by $e^{0.33667}$; its slope is $-1.47253-0.12534$. These are multiplicative changes in counts, not changes in probabilities.

The maximum-likelihood estimates use the Poisson regression likelihood. With design matrix $X$, estimated information $X^T\mathrm{diag}({\hat{\mu }}_j)X$ gives the inverse covariance matrix; diagonal square roots give the standard errors. The printed Wald tests use approximately standard-normal coefficient-to-error ratios. Fuel and strain effects are strong, whereas the interaction has $p=0.182$, suggesting a common slope.

$H_2$ removes just the interaction; $H_3$ also removes strain. Their parameter counts are four, three and two. Under the usual large-sample likelihood-ratio test, $D_2-D_1=1.78089$ is compared with ${\chi }_1^2$ and is below $3.841459$, so the interaction is not required at 5%. By contrast, $D_3-D_2=32.61857$ is far above that threshold, so the strain main effect should remain. Comparing $H_3$ directly with $H_1$ uses $D_3-D_1$ and two degrees of freedom.

For an approximate absolute goodness-of-fit test, compare $D_1=84.59557$ with ${\chi }_{70-4}^2$; its upper-tail probability is about $0.061$. For $H_2$, $D_2=86.37646$ on $67$ degrees of freedom has $p\approx 0.056$. The additive model $H_2$ is the preferred parsimonious fit at 5%, subject to the independent Poisson distribution and large-sample assumptions. The printed nine rows are only a subset; the supplied fitted output refers to all 70 observations.

The second analysis is an exposure-adjusted Poisson regression for case counts:Cell counts are independent, and their conditional variance equals their conditional mean. The population term is an offset with coefficient fixed at one. Thus the fitted case rate is ${\lambda }_{gj}/N_{gj}=\exp ({\alpha }_P+a_{P,j}+{\gamma }_{P}g)$; treating $\log N_{gj}$ as a free covariate would fit a different model. Age multipliers and the adjusted city rate ratio are exponentials of the corresponding regression coefficients.

The fitted model again has nine parameters and six residual degrees of freedom. Sequential additions of age and city reduce the Poisson deviance by $2095.56$ on seven degrees of freedom and $226.52$ on one, respectively. The residual deviance is $5.21$, close to the six residual degrees of freedom. The city likelihood-ratio test is overwhelmingly significant, as in the binomial analysis. The table provides no Poisson coefficient estimates, so an exact Poisson city rate ratio cannot be read from it.

Model the count relative to exposure, for exampleHere $E_i$ is a measured exposure such as patient-days, $x_i$ contains hospital predictors, and $b_i$ is a random effect. This Poisson regression separates hospital size from rate variation and supplies partial pooling. Use proper, scale-appropriate priors on coefficients and variation; add regional or temporal effects if the data demand them. A gamma mixing law with an estimated shape is another way to allow more flexible overdispersion. Compare the fitted model's counts and dispersion through posterior predictive checks.

The mean ratio is $R=(\lambda e^{\beta })/\lambda =e^{\beta }$, so its logarithm is exactly $\beta$; the baseline coefficient $\alpha$ cancels. Under the stated asymptotic normal approximation, the Wald confidence interval for $\beta$ is $0.69\pm 1.96(0.15)=(0.396,0.984)$. Exponentiation is monotone and therefore preserves the interval's coverage event. HenceThe standard error of $\alpha$ is not needed for this confidence interval: the Poisson regression already supplies the standard error of the contrast $\beta$. It would be incorrect to combine the two reported standard errors as if the mean ratio depended on both coefficients independently.

The solid Poisson regression profile and dashed negative binomial regression profile peak at essentially the same $\beta \simeq 0.69$. The dashed profile is slightly wider and has heavier shoulders: it gives larger relative likelihood to values farther from the common maximizer. Thus the estimated mean ratio is stable, while the negative binomial model expresses slightly greater uncertainty.

The Poisson distribution constrains variance to equal the mean. A common negative binomial parametrization instead has $\mathrm{Var}(Y)=\mu +{\mu }^2/k$, with additional shape $k>0$; it permits overdispersion due, for example, to unexplained between-physician heterogeneity. Profiling this extra parameter lets the data support greater count variability and usually reduces information about the mean contrast. In this two-group model, the fitted group means remain the sample means for fixed $k$, which explains the nearly identical maximizers. The small separation of the profiles indicates a modest effect here, not overwhelming evidence for strong overdispersion. The comparison concerns the original PDF figure; no figure is reproduced.