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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.

Let $X$ have columns $1,t$, and let $Z$ have one column per incubator, with entry $t_i$ in the column for observation $i$'s incubator and zeros elsewhere. Integrating the Gaussian random effects givesConsequently its marginal log-likelihood isThe likelihood-ratio test of $H_0:{\tau }^2=0$ against ${\tau }^2>0$ usesFor each covariance, the optimizing coefficients are the generalized least squares estimator $(X^T{V}^{-1}X{)}^{-1}{X}^T{V}^{-1}Y$. Both fits should use the same marginal likelihood convention, for example fitting the alternative with update(fly.model, REML=FALSE) and the null as a Gaussian linear model. The displayed alternative REML criterion alone cannot supply $T$.

The variance component is constrained to be nonnegative, and zero lies on the boundary. Thus the interior-parameter hypothesis of Wilks theorem fails; the usual ${\chi }_1^2$ approximation is inappropriate. A $\frac{1}{2}{\delta }_0+\frac{1}{2}{\chi }_1^2$ limit sometimes applies to a single variance component under additional asymptotic conditions, but four incubators do not give a persuasive large-number-of-groups approximation.

Use a parametric bootstrap under the null: estimate $\beta ,{\sigma }^2$ there; simulate independent Gaussian errors with that fitted variance at exactly the observed times and grouping labels; refit both models by maximum likelihood for each simulated sample; and calculate $T^{\ast }$. The upper-tail proportion, conventionally $(1+\#\{T^{\ast }\ge T\})/(B+1)$, estimates the p-value. Boundary fits with estimated ${\tau }^2=0$ must remain in the simulation distribution. A restricted-likelihood ratio and a correspondingly calibrated bootstrap are another option when the fixed-effect design is identical, but one must not mix ordinary and restricted likelihoods in a single statistic.

Use one joint hypothesis test for $H_0:{\beta }_{\lambda }={\beta }_{\mu }=0$. Fit the unrestricted model with statistical parameters $({\lambda }_0,{\mu }_0,{\beta }_{\lambda },{\beta }_{\mu })$, where ${\lambda }_i={\lambda }_0{e}^{{\beta }_{\lambda }{z}_i}$ and ${\mu }_i={\mu }_0{e}^{{\beta }_{\mu }{z}_i}$ for treatment indicator $z_i$. Refit under $H_0$, estimating the two common baseline intensities; do not fix them at their unrestricted estimates. The likelihood-ratio test statistic isUnder regular identifiable positive-rate models, sufficient independent patient trajectories, and the null hypothesis, Wilks theorem gives two degrees of freedom because two coefficients are constrained. Reject if $D$ exceeds the chosen upper chi-squared distribution quantile. Alternatively a two-dimensional Wald test uses ${\hat{\beta }}^T\widehat{\mathrm{Cov}}(\hat{\beta }{)}^{-1}\hat{\beta }$, including the off-diagonal covariance. Two separate tests do not provide this single calibrated assessment. The point estimates alone cannot produce a numerical p-value without fitted likelihoods or a coefficient covariance matrix.

Use the convention that the likelihood ratio is maximized alternative likelihood divided by null likelihood. The alternative excludes zero, but its supremum agrees with the unrestricted maximum, even on the exceptional sample $\overline{X}=0$. Completing the square givesUnder the null, $\sqrt{n}\overline{X}\sim {\mathcal{N}}_d(0,I_d)$, so $2\log \Lambda \sim {\chi }_d^2$ exactly for every $n$. Equivalently $\Lambda$ is distributed as $e^{Y/2}$ for $Y\sim {\chi }_d^2$. With the reciprocal likelihood-ratio convention the statistic is $-2\log \Lambda$ instead.

Wilks theorem predicts a limiting chi-squared distribution with degrees of freedom equal to the difference of parameter dimensions, here $d-0=d$, under regularity and an interior null point of the unrestricted model. This example strengthens the asymptotic conclusion to an exact finite-sample identity.

There are two defects. The mixed fit supplies a restricted maximum likelihood, whereas the ordinary fit supplies an ordinary log-likelihood, so their difference is not a likelihood-ratio test statistic. Also, $H_0:{\tau }^2=0$ lies on the boundary of ${\tau }^2\ge 0$; the usual Wilks theorem does not give a ${\chi }_1^2$ null law.

A valid approach first refits the mixed model by ordinary maximum likelihood estimation, setting REML to false, and fits the same fixed effects under $H_0$. For $V={\tau }^{2}Z{Z}^T+{\sigma }^{2}I$, maximizeover $\beta ,{\sigma }^2>0,{\tau }^2\ge 0$, and separately over $\beta ,{\sigma }^2>0$ with ${\tau }^2=0$. Set $T=2({\hat{\ell }}_1-{\hat{\ell }}_0)\ge 0$.

For finite-sample calibration, use the location-scale invariant simulation test for a Gaussian variance component. With the actual $X,Z$ fixed, simulate $B$ independent vectors $Y^{(b)}\sim N(0,I_{176})$, refit both models by ordinary maximum likelihood estimation to each, and calculate $T_b$ in exactly the same way. Under $H_0$, $Y=X\beta +\sigma z$. Translating by a vector in the column space of $X$ and multiplying by a positive scalar preserves the statistic: both maximized log-likelihoods acquire the same scale constant. Thus this simulation has the correct null law without knowing $\beta$ or $\sigma$. A conservative Monte Carlo p-value isIn the usual regular limit with increasing independent groups, a variance-component likelihood-ratio test at a boundary instead uses $\frac{1}{2}{\delta }_0+\frac{1}{2}{\chi }_1^2$: for $T>0$, its approximate tail probability is $\frac{1}{2}\mathrm{Pr}({\chi }_1^2\ge T)$, and at $T=0$ the nonrandomized p-value is one. This approximation is not an exact guarantee for sixteen rats. Alternatively one can compare consistently defined restricted likelihoods with the same $X$ and simulate their restricted-ratio null distribution, as in the RLRsim documentation.

Write $Y_i$ for the request count on day $i$. Both fits are Poisson regression models with independentIn model 1,where the variables are indicator variables for winter, weekend, bank holiday, and low or medium pollution; non-winter weekdays that are not bank holidays and have high pollution form the baseline. Model 2 imposesretaining only winter and weekend effects.

The models are nested, and the likelihood-ratio test statistic is the decrease in residual deviance,There are three restrictions. By Wilks theorem, under $H_0$ this statistic is asymptotically a chi-squared distribution with three degrees of freedom. Sincewe reject model 2 at the $1\%$ level in favour of model 1.

The null hypothesis is that price and score are independent, equivalently that the independence log-linear model for a two-way contingency table is correct. Against the saturated alternative, the reported Poisson deviance is the likelihood-ratio test statisticUnder the null and the usual large-sample regularity assumptions, Wilks theorem givesThe observations must be independent, the cell probabilities must not lie on the boundary of the parameter space, and the expected counts must be large enough for the chi-squared asymptotic approximation. The fitted counts are either $31/3$ or $28/3$, so the customary expected-count check is comfortably satisfied. Since(equivalently, the p-value is about $0.329$), we do not reject the null at the $1\%$ significance level. The data provide no significant lack of fit for independence.

Model 2 is obtained from model 3 by setting ${\beta }_0=0$. Model 3 is contained in model 1 by choosing the symmetric block parametersHence

Assuming every college-pair cell is observed and the models are identifiable, model 1 has $C(C+1)/2$ parameters and model 3 has $C+1$. The likelihood-ratio test statistic for model 3 against model 1 isUnder model 3 and the regularity conditions of Wilks theorem,

Letbe the block-saturated mean space, and letbe the fitted GLM mean space. Full column rank and the regularity assumptions give dimensions $n$ and $p$. The factorization in part (ii) makes the nuisance factor cancel from their likelihood ratio, while $\bar{Y}\stackrel{d}{=}Y$ givesAs ${\sigma }^2\to 0$, the replicate count $m=1/{\sigma }^2$ tends to infinity. Wilks theorem applied to these nested regular models makes twice the log-likelihood ratio converge in distribution to a chi-squared variable with the difference in dimensions:

The null value $\pi =1/2$ is interior and the model is regular. Wilks theorem therefore givesunder the null, because the alternative adds one free parameter. An asymptotic size-$\alpha$ test rejects when

For a null parameter space ${\Theta }_0\subseteq \Theta$, the generalized likelihood ratio isThe Wilks theorem states, under standard regularity conditions and under the null, that

The variance $\mu +{\mu }^2/\theta$ approaches the Poisson distribution variance $\mu$ when $\theta \to \infty$. Equivalently, with $\alpha =1/\theta \ge 0$, the Poisson model is the boundary value $\alpha =0$. The usual Wilks theorem for a likelihood-ratio test assumes that the null parameter is an interior point of a smooth parameter space, so comparing the statistic with an ordinary ${\chi }_1^2$ law is invalid here.

A valid parametric bootstrap proceeds as follows. First fit the null Poisson model and retain its fitted means ${\hat{\mu }}_i$. For each bootstrap repetition $b=1,\dots ,B$, independently simulateusing the original doses, refit both the Poisson and negative-binomial models to that simulated data, and calculateFor the observations calculate the analogous $T_{obs}$. The bootstrap $p$-valueuses the null distribution with its boundary and finite-sample fitting behaviour automatically reproduced. A large $B$ controls the Monte Carlo error of this estimate.