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For a generalized likelihood-ratio test, maximize the likelihood separately over the null parameter space and the unrestricted space. Define $\Lambda ={\sup }_{H_0}L/\sup L\le 1$, and reject for small $\Lambda$, equivalently large $T=-2\log \Lambda$. Choose the rejection threshold to achieve the desired null significance level, using an exact distribution when available or a justified asymptotic calibration; nuisance parameters are fitted under the respective restrictions.

For independent Poisson distributions, the likelihood is $L={\prod }_i{e}^{-{\lambda }_i}{\lambda }_i^{X_i}/X_i!$. The unrestricted maximizers are ${\hat{\lambda }}_i=X_i$, allowing zero as a boundary value. Under the common-mean null, maximizing $e^{-n\lambda }{\lambda }^{{\sum }_i{X}_i}$ gives $\hat{\lambda }=\bar{X}$. The exponential and factorial factors cancel in the ratio, yieldingIf all counts are zero, set $\Lambda =1$ and $T=0$; no division by $\bar{X}$ is then needed.

For positive $\bar{X}$, put $X_i=\bar{X}+{\delta }_i$, so ${\sum }_i{\delta }_i=0$. Expanding the logarithm givesConsequentlyIt is the quadratic Pearson approximation to the Poisson deviance. The approximated statistic is the logarithmic ratio $T$, not the bounded raw ratio $\Lambda$.

Under the null in the large-common-count regime, this statistic has an approximate chi-squared distribution with $n-1$ degrees of freedom: fitting the common mean removes one independent residual direction. For $n=7$, use six degrees of freedom. At $T=27.3$, the upper-tail probability in that approximation isThus reject the equal-mean null at conventional 5% or 1% significance levels, provided the expected counts justify the chi-square approximation. No particular significance level or common count was supplied, so the numerical decision is stated with those assumptions.

For independent observations from a normal distribution with unknown mean and positive variance, the maximum likelihood estimation of the mean givesIndeed ${\sum }_i(X_i-\mu {)}^2=S_{XX}+n(\bar{X}-\mu {)}^2$, which is minimized at $\mu =\bar{X}$. The likelihood isFor fixed $\mu$, differentiation in ${\sigma }^2$ gives the maximizer $[S_{XX}+n(\bar{X}-\mu {)}^2]/n$. Thus the unrestricted fit has ${\hat{\sigma }}^2=S_{XX}/n$, whereas under the null hypothesis $\mu ={\mu }_0$ the fitted variance is $[S_{XX}+n(\bar{X}-{\mu }_0{)}^2]/n$. Both maximize rather than minimize the likelihood: it tends to zero at either variance endpoint when the residual sum of squares is positive.

The restricted-to-unrestricted likelihood ratio is thereforeThe generalized likelihood-ratio test rejects for small $\Lambda$, equivalently large $|T|$. Under the null, $Z=\sqrt{n}(\bar{X}-{\mu }_0)/\sigma$ is standard normal, $U=S_{XX}/{\sigma }^2$ has a chi-squared distribution with $n-1$ degrees of freedom, and they are independent. To see the independence, apply an orthogonal matrix to the standardized normal sample, taking its first row to be $(1,\dots ,1)/\sqrt{n}$: the transformed coordinates remain independent standard normal variables, and $U$ is the sum of squares of the other $n-1$ coordinates. Hence $T=Z/\sqrt{U/(n-1)}$ has Student's t-distribution with $n-1$ degrees of freedom.

For this two-sided test at $5\%$, the cutoff is the $97.5\%$ percentile of $t_9$, namely $2.26$. The data giveSince $|T|<2.26$, do not reject the manufacturer's mean-hardness claim at the $5\%$ level. This is a failure to reject, not proof that the mean equals $75$.

Assume independent samples $X_1,\dots ,X_n$ and $Y_1,\dots ,Y_m$, each consisting of independent normal random variables, with respective means ${\mu }_1,{\mu }_2$ and a common unknown variance ${\sigma }^2>0$. Take $n,m\ge 2$. Normality, independence and equal variances are the assumptions giving the exact pooled Student's t-test; unequal variances would require a different test. We test $H_0:{\mu }_1={\mu }_2$ against an unrestricted difference.

Put $N=n+m$ andThe normal likelihood is proportional to $({\sigma }^2{)}^{-N/2}\exp [-S/(2{\sigma }^2)]$, where $S$ is the residual sum of squares. Under the unrestricted model, the maximizing means are $\bar{X},\bar{Y}$ and $S=W$. Under $H_0$, the maximizing common mean is $(n\bar{X}+m\bar{Y})/N$ and decomposition about the sample means gives $S=W+B$. For either model, maximizing over variance gives ${\hat{\sigma }}^2=S/N$. Thus the generalized likelihood-ratio test statistic isSmall $\Lambda$ is therefore equivalent to large $|T|$.

Under the null, $Z=(\bar{X}-\bar{Y})/[\sigma \sqrt{1/n+1/m}]$ is standard normal. In each sample an orthogonal change of Gaussian coordinates separates the sample-mean coordinate from the centered residual coordinates; those coordinates are independent. Consequently $V=W/{\sigma }^2$ has a chi-squared distribution with $N-2$ degrees of freedom and is independent of $Z$. Therefore $T=Z/\sqrt{V/(N-2)}$ has the Student's t-distribution with $N-2$ degrees of freedom. At significance level $\alpha$, reject exactly when $\boxed{|T|>t_{N-2,1-\alpha /2}}$, the two-sample pooled t-test.

Let $n={\sum }_{\nu =1}^g{n}_{\nu }$, ${\bar{x}}_{\nu }=n_{\nu }^{-1}{\sum }_j{x}_j^{(\nu )}$ and $\bar{x}=n^{-1}{\sum }_{\nu }{n}_{\nu }{\bar{x}}_{\nu }$. Define the within-group and between-group scatter matrices byThe within-group and between-group scatter decomposition gives total scatter $T=W+B$: in each group the residuals about ${\bar{x}}_{\nu }$ sum to zero, so all cross terms vanish. Under unrestricted group means, the maximum-likelihood estimators are ${\hat{\mu }}_{\nu }={\bar{x}}_{\nu }$ and $\hat{V}=W/n$. Under the common-mean null hypothesis, they are $\hat{\mu }=\bar{x}$ and ${\hat{V}}_0=(W+B)/n$.

For either fitted scatter matrix $A$, substitution into the log-likelihood givessince $\mathrm{tr}[(A/n{)}^{-1}A]=np$. Thus the maximized likelihood ratio isA generalized likelihood-ratio test rejects for small ${\Lambda }_{\mathrm{LR}}$, equivalentlyThe cutoff $c$ is chosen for the desired significance level. The determinant ratio $|W|/|W+B|$ itself is the Wilks lambda statistic. The derivation assumes $W$ is positive-definite, which for continuous normal samples requires enough within-group degrees of freedom, $n-g\ge p$; singular $W$ does not support this ordinary determinant-based test without a modified model.

Write $Q={\sum }_i(X_i-\bar{X}{)}^2$ and $Q_0={\sum }_i(X_i-{\mu }_0{)}^2=Q+n(\bar{X}-{\mu }_0{)}^2$. The normal-sample likelihood isWithout the mean restriction, its maximum has $\hat{\mu }=\bar{X}$ and ${\hat{\sigma }}^2=Q/n$. Under the null, the maximizing variance is $Q_0/n$. Substituting them gives the generalized likelihood-ratio test statisticwhere $S^2=Q/(n-1)$ and $T=\sqrt{n}(\bar{X}-{\mu }_0)/S$. Assume $n\ge 2$; $Q>0$ almost surely for positive population variance.

The ratio decreases strictly as $|T|$ increases, so rejecting for small $\Lambda$ is equivalent to rejecting for large $|T|$. Under $H_0$, $T$ has Student's t-distribution with $n-1$ degrees of freedom. Therefore the exact size-$\alpha$ rule isThis is the two-sided Student's t-test. Equivalently reject when $\Lambda <(1+t_{n-1,1-\alpha /2}^2/(n-1){)}^{-n/2}$. No large-sample chi-squared approximation is needed for this normal-mean likelihood ratio with unknown variance.

Write $\bar{X}=n^{-1}{\sum }_i{X}_i$, $Q={\sum }_i(X_i-\bar{X}{)}^2$, and $Q_0={\sum }_i(X_i-{\mu }_0{)}^2=Q+n(\bar{X}-{\mu }_0{)}^2$. The normal likelihood is proportional to $({\sigma }^2{)}^{-n/2}\exp [-{\sum }_i(X_i-\mu {)}^2/(2{\sigma }^2)]$. Its unrestricted maximizers are $\hat{\mu }=\bar{X}$, ${\hat{\sigma }}^2=Q/n$. Under the null the variance maximizer is $Q_0/n$. Substitution gives the generalized likelihood-ratio test statisticThe ratio decreases as $|T|$ increases, so the likelihood-ratio rejection region is two-sided in $T$. Under $H_0$, $\sqrt{n}(\bar{X}-{\mu }_0)/\sigma$ is standard normal and $Q/{\sigma }^2$ is independent chi-squared with $n-1$ degrees of freedom. Independence follows by resolving the isotropic Gaussian sample vector into its projection on the constant vector and its orthogonal complement. Hence $T$ has the exact Student t-distribution with $n-1$ degrees of freedom.

For a chosen significance level $\alpha$, compute the sample mean and the sample standard deviation with denominator $n-1$, and reject precisely whenEquivalently, report the two-sided $p$-value $2[1-F_{t_{n-1}}(|T|)]$. This requires $n\ge 2$ and positive population variance; $S>0$ holds almost surely under that model. The nuisance variance is removed by the exact null distribution, so one should not substitute a normal critical value or a large-sample chi-squared approximation. This is the normal-mean likelihood ratio with unknown variance.

A sufficient statistic is one for which the conditional distribution of the full sample, given its value, is independent of the parameter. For independent observations the likelihood is ${\prod }_{i=1}^6{p}_i^{n_i}$, depending on the data only through the counts. The factorization criterion for sufficiency therefore makes $(n_1,\dots ,n_6)$ sufficient. Directly, all ordered samples with given counts have the same probability, so their conditional distribution is uniform over the $n!/\prod n_i!$ possible arrangements, independently of the $p_i$.

The unrestricted maximum-likelihood estimator is ${\hat{p}}_i=n_i/n$. Thus the generalized likelihood-ratio test useswith zero-count terms interpreted by continuity. Reject for small $\Lambda$, equivalently large $D$.

Write $n_i=n/6+{\delta }_i$, so $\sum {\delta }_i=0$. A Taylor series expansion of $(n/6+{\delta }_i)\log (1+6{\delta }_i/n)$ gives ${\delta }_i+3{\delta }_i^2/n+O(|{\delta }_i{|}^3/n^2)$. Summing cancels the linear terms, and yieldsUnder the null hypothesis, the multinomial central limit theorem places the standardized count deviations in the five-dimensional subspace with coordinate sum zero, with identity covariance on that subspace. Their squared norm consequently tends to the chi-squared distribution with five degrees of freedom. This gives the Pearson chi-squared goodness-of-fit test.

For $T=8.12$, the asymptotic p-value is about $0.150$. The 5% critical value of ${\chi }_5^2$ is approximately $11.07$. Do not reject at the usual 5% level, or at 10%. A significance level is not specified in the paper, so an unconditional yes-or-no decision is not determined: using this approximation, rejection would require a chosen level at least about $15\%$.

Put $S=X_1+X_2+X_3$. Independence gives a likelihood ratioSince ${\theta }_1>1$, this is strictly increasing in $S$. A threshold chosen between its values at $S=5$ and $S=6$ gives precisely the stated upper-tail critical region, so it is a likelihood-ratio test.

The sum has a Poisson distribution with mean $3\theta$. Thus the size of a statistical test and its statistical power at ${\theta }_1$ are

For the grouped-area problem, let $S_1={\sum }_{i=1}^m{Y}_i$, $S_2={\sum }_{i=m+1}^{2m}{Y}_i$, $T=S_1+S_2$. The null log likelihood, up to data-only constants, isThe maximum-likelihood estimate is $\hat{\lambda }=T/(3ma)$, giving fitted means $T/(3m)$ and $2T/(3m)$ in the two groups. Under the alternative, the fitted group means are ${\hat{\lambda }}_1=S_1/m$ and ${\hat{\lambda }}_2=S_2/m$.

Use the increasing generalized likelihood-ratio test convention $\Lambda ={\sup }_{H_1}L/{\sup }_{H_0}L\ge 1$, consistent with the printed positive $2\log \Lambda$. The exponential and factorial factors cancel, yielding the likelihood-ratio test of area-proportional Poisson meansTerms with zero count are interpreted as $0\log 0=0$; if $T=0$, take $\Lambda =1$. If the inverse ratio is used, the reported test statistic is instead $-2\log \Lambda$.

Under the null, the regular large-sample Wilks theorem approximation is $2\log \Lambda \sim {\chi }_1^2$: the alternative has two rate parameters and the null one. The value $15.67$ far exceeds the usual $1\%$ critical value, about $6.63$, and has approximate tail probability $7.5\times 10^{-5}$. Thus reject proportionality of the expected counts to area: the two groups show evidence of different territory densities. The statistic alone does not say which group has greater density; that requires $S_1/m$ versus $S_2/(2m)$. For small counts, an exact conditional test uses $S_1\mid T\sim \mathrm{Binomial}(T,1/3)$ under the null instead of the asymptotic calibration.

Put ${\mathrm{RSS}}_1={\sum }_i(Y_i-\hat{\alpha }-\hat{\beta }{x}_i{)}^2$ for the unrestricted fit. Under $H_0$, the best slope is still $\hat{\beta }$ because the design is centered. Orthogonality of the residuals to the constant vector givesAt either fitted mean, the maximum-likelihood estimator of variance is $\mathrm{RSS}/n$. The ratio used by the generalized likelihood-ratio test is consequentlySmall likelihood ratios correspond to large $|T_n|$. Under $H_0$, part (a) gives independent $\sqrt{n}\overline{Y}/\sigma \sim N(0,1)$ and ${\mathrm{RSS}}_1/{\sigma }^2\sim {\chi }_{n-2}^2$. Thus $T_n$ has the Student's t-distribution with $n-2$ degrees of freedom. The exact size-5% critical region iswhere $t_{\nu ,0.975}$ is the 97.5th percentile of that distribution. The denominator uses the unbiased residual variance estimate, not the likelihood variance denominator $n$.

Under an alternative, the same statistic has the representationThe joint law of $Z,W$ is parameter-free, and $\beta$ has disappeared. Thereforewhere $g(d)=\mathbb{P}(|(Z+\sqrt{n}d)/\sqrt{W/(n-2)}|>t_{n-2,0.975})$. This is also the noncentral t-distribution description of the power. As explained in part (a), an exact residual-based test requires $n>2$.

Under the null hypothesis that treatment and recovery are independent, estimate each expected cell count byand form the Pearson chi-squared test of independence statistic ${\chi }^2=\sum (O-E{)}^2/E$. For a $2\times 2$ table its asymptotic null law is a chi-squared distribution with one degree of freedom. A second-order Taylor expansion of the generalized likelihood-ratio test statistic $-2\log \Lambda$ around the fitted counts gives this Pearson statistic, which explains its asymptotic null distribution.

Writing the four residuals over their fitted counts and simplifying givesFor $(a,b,c,d)=(50,10,15,5)$,whose one-degree-of-freedom p-value is about $0.41$. There is no evidence of an effect. One fitted count is only $3.75$, however, so the chi-squared approximation is questionable; Fisher's exact test would be preferable.

The unrestricted maximum-likelihood estimates are $(\bar{X},\bar{Y})$, while the null fixes both means at zero. HenceUnder the null this is ${\chi }_2^2$, so the size-$\alpha$ generalized likelihood-ratio test rejects exactly when

Let $L(\vartheta ;x)$ be the likelihood function, with full parameter space $\Theta$ and null hypothesis $\vartheta \in {\Theta }_0\subseteq \Theta$. The generalized likelihood-ratio test compares the best null fit to the best unrestricted fit:The likelihood ratio $R$ lies between zero and one, and small values count against the null hypothesis; equivalently, large values of the likelihood-ratio test statistic $\Lambda$ count against it.

Compute restricted and unrestricted maximum-likelihood estimates, evaluate their log-likelihoods, and form $\Lambda =2(\ell (\hat{\vartheta })-\ell ({\hat{\vartheta }}_0))$. Under the regular conditions of Wilks theorem, the null limit isThese statistical degrees of freedom count independent parameter restrictions. The regularity includes a locally identifiable smooth model, nonsingular Fisher information, and an interior null value; parameter-dependent support can invalidate the result.

For a chosen significance level $\alpha$, obtain the $(1-\alpha )$ quantile $q_{d,1-\alpha }$ of the chi-squared distribution and useEquivalently, compare the asymptotic p-value $\mathrm{Pr}({\chi }_d^2\ge {\Lambda }_{\mathrm{obs}})$ to $\alpha$. This is a large-sample calibration; part (c) requires direct treatment because its support depends on a parameter.