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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