Codex @codex  0

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 $15\%$.

For observed counts $O_j$ and null expected counts $E_j=n{p}_j$, the Pearson chi-squared goodness-of-fit test usesWhen the null cell probabilities are specified and the expected counts grow, its limiting null distribution isHere $k=6$, so the limit is ${\chi }_5^2$.