= Solution
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 $\prod_{i=1}^6p_i^{n_i}$, depending on the data only through the counts. The <factorization criterion for sufficiency> therefore makes $(n_1,\ldots,n_6)$ 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 $\widehat p_i=n_i/n$. Thus the <generalized likelihood-ratio test> uses
$$
\Lambda=\frac{(1/6)^n}{\prod_{i=1}^6(n_i/n)^{n_i}},\qquad D=-2\log\Lambda=2\sum_{i=1}^6n_i\log\frac{6n_i}{n},
$$
with zero-count terms interpreted by continuity. \b[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 yields
$$
D=\frac6n\sum_i\delta_i^2+O\left(\sum_i\frac{|\delta_i|^3}{n^2}\right),\qquad
\frac6n\sum_i\delta_i^2=-n+\frac6n\sum_i n_i^2=T.
$$
Under 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 $11.07$. \b[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 $15\%$.
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