= Solution
The log <likelihood ratio> statistic comparing the unrestricted fit with the <independence> fit is
$$
\begin{aligned}
\Lambda&=2\{\ell(\widehat\theta)-\ell(\widehat\theta^{\,0})\}\\
&=2\sum_{i,j}n_{ij}\log\left(\frac{n_{ij}/N}{n_{i.}n_{.j}/N^2}\right)
=\boxed{2\sum_{i,j}n_{ij}\log\left(\frac{n_{ij}N}{n_{i.}n_{.j}}\right)}.
\end{aligned}
$$
The unrestricted model has $rc-1=8$ free parameters. <Independence> has $(r-1)+(c-1)=4$, so the difference is $(r-1)(c-1)=4$. By <Wilks theorem>, under the <null hypothesis> the statistic has asymptotically a <chi-squared distribution> with $4$ degrees of freedom, rather than $6$, $8$, or $9$.
For the supplied statistic, the approximate <p-value> is $1-F_4(11.8)=0.02$. Thus \b[reject <independence> at the 5% level], but not at the 1% level: the data provide evidence of an association. The approximation is reasonable here; the smallest fitted cell count is $27\cdot99/299\approx8.94$, so all fitted counts exceed $5$. The <statistical hypothesis test> concerns association, not a causal effect of personality.
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