= Solution
{c}
Under the <independence log-linear model for a two-way contingency table>, the six cell probabilities factor as
$$
H_0:\quad\theta_{ij}=r_i c_j,\qquad
\sum_{i=1}^2r_i=\sum_{j=1}^3c_j=1.
$$
The row distribution has one free parameter and the column distribution two, giving \b[three free parameters under $H_0$]. Under the unrestricted alternative, six cell probabilities sum to one, giving \b[five free parameters under $H_1$]. The difference is two, which is the degrees of freedom of the resulting chi-squared test.
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