= Solution
{c}
Under the <null hypothesis>, the full model has two free positive <statistical parameters> and the constrained model has one. The <Wilks theorem> therefore gives
$$
W=-2\log\Lambda\xrightarrow d\chi^2_1
$$
as both sample sizes tend to infinity, with $n/(n+m)$ tending to a number strictly between zero and one. For the observed $t$, calculate
$$
w_{\rm obs}=2\left[n\log\frac{n}{(n+m)t}+m\log\frac{m}{(n+m)(1-t)}\right].
$$
The large-sample <p-value> is the upper-tail <probability>
$$
\boxed{p_{\rm approx}=1-F_{\chi^2_1}(w_{\rm obs})
=2\{1-\Phi(\sqrt{w_{\rm obs}})\},}
$$
where $\Phi$ is the <cumulative distribution function> of the <standard normal distribution>. The last equality uses the fact that the square of a <standard normal random variable> has a <chi-squared distribution> with one <statistical degree of freedom>. Values of $T$ on either side of $n/(n+m)$ are already included by the transformation to $W$.
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