= Solution
Test the <null hypothesis> $H_0:\theta_m=\theta_f=\theta$ against the <alternative hypothesis> that the two rate parameters differ. Under $H_0$, the pooled <maximum-likelihood estimate> is $\widehat\theta_0=16/36.8\simeq0.434783$. Under the alternative, the estimates are $\widehat\theta_m=8/22\simeq0.363636$ and $\widehat\theta_f=8/14.8\simeq0.540541$.
In the <likelihood-ratio test statistic>, the data-only terms cancel, and both fitted models have total $\widehat\theta S=16$. Consequently,
$$
T=2(\ell_1-\ell_0)
=16\log\frac{\widehat\theta_m\widehat\theta_f}{\widehat\theta_0^2}
=16\log\frac{36.8^2}{4(22)(14.8)}\simeq0.6245.
$$
The <alternative hypothesis> adds one parameter, so <Wilks theorem> gives an approximate <chi-squared distribution> with one <degree of freedom>. \b[Do not reject the common-rate model at the 5% level], since $T<3.841$. The corresponding approximate <p-value> is $0.43$. The fitted <expected values> are $5.5$ and $3.7$ days, but four observations per group supply little evidence that the underlying <expected values> differ; failure to reject does not establish equality. The <chi-squared distribution> calibration is asymptotic and should be treated cautiously for these small samples.
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