= Solution
A longer <neutron> <half-life> means a larger mean lifetime $\tau_n=\tau_{1/2}/\ln2$. Holding the initial <neutron-to-proton ratio> and the nuclear-burning delay fixed, the derivative of the <neutron> survival factor is positive:
$$
\frac{\partial\ln Y_4}{\partial\tau_n}=\frac{\Delta t}{\tau_n^2}>0.
$$
Thus fewer <neutrons> disappear during the <deuterium bottleneck>, and \b[the final helium-4 mass fraction increases].
There can be a second effect if the increased lifetime represents a uniform reduction of the relevant <weak interaction> strength. With fixed masses and other couplings, <neutron> decay and the conversion reactions both scale as $G_F^2$. Thus $\tau_n\propto G_F^{-2}$, whereas $T_F\propto G_F^{-2/3}\propto\tau_n^{1/3}$. Weaker conversion then freezes out earlier and increases $r_F$, reinforcing the survival effect. The survival argument alone already establishes the requested <neutron half-life effect on primordial helium>; linking lifetime to all conversion rates additionally requires this microscopic assumption.
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