= Solution
The symmetric continuous <prior distribution> gives $\Pr(\theta<0)=\Pr(\theta\geq0)=1/2$, so the <prior odds> are one. Therefore the <Bayes factor> equals the <posterior odds>. Standardizing the normal <posterior distribution>, put
$$
z=\frac{\mu_n}{\sqrt{v_n}}
=\frac{n\tau\overline x}{\sigma\sqrt{\sigma^2+n\tau^2}}.
$$
Then $\Pr(H_0\mid x)=\Phi(-z)$ and $\Pr(H_1\mid x)=\Phi(z)$. The endpoint $\theta=0$ has zero posterior probability, so the strict/non-strict split causes no additional term. Hence
$$
\boxed{B=\frac{\Phi(-z)}{\Phi(z)}
=\frac{1-\Phi(z)}{\Phi(z)},\qquad
z=\frac{n\tau\overline x}{\sigma\sqrt{\sigma^2+n\tau^2}}.}
$$
In particular, positive values of the <sample mean> make this <Bayes factor> smaller, favouring the positive-mean hypothesis.
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