Past exam of the statistics course of the University of Oxford / 2015 / 3 / b / ii

The symmetric continuous prior distribution gives $\mathrm{Pr}(\theta <0)=\mathrm{Pr}(\theta \ge 0)=1/2$, so the prior odds are one. Therefore the Bayes factor equals the posterior odds. Standardizing the normal posterior distribution, putThen $\mathrm{Pr}(H_0\mid x)=\Phi (-z)$ and $\mathrm{Pr}(H_1\mid x)=\Phi (z)$. The endpoint $\theta =0$ has zero posterior probability, so the strict/non-strict split causes no additional term. HenceIn particular, positive values of the sample mean make this Bayes factor smaller, favouring the positive-mean hypothesis.