 Solution

A simple hypothesis specifies a single probability law, with no unknown parameter remaining. For a possibly randomized hypothesis test, let $\phi (x)\in [0,1]$ be its conditional rejection probability. Its test size is $E_0\phi (X)$ and its statistical power against the specified alternative is $E_1\phi (X)$.

The Neyman-Pearson lemma states that for two densities $f_0,f_1$ with respect to a common measure, a likelihood-ratio test rejecting when $f_1/f_0>k$, accepting when it is below $k$, and randomizing on equality to obtain size $\alpha$, maximizes power among all tests of size at most $\alpha$.

Here the likelihood ratio isWriting $r=|x|$, its logarithm apart from the constant is $(r^2-r)/2$. This decreases until $r=1/2$ and then increases. In particular it is nonpositive for $0\le r\le 1$ and positive for $r>1$. The supplied normal-tail value and $\alpha <1/4$ imply that the quantile $t$ defined by $2[1-\Phi (t)]=\alpha$ satisfies $t>1$. Hence the threshold $\Lambda (t)$ is above every central value on $r\le 1$; outside that interval the ratio is strictly increasing. The Neyman-Pearson lemma therefore selects precisely the two tails:The equality boundary has probability zero under both continuous laws, so no randomization is needed. Under the alternative density, the statistical power isThe central dip in the likelihood ratio is why checking the threshold against the entire central region, rather than assuming monotonicity in $|x|$ everywhere, is necessary.