= Solution
For a one-sided <Wald test> use the <signed normal Wald statistic>, rather than its square. The <maximum-likelihood estimate> is the <sample mean> $\bar Y_n$, with exact <variance> $\sigma^2/n$, so
$$
\boxed{W_n=\frac{\bar Y_n}{\sigma/\sqrt n}=\frac{\sum_{i=1}^nY_i}{\sigma\sqrt n}.}
$$
A sum of independent <normal random variables> is normal, giving $\bar Y_n\sim N(\delta,\sigma^2/n)$ and hence $W_n\sim N(\sqrt n\delta/\sigma,1)$. Under the <null hypothesis> its mean is zero:
$$
\boxed{W_n\mid\delta=0\sim N(0,1).}
$$
Thus its null distribution is exactly the <standard normal distribution>, without an asymptotic approximation. If the squared <Wald statistic> convention is used, $W_n^2$ has a <chi-squared distribution> with one degree of freedom; the signed form is needed to distinguish the two directions.
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