 Solution

An independent sum of normal variables is normal, with the means and variances added. Therefore $\overline{X}\sim N(\mu ,{\sigma }^2/n)$ and $Z\sim N(0,1)$ exactly.

Under the null, $Z_0=\sqrt{n}(\overline{X}-{\mu }_0)/\sigma$ is standard normal. Reject for $Z_0>w$, where $w={\Phi }^{-1}(1-\alpha )$, giving size $\alpha$. For $\alpha =0.05$, the supplied table gives the critical region $\overline{X}>{\mu }_0+1.64\sigma /\sqrt{n}$, using its rounded quantile.

The power function is the probability of rejection when the true parameter is $\mu$. Recenter by that true mean:For $n=100$, $\sigma =2$, power at least $0.70$ means $w-5(\mu -{\mu }_0)\le -0.52$. Hence, with the supplied rounding,The exact unrounded threshold is ${\mu }_0+({\Phi }^{-1}(0.95)-{\Phi }^{-1}(0.30))/5$.