Answer
$β=0.0016$
Power of the test: 0.9984
Work Step by Step
$z_α=z_{0.10}$
If the area of the standard normal curve to the right of $z_{0.10}$ is 0.10, then the area of the standard normal curve to the left of $z_{0.10}$ is $1−0.10=0.90$
According to Table V, the z-score which gives the closest value to 0.90 is 1.28.
$p ̂=p_0-z_α.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂=0.38-1.28\sqrt {\frac{0.38(1-0.38)}{1122}}=0.361$
$β=P(Type~II~error)=P(p ̂\gt0.361~given~that~p=0.32)$
$β=P(z\gt\frac{0.361-0.32}{\sqrt {\frac{0.32(1-0.32)}{1122}}})=P(z\gt2.94)=1-0.9984=0.0016$
Power of the test: $1-β=0.9984$
