Answer
$β=0.0268$
Power of the test: 0.9732
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.
$p ̂=p_0-z_α.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂=0.47-1.28\sqrt {\frac{0.47(1-0.47)}{1013}}=0.450$
$β=P(Type~II~error)=P(p ̂\gt0.450~given~that~p=0.42)$
$β=P(z\gt\frac{0.450-0.42}{\sqrt {\frac{0.42(1-0.42)}{1013}}})=P(z\gt1.93)=1-0.9732=0.0268$
Power of the test: $1-β=0.9732$
