Answer
$β=0.7389$
Power of the test: 0.2611
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.
$p ̂=p_0-z_α.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂=0.47-1.28\sqrt {\frac{0.47(1-0.47)}{1013}}=0.450$
$β=P(Type~II~error)=P(p ̂\gt0.450~given~that~p=0.46)$
$β=P(z\gt\frac{0.450-0.46}{\sqrt {\frac{0.46(1-0.46)}{1013}}})=P(z\gt-0.64)=1-0.2611=0.7389$
Power of the test: $1-β=0.2611$
