Answer
$β=0.7710$
Power of the test: 0.2290
Work Step by Step
$z_{\frac{α}{2}}=z_{0.05}$
If the area of the standard normal curve to the right of $z_{0.05}$ is 0.05, then the area of the standard normal curve to the left of $z_{0.05}$ is $1−0.05=0.95$
According to Table V, there are 2 z-scores which give the closest value to 0.95: 1.64 and 1.65. So, let's find the mean of these z-scores: $\frac{1.64+1.65}{2}=1.645$
$p ̂_L=p_0-z_{\frac{α}{2}}.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂_L=0.45-1.645\sqrt {\frac{0.45(1-0.45)}{500}}=0.413$
$p ̂_U=p_0+z_{\frac{α}{2}}.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂_U=0.45+1.645\sqrt {\frac{0.45(1-0.45)}{500}}=0.487$
$β=P(Type~II~error)=P(0.413\lt p ̂\lt0.487~given~that~p=0.47)$
$β=P(\frac{0.413-0.47}{\sqrt {\frac{0.47(1-0.47)}{500}}}\lt z\lt\frac{0.487-0.47}{\sqrt {\frac{0.47(1-0.47)}{500}}})=P(-2.55\lt z\lt0.76)=0.7764-0.0054=0.7710$
Power of the test: $1-β=0.2290$
