Answer
$β=0.8186$
Power of the test: 0.1814
Work Step by Step
$z_α=z_{0.05}$
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: $\frac{1.64+1.65}{2}=1.645$
$p ̂=p_0+z_α.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂=0.5+1.645\sqrt {\frac{0.5(1-0.5)}{150}}=0.567$
$β=P(Type~II~error)=P(p ̂\lt0.567~given~that~p=0.53)$
$β=P(z\lt\frac{0.567-0.53}{\sqrt {\frac{0.53(1-0.53)}{150}}})=P(z\lt0.91)=0.8186$
Power of the test: $1-β=0.1814$
