Answer
$β=0.8238$
Power of the test: $0.1762$
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: $\frac{1.64+1.65}{2}=1.645$
$p ̂=p_0-z_α.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂=0.3-1.645\sqrt {\frac{0.3(1-0.3)}{300}}=0.256$
$β=P(Type~II~error)=P(p ̂\gt0.256~given~that~p=0.28)$
$β=P(z\gt\frac{0.256-0.28}{\sqrt {\frac{0.28(1-0.28)}{300}}})=P(z\gt-0.93)=1-0.1762=0.8238$
Power of the test: $1-β=0.1762$
