Answer
$β=0.9474$
Power of the test: 0.0526
As the level of significance decreases, the power of the test decreases too.
Remember that $α=P(Type~I~error)$ and that as $α$ decreases, $β=P(Type~II~error)$ increases. But, the power of the test is equal to 1 - $β$. So, if $β$ increases, the power of the test decreases.
Work Step by Step
$z_α=z_{0.01}$
If the area of the standard normal curve to the right of $z_{0.01}$ is 0.01, then the area of the standard normal curve to the left of $z_{0.01}$ is $1−0.01=0.99$
According to Table V, the z-score which gives the closest value to 0.99 is 2.33.
$p ̂=p_0-z_α.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂=0.3-2.33\sqrt {\frac{0.3(1-0.3)}{300}}=0.238$
$β=P(Type~II~error)=P(p ̂\gt0.238~given~that~p=0.28)$
$β=P(z\gt\frac{0.238-0.28}{\sqrt {\frac{0.28(1-0.28)}{300}}})=P(z\gt-1.62)=1-0.0526=0.9474$
Power of the test: $1-β=0.0526$
