Answer
$β=0.8790$
Power of the test: $1-β=0.1210$
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.
$p ̂=p_0+z_α.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂=0.40+2.33\sqrt {\frac{0.40(1-0.40)}{200}}=0.481$
$β=P(Type~II~error)=P(p ̂\lt0.481~given~that~p=0.44)$
$β=P(z\lt\frac{0.481-0.44}{\sqrt {\frac{0.44(1-0.44)}{200}}})=P(z\lt1.17)=0.8790$
Power of the test: $1-β=0.1210$
