Answer
$β=0.1949$
Power of the test: 0.8051
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.
$p ̂=p_0-z_α.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂=0.75-2.33\sqrt {\frac{0.75(1-0.75)}{400}}=0.700$
$β=P(Type~II~error)=P(p ̂\gt0.700~given~that~p=0.68)$
$β=P(z\gt\frac{0.700-0.68}{\sqrt {\frac{0.68(1-0.68)}{400}}})=P(z\gt0.86)=1-0.8051=0.1949$
Power of the test: $1-β=0.8051$
