Answer
$β=0.8124$
Power of the test: 0.1876
Work Step by Step
$z_{\frac{α}{2}}=z_{0.025}$
If the area of the standard normal curve to the right of $z_{0.025}$ is 0.025, then the area of the standard normal curve to the left of $z_{0.025}$ is $1−0.025=0.975$
According to Table V, the z-score which gives the closest value to 0.975 is 1.96.
$p ̂_L=p_0-z_{\frac{α}{2}}.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂_L=0.22-1.96\sqrt {\frac{0.22(1-0.22)}{500}}=0.184$
$p ̂_U=p_0+z_{\frac{α}{2}}.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂_U=0.22+1.96\sqrt {\frac{0.22(1-0.22)}{500}}=0.256$
$β=P(Type~II~error)=P(0.184\lt p ̂\lt0.256~given~that~p=0.20)$
$β=P(\frac{0.184-0.20}{\sqrt {\frac{0.20(1-0.20)}{500}}}\lt z\lt\frac{0.256-0.20}{\sqrt {\frac{0.20(1-0.20)}{500}}})=P(-0.89\lt z\lt3.13)=0.9991-0.1867=0.8124$
Power of the test: $1-β=0.1876$
