Answer
$β=0.73315$
Power of the test: 0.26685
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.
$p ̂_L=p_0-z_{\frac{α}{2}}.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂_L=0.25-1.96\sqrt {\frac{0.25(1-0.25)}{350}}=0.205$
$p ̂_U=p_0+z_{\frac{α}{2}}.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂_U=0.25+1.96\sqrt {\frac{0.25(1-0.25)}{350}}=0.295$
$β=P(Type~II~error)=P(0.205\lt p ̂\lt0.295~given~that~p=0.28)$
$β=P(\frac{0.205-0.28}{\sqrt {\frac{0.28(1-0.28)}{350}}}\lt z\lt\frac{0.295-0.28}{\sqrt {\frac{0.28(1-0.28)}{350}}})=P(-3.125\lt z\lt0.625)=\frac{0.7324+0.7357}{2}-\frac{0.0009+0.0009}{2}=0.73315$
Power of the test: $1-β=0.26685$
