Answer
$β=0.3897$
Power of the test: 0.6103
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.
$p ̂_L=p_0-z_{\frac{α}{2}}.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂_L=0.52-1.96\sqrt {\frac{0.52(1-0.52)}{800}}=0.485$
$p ̂_U=p_0+z_{\frac{α}{2}}.\sqrt {\frac{p_0(1-p_0)}{n}}$
$p ̂_U=0.52+1.96\sqrt {\frac{0.52(1-0.52)}{800}}=0.555$
$β=P(Type~II~error)=P(0.485\lt p ̂\lt0.555~given~that~p=0.48)$
$β=P(\frac{0.485-0.48}{\sqrt {\frac{0.48(1-0.48)}{800}}}\lt z\lt\frac{0.555-0.48}{\sqrt {\frac{0.48(1-0.48)}{800}}})=P(0.28\lt z\lt4.24)=1-0.6103=0.3897$
Power of the test: $1-β=0.6103$
