Answer
$n=1708$
Work Step by Step
$level~of~confiance=(1-α).100$%
$99$% $=(1-α).100$%
$0.99=1-α$
$α=0.01$
$z_{\frac{α}{2}}=z_{0.005}$
If the area of the standard normal curve to the right of $z_{0.005}$ is 0.005, then the area of the standard normal curve to the left of $z_{0.005}$ is $1−0.005=0.995$
According to Table V, there are 2 z-scores which give the closest value to 0.995: 2.57 and 2.58. So, let's find the mean of these z-scores: $\frac{2.57+2.58}{2}=2.575$
Now, the sample size:
$n=p ̂(1-p ̂)(\frac{z_{\frac{α}{2}}}{E})^2$
$n=0.635(1-0.635)(\frac{2.575}{0.03})^2$
$n=1707.57$
Round up:
$n=1708$
