Answer
$n=3007$
Work Step by Step
$level~of~confidence=(1-α).100$%
$90$% $=(1-α).100$%
$0.90=1-α$
$α=0.1$
$z_{\frac{α}{2}}=z_{0.05}$
If the area of the standard normal curve to the right of $z_{0.05}$ is 0.05, then the area of the standard normal curve to the left of $z_{0.05}$ is $1−0.05=0.95$
According to Table V, there are 2 z-scores which give the closest value to 0.95: 1.64 and 1.65. So, let's find the mean of these z-scores: $\frac{1.64+1.65}{2}=1.645$
Now, the sample size:
$E=0.015$ (within 1.5 percentage points)
$z_{\frac{α}{2}}=z_{0.05}=1.645$
$n=0.25(\frac{z_{\frac{α}{2}}}{E})^2$
$n=0.25(\frac{1.645}{0.015})^2$
$n=3006.69$
Round up:
$n=3007$
