Answer
$n=1731$
Work Step by Step
$level~of~confiance=(1-α).100$%
$98$% $=(1-α).100$%
$0.98=1-α$
$α=0.02$
$z_{\frac{α}{2}}=z_{0.01}$
If the area of the standard normal curve to the right of $z_{0.01}$ is 0.01, then the area of the standard normal curve to the left of $z_{0.01}$ is $1−0.01=0.99$
According to Table V, the z-score which gives the closest value to 0.99 is 2.33.
Now, the sample size:
$E=0.02$ (within 2 percentage points)
$p ̂=0.15$ (estimate of 15%)
$n=p ̂(1-p ̂)(\frac{z_{\frac{α}{2}}}{E})^2$
$n=0.15(1-0.15)(\frac{2.33}{0.02})^2$
$n=1730.46$
Round up:
$n=1731$
