Answer
1060 is close to 1068.
Notice that, in order to find a sample size of 1068, we made $p ̂=0.5$ in $n=p ̂(1-p ̂)(\frac{z_{\frac{α}{2}}}{E})^2$. This value, $p ̂=0.5$, gives the highest sample size for a 95% confidence interval with a margin of error of 3%. Probably, a smaller sample is required. So, a sample of 1060 individuals is a good choice.
Work Step by Step
$level~of~confiance=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$z_{\frac{α}{2}}=z_{0.025}$
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.
Now, the sample size (no prior estimate of $p ̂$) :
$E=0.03$ (margin of error of 3%)
$n=0.25(\frac{z_{\frac{α}{2}}}{E})^2$
$n=0.25(\frac{1.96}{0.03})^2$
$n=1067.11$
Round up:
$n=1068$
