Answer
$\approx0.0226$
Work Step by Step
$\hat{p}=\frac{848}{1010}\approx0.8396$
The z-value belonging to the $95\%$ confidence interval according to the table is $z=1.96$, thus the margin of error is: $z\sqrt{\frac{p(1-p)}{n}}$, which here is: $1.96\sqrt{\frac{0.8396\cdot(1-0.8396)}{1010}}\approx0.0226$
