given the information on the picture, since we have that n*p is greater than 10, we can use a normal distribution.
We have that the critical value for 99% confidence level is 2.58.
Next, we can find the Standard Error with the following expression:
[tex]SE=\sqrt[]{\frac{p(1-p)}{n}}[/tex]
in this case we have the following:
[tex]SE=\sqrt[]{\frac{0.6\cdot0.4}{160}}=00387[/tex]
then, according to the margin of error formula:
[tex]ME=CV\cdot SE[/tex]
where CV is the critical value, we get:
[tex]ME=2.58\cdot0.0387=0.099[/tex]
therefore, the margin of error is 0.099 which is a 9.9% error
