I was given the following question:

A survey found that 89% of a random sample of 1024 American adults approved of cloning endangered animals. Find the margin of error for this survey if we want 90% confidence in our estimate of the percent of American adults who approve of cloning endangered animals.

I know that for 90% Confidence, ME ~ 0.82/sqrt(n).

I attempted using this formula with n equal to both 1024 and (.89)1024. I got 0.025625 and 0.02716, respectively. The answer given for the problem is 1.61%. I do not understand where I went wrong. Perhaps I am using the Margin of Error formula incorrectly?

Thanks. :)