Surveys: In a survey, 93% of respondants said 'yes'....

In a survey of 1004 adults, 93% said yes.
If you were to conduct a poll confirm that the percentage continues to be correct, how many adults would you have to survey if you want 98% confidence that the margin of error is four percentage points?

You should have a formula for that.

Something like: \(\displaystyle \L\,n = \frac{t^{2}s^{2}}{d^{2}}\)

The margin of error used to construct a symmetric confidence interval for the binomial proportion is given by \(\displaystyle E = z\sigma/\sqrt{n}\), where \(\displaystyle \sigma = sqrt{p(1-p)}\) is an estimate of the standard deviation, \(\displaystyle z\) is the standard normal deviate for the confidence level, and \(\displaystyle n\) is the sample size.