Suppose we want to conduct a study to estimate the percentage of snowboarders (as opposed to skiers) at resorts in America. If we want to find a $90 \%$ confidence interval, determine the sample size required to produce a margin of error of at most 3.5 percentage points if we have no idea what proportion of American riders are snowboarders. Enter your answer as a whole number.

Step 1 ：We are given that the desired margin of error (E) is 3.5% or 0.035, the Z-score for a 90% confidence interval (Z) is approximately 1.645, and since we don't know the proportion of snowboarders, we'll use the most conservative estimate, which is p = 0.5.

Step 2 ：We want to find the sample size (n), so we can use the rearranged formula for the margin of error in a proportion: \(n = \frac{Z^2p(1-p)}{E^2}\).

Step 3 ：Substituting the given values into the formula, we get: \(n = \frac{(1.645)^2 * 0.5 * 0.5}{(0.035)^2}\).

Step 4 ：Calculating the above expression, we find that n is approximately 552.7.

Step 5 ：Since we can't have a fraction of a person, we'll round up to the nearest whole number.

Step 6 ：Final Answer: The required sample size to produce a margin of error of at most 3.5 percentage points with a 90% confidence interval is \(\boxed{553}\).