Suppose that, from a previous sample, we were $95 \%$ confident that the true population proportion was in the interval $(0.0622,0.1844)$. Determine the sample size required in a future study to produce a margin of error of at most 3 percentage points. Enter your answer as a whole number.

Step 1 ：We are given a confidence interval of \((0.0622,0.1844)\) for the population proportion. The midpoint of this interval can be used as an estimate for the population proportion. The midpoint is calculated as \((0.0622 + 0.1844) / 2 = 0.1233\).

Step 2 ：We are given that the margin of error should be at most 3 percentage points, or 0.03.

Step 3 ：The Z-score for a 95% confidence interval is 1.96.

Step 4 ：We can use these values in the formula for the sample size in a confidence interval for a proportion, which is \(n = \frac{{Z^2 \cdot p \cdot (1-p)}}{E^2}\).

Step 5 ：Substituting the given values into the formula, we get \(n = \frac{{1.96^2 \cdot 0.1233 \cdot (1-0.1233)}}{0.03^2}\).

Step 6 ：Calculating the above expression, we find that the sample size needed is 462.

Step 7 ：Since we can't have a fraction of a sample, we round up to the nearest whole number if necessary. In this case, the sample size is already a whole number.

Step 8 ：Final Answer: The required sample size in a future study to produce a margin of error of at most 3 percentage points is \(\boxed{462}\).