Assuming the population has a bell-shaped distribution, if a sample size $n=25$ has a sample mean $\bar{x}=32$ with a sample standard deviation $s=2$, find the margin of error (MOE) at a $99 \%$ confidence level. Round the answer to two decimal places. \[ \mathrm{MOE}= \]

Step 1 ：Given a sample size \(n = 25\), a sample mean \(\overline{x} = 32\), and a sample standard deviation \(s = 2\). We are asked to find the margin of error (MOE) at a 99% confidence level. The Z-score for a 99% confidence level is \(Z = 2.576\).

Step 2 ：The formula for the margin of error (MOE) is given by \(MOE = Z \cdot \frac{s}{\sqrt{n}}\).

Step 3 ：Substitute the given values into the formula: \(MOE = 2.576 \cdot \frac{2}{\sqrt{25}}\).

Step 4 ：Solving the above expression gives \(MOE = 1.0304\).

Step 5 ：Rounding to two decimal places, the margin of error (MOE) at a 99% confidence level is approximately \(\boxed{1.03}\).