If $\mathrm{n}=300$ and $\widehat{p}$ (p-hat) $=0.39$, find the margin of error at a $99 \%$ confidence level Give your answer to three decimals

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Step 1 ：Given that the sample size (n) is 300, the sample proportion (\(\widehat{p}\)) is 0.39, and the Z-score for a 99% confidence level is approximately 2.576.

Step 2 ：We can calculate the margin of error (E) using the formula: \[E = Z * \sqrt{\frac{{\widehat{p}(1-\widehat{p})}}{n}}\]

Step 3 ：Substitute the given values into the formula: \[E = 2.576 * \sqrt{\frac{{0.39(1-0.39)}}{300}}\]

Step 4 ：Solving the above expression gives us E = 0.07254081863337358

Step 5 ：Rounding to three decimal places, we get E = 0.073

Step 6 ：Thus, the margin of error at a 99% confidence level is approximately \(\boxed{0.073}\).