If $n=140$ and $\hat{p}$ ( $p$-hat) $=0.25$, find the margin of error at a $90 \%$ confidence level Give your answer to three decimals

Step 1 ：Given that the sample size \(n = 140\), the sample proportion \(\hat{p} = 0.25\), and the z-score for a 90% confidence level is approximately \(z = 1.645\).

Step 2 ：We can calculate the margin of error using the formula: \[\text{Margin of Error} = Z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\]

Step 3 ：Substitute the given values into the formula: \[\text{Margin of Error} = 1.645 \times \sqrt{\frac{0.25(1 - 0.25)}{140}}\]

Step 4 ：Simplify the expression inside the square root: \[\text{Margin of Error} = 1.645 \times \sqrt{\frac{0.1875}{140}}\]

Step 5 ：Calculate the square root: \[\text{Margin of Error} = 1.645 \times 0.040\]

Step 6 ：Finally, multiply the z-score by the result from the square root to find the margin of error: \[\text{Margin of Error} = 0.06\]

Step 7 ：\(\boxed{\text{The margin of error at a 90% confidence level is 0.06}}\)