Use the sample data and confidence level given below
A research institute poll asked respondents if they felt vulnerable to identity theit. In the poil, $n=987$ and $x=567$ who said " yes. " Use a $90 \%$ confidence level

Identify the value of the margin of error $E$
\[
E=\square
\]
(Round to four decimal places as needed)

Step 1 ：Given that the number of respondents, \(n = 987\), and the number of respondents who said 'yes', \(x = 567\). The z-score for a 90% confidence level, \(Z = 1.645\).

Step 2 ：First, calculate the sample proportion, \(p\), using the formula \(p = \frac{x}{n}\).

Step 3 ：Substitute the given values into the formula to get \(p = \frac{567}{987} = 0.574468085106383\).

Step 4 ：Next, calculate the margin of error, \(E\), using the formula \(E = Z \sqrt{\frac{p(1 - p)}{n}}\).

Step 5 ：Substitute the values into the formula to get \(E = 1.645 \sqrt{\frac{0.574468085106383(1 - 0.574468085106383)}{987}} = 0.02588846781788074\).

Step 6 ：Round the value of \(E\) to four decimal places to get \(E = 0.0259\).

Step 7 ：Final Answer: The value of the margin of error \(E\) is \(\boxed{0.0259}\).