Use the sample data and confidence level given below to complete parts (a) through (d).
A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, $n=1094$ and $x=564$ who said "yes." Use a $90 \%$ confidence level.
b) Identify the value of the margin of error $E$.
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E=
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Answer
Thus, the value of the margin of error \(E\) is approximately \(\boxed{0.0249}\).
Step 1 ：Given that the sample size \(n = 1094\), the number of respondents who said 'yes' \(x = 564\), and the z-score \(Z = 1.645\) for a 90% confidence level.
Step 2 ：First, calculate the sample proportion \(p\) as \(x/n\).
Step 3 ：Substitute \(n = 1094\) and \(x = 564\) into the formula to get \(p = 564/1094 = 0.5155\).
Step 4 ：Next, calculate the margin of error \(E\) using the formula \(E = Z * \sqrt{\frac{p(1-p)}{n}}\).
Step 5 ：Substitute \(Z = 1.645\), \(p = 0.5155\), and \(n = 1094\) into the formula to get \(E = 1.645 * \sqrt{\frac{0.5155(1-0.5155)}{1094}} = 0.0249\).
Step 6 ：Thus, the value of the margin of error \(E\) is approximately \(\boxed{0.0249}\).
