$$ 1 \leftarrow $$ 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, $$ \mathrm{n}=933 $$ and $$ \mathrm{x}=571 $$ who said "yes. " Use a $$ 99 \% $$ confidence level. a) Flick the icon to view a table of $$ z $$ scores. 0.612 (Round to three decimal places as needed.) b) Identify the value of the margin of error E . E= 0.041 (Round to three decimal places as needed.) c) Construct the confidence interval. $$ \square

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**(a)**  
Calculate the sample proportion:  
$$
\hat{p}=\frac{x}{n}=\frac{571}{933}\approx 0.612
$$

**(b)**  
The margin of error is given by:  
$$
E=z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}
$$  
For a $$ 99\% $$ confidence level, the $$ z $$-score is approximately $$ 2.576 $$. Thus,  
$$
E \approx 2.576\sqrt{\frac{0.612(1-0.612)}{933}}=2.576\sqrt{\frac{0.612\times0.388}{933}}
$$
$$
\approx 2.576\sqrt{\frac{0.2373}{933}}\approx 2.576\sqrt{0.0002545}\approx 2.576(0.01595)\approx 0.041
$$

**(c)**  
The $$ 99\% $$ confidence interval for the population proportion is:  
$$
\left(\hat{p}-E,\ \hat{p}+E\right)=\left(0.612-0.041,\ 0.612+0.041\right)=(0.571,\ 0.653)
$$

**(d)**  
We can interpret the confidence interval as follows: We are $$ 99\% $$ confident that the true proportion of respondents who feel vulnerable to identity theft lies between $$ 57.1\% $$ and $$ 65.3\% $$.
**(a)** Sample proportion: $$ \hat{p} = 0.612 $$ **(b)** Margin of error: $$ E = 0.041 $$ **(c)** Confidence interval: $$ 0.571 < p < 0.653 $$ **(d)** We are $$ 99\% $$ confident that the true proportion of respondents who feel vulnerable to identity theft is between $$ 57.1\% $$ and $$ 65.3\% $$.

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