\textbf{$\hat{p} \pm z^{*}\sigma\_{\hat{p}}=\hat{p} \pm Z\_{\frac{\alpha}{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}= 0.32 \pm Z\_{0.05}\sqrt{\frac{(0.32)(0.68)}{1005}}$} 
\textbf{From a random sample of 1,005 adults in the United States, it was found that 32 percent own an e-reader. Which of the following is the appropriate 90 percent confidence interval to estimate the proportion of all adults in the United States who own an e-reader?~} 

  \textbf{$0.32 \pm 1.960(\frac{(0.32)(0.68)}{\sqrt{1,005}})$} 
  \textbf{$0.32 \pm 1.645(\frac{(0.32)(0.68)}{\sqrt{1,005}})$} 
  \textbf{$0.32 \pm 2.575\sqrt{(\frac{(0.32)(0.68)}{1,005})}$} 
  \textbf{$0.32 \pm 1.960\sqrt{(\frac{(0.32)(0.68)}{1,005})}$} 
  \textbf{$0.32 \pm 1.645\sqrt{(\frac{(0.32)(0.68)}{1,005})}$} 