其它
\textbf{A news organization conducted a survey about preferred methods for obtaining the news. A random sample of 1,605 adults living in a certain state was selected, and 16.2 percent of the adults in the sample reported that television was their preferred method. Which of the following is an appropriate margin of error for a 90 percent confidence interval to estimate the population proportion of all adults living in the state who would report that television is their preferred method for obtaining the news?} 
Overseas Competition->Knowledge Point->Calculation Modules
\textbf{$\hat{p} = 0.162$}  \textbf{MOE = $Z\_{0.05}\sqrt{\frac{0.162*(1-0.162)}{1605}}$ = 1.65*$\sqrt{\frac{0.162*(1-0.162)}{1605}}$} 