Trivial AP Statistics Problem 2012: Confidence Intervals of a survey

Question

I assume that this problem is trivial but I am not sure what to do: In a survey of 900 people in the US a journalist says that 60% of people support a new law. If the margin of error is 2.7% for the percentage, what is the level of confidence?

Answer 1

By definition, \begin{equation*} \text{ margin or error } =\text{ critical value } \times \text {standard error} \end{equation*}

Given $p=.6$

Then standard error is $\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.6\times (1-0.6)}{900}}=0.01632993$

Given that the margin of error = 2.7% = 0.027. Then, \begin{eqnarray*} 0.027 &=& \text{ crtical value }\times SE =\text{ crtical value }\times 0.01632993\\ \text{ critical value} &=& \dfrac{0.027}{0.01632993}= 1.653406 \end{eqnarray*} As the sample size is large, using normal distribution, we observe that the critical value $ 1.653406$ corresponds to $95\%$ confidence level.