1) The first thing to consider when we tackle this question, is to make use of a z-score to construct that required interval of confidence: 90%
2) Let's gather from the data and write out the following:
[tex]\begin{gathered} n=700,\pi=\frac{350}{700},\alpha=0.9 \\ p=\frac{1+\alpha}{2}=\frac{1+0.9}{2}=.95 \\ z=1.645 \\ \end{gathered}[/tex]Note that we resorted to a table to find the z score.
2.2. The next step is to figure out the Lower and the Upper Limits :
[tex]\begin{gathered} Lower\: Limit\colon \\ \pi-z\sqrt[]{\frac{\pi(1-\pi)}{n}}=\frac{350}{700}-1.645\sqrt[]{\frac{0.5(1-0.5)}{700}}=.4689 \\ Upper\: Limit\colon \\ \pi+z\sqrt[]{\frac{\pi(1-\pi)}{n}}=\frac{350}{700}+1.645\sqrt[]{\frac{0.5(1-0.5)}{700}}=.5311 \end{gathered}[/tex]Once we have the lower and the upper limit, we can state that the answer is:
[tex]A)0.469,0.531[/tex]
