We have the following data:
[tex]\begin{gathered} n=560 \\ p=0.53 \end{gathered}[/tex]
From the normal distribution curve, the z-score corresponding to 95% confidence interval to
95 % confidence interval, is 1.96
To construct the confidence interval, we will use the formula below
[tex]\begin{gathered} p\pm z_{\frac{\alpha}{2}}\sqrt[]{\frac{p(1-p)}{n}} \\ \text{where} \\ z_{\frac{\alpha}{2}}=1.96 \\ p=0.53 \\ n=560 \end{gathered}[/tex]
[tex]\begin{gathered} 0.53\pm1.96\sqrt[]{\frac{0.53(1-0.53)}{560}} \\ 0.53\pm1.96\sqrt[]{\frac{0.53(0.47)}{560}} \\ 0.53\pm0.0413 \end{gathered}[/tex]
Thus the interval will be
[tex]\begin{gathered} 0.53-0.0413=0.4887 \\ 0.53+0.0413=0.5713 \end{gathered}[/tex]
The values to 3 decimal places are
[tex]\begin{gathered} 0.489 \\ \text{and} \\ 0.571 \end{gathered}[/tex]
