We have the following data:

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
![\begin{gathered} p\pm z_{(\alpha)/(2)}\sqrt[]{(p(1-p))/(n)} \\ \text{where} \\ z_{(\alpha)/(2)}=1.96 \\ p=0.53 \\ n=560 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ar054biyvm836sko27yvhwltqmmjlie9nx.png)
![\begin{gathered} 0.53\pm1.96\sqrt[]{(0.53(1-0.53))/(560)} \\ 0.53\pm1.96\sqrt[]{(0.53(0.47))/(560)} \\ 0.53\pm0.0413 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4fubvxuuo8o57jec2g0zbp7d4awwe8oy7l.png)
Thus the interval will be

The values to 3 decimal places are
