Given;
[tex]x=180,n=300[/tex]Then, we can find the point estimation as;
[tex]\begin{gathered} \hat{p}=\frac{x}{n} \\ \hat{p}=\frac{180}{360}=0.60 \end{gathered}[/tex][tex]Z_{\frac{\alpha}{2}}=Z_{0.05}=1.96[/tex]Thus, the margin of error E is;
[tex]\begin{gathered} E=Z_{\frac{\alpha}{2}}\sqrt[]{\frac{\hat{p}(1-\hat{p})}{n}} \\ E=1.96\sqrt[]{\frac{0.60(0.40)}{300}} \\ E=1.96\sqrt[]{0.0008} \\ E=0.055 \end{gathered}[/tex]A 95% confidence interval for population proportion p is;
[tex]\hat{p}\pm E=0.60\pm0.055[/tex]The lower bound is;
[tex]0.60-0.055=0.545[/tex]The upper bound is;
[tex]0.60+0.055=0.655[/tex]