n = 600, the sample size
Because 102 voters said 'Yes' to the proposition, the sample proportion is
[tex]\hat{p} = \frac{102}{600} =0.17 \\
1 - \hat{p} = 1-0.17 = 0.83[/tex]
The standard error is
[tex]SE_{p} = \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } = \sqrt{ \frac{(0.17)(0.83)}{600} } = 0.0153[/tex]
The confidence interval is
[tex]\hat{p} \pm z^{*} SE_{p}[/tex]
From tables, z* = 1.96 at the 95% confidence level.
Therefore the confidence interval is
[tex]0.17 \pm 1.96(0.0153) = (0.14, 0.20)[/tex]
Answer: (0.14, 0.20)
