contestada

A random sample of 600 voters in a particular city found 102 voters who voted yes on proposition 200. find a 95% confidence interval for the true percent of voters in this city who voted yes on proposition 200. express your results to the nearest hundredth of a percent.

Respuesta :

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)