Question

1. a survey of 866 voters in one state reveals that 47.1% of them favor approval of an issue before the legislature. construct a 95% confidence interval for the true proportion of all voters in the state who favor approval. express the c.i. as (__ ±__). (0.438, 0.504). (47.1 ± 3.3)

Explanation:

Step1: Determine the sample proportion \( \hat{p} \) and sample size \( n \)

The sample proportion \( \hat{p}=0.471 \), and the sample size \( n = 866 \).

Step2: Find the critical value \( z_{\alpha/2} \)

For a 95% confidence interval, \( \alpha=1 - 0.95=0.05 \), so \( \alpha/2=0.025 \). From the standard normal distribution table, \( z_{\alpha/2}=z_{0.025}=1.96 \).

Step3: Calculate the margin of error \( E \)

The formula for the margin of error for a proportion is \( E = z_{\alpha/2}\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \).
Substitute the values: \( \hat{p}=0.471 \), \( n = 866 \), \( z_{\alpha/2}=1.96 \).
First, calculate \( \hat{p}(1 - \hat{p})=0.471\times(1 - 0.471)=0.471\times0.529 = 0.249159 \).
Then \( \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{\frac{0.249159}{866}}\approx\sqrt{0.0002877}\approx0.01696 \).
\( E=1.96\times0.01696\approx0.033 \).

Answer:

\((0.471 - 0.033,0.471+ 0.033)\)