A poll of 1391 voters asked what the most significant issue was in the upcoming election. 932 of those polled answered the economy. Construct a \( 90 \% \) confidence interval to estimate the proportion of registered voters who feel that the economy is the most significant issue. \( \widehat{p}= \) \( \frac{\alpha}{2}= \) Margin of Error: \( E= \) We are \( 90 \% \) confident that the proportion of voters who feel that the economy is the most significant issue is between (round to 4 decimal places)

Let's construct a \( 90\% \) confidence interval to estimate the proportion of registered voters who believe the economy is the most significant issue in the upcoming election. ### Given Data: - **Sample Size (\( n \))**: 1391 voters - **Number of Successes (\( x \))**: 932 voters believe the economy is the most significant issue ### Step 1: Calculate the Sample Proportion (\( \widehat{p} \)) \[ \widehat{p} = \frac{x}{n} = \frac{932}{1391} \approx 0.6690 \] ### Step 2: Determine \( \frac{\alpha}{2} \) For a \( 90\% \) confidence level: \[ \alpha = 1 - 0.90 = 0.10 \\ \frac{\alpha}{2} = 0.05 \] ### Step 3: Find the Z-Score for \( \frac{\alpha}{2} \) For \( \frac{\alpha}{2} = 0.05 \), the corresponding Z-score (\( z_{\alpha/2} \)) is: \[ z_{0.05} = 1.645 \] ### Step 4: Calculate the Standard Error (SE) \[ SE = \sqrt{\frac{\widehat{p}(1 - \widehat{p})}{n}} = \sqrt{\frac{0.6690 \times 0.3310}{1391}} \approx \sqrt{\frac{0.221439}{1391}} \approx \sqrt{0.0001592} \approx 0.0126 \] ### Step 5: Compute the Margin of Error (\( E \)) \[ E = z_{\alpha/2} \times SE = 1.645 \times 0.0126 \approx 0.0207 \] ### Step 6: Construct the Confidence Interval \[ \widehat{p} \pm E = 0.6690 \pm 0.0207 \] \[ \text{Lower Bound} = 0.6690 - 0.0207 = 0.6483 \\ \text{Upper Bound} = 0.6690 + 0.0207 = 0.6897 \] ### Final Confidence Interval We are \( 90\% \) confident that the proportion of voters who feel the economy is the most significant issue is between **0.6483** and **0.6897**. \[ \boxed{ \begin{aligned} \widehat{p} &= \frac{932}{1391} \approx 0.6690 \\ \frac{\alpha}{2} &= 0.05 \\ \text{Margin of Error: } E &= 0.0207 \\ \text{Confidence Interval: } &\ 0.6483 \leq p \leq 0.6897 \end{aligned} } \]