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)

To construct a \( 90\% \) confidence interval for the proportion of registered voters who feel that the economy is the most significant issue, we can follow these steps: 1. **Calculate \(\widehat{p}\):** \[ \widehat{p} = \frac{x}{n} = \frac{932}{1391} \approx 0.6691 \] 2. **Find \(\frac{\alpha}{2}\):** For a \( 90\% \) confidence level, \(\alpha\) is \( 0.10\), thus \(\frac{\alpha}{2} = 0.05\). 3. **Calculate the Margin of Error (E):** First, we find the critical value \( z_{\alpha/2} \) for a \( 90\% \) confidence level, which is approximately \( 1.645 \). Then calculate the standard error (SE): \[ SE = \sqrt{\frac{\widehat{p}(1 - \widehat{p})}{n}} = \sqrt{\frac{0.6691(1 - 0.6691)}{1391}} \approx 0.0270 \] Now calculate the margin of error: \[ E = z_{\alpha/2} \times SE = 1.645 \times 0.0270 \approx 0.0443 \] 4. **Construct the Confidence Interval:** Now we can calculate the confidence interval: \[ \text{Lower limit} = \widehat{p} - E \approx 0.6691 - 0.0443 \approx 0.6248 \] \[ \text{Upper limit} = \widehat{p} + E \approx 0.6691 + 0.0443 \approx 0.7134 \] Therefore, we are \( 90\% \) confident that the proportion of voters who feel that the economy is the most significant issue is between **0.6248 and 0.7134** (rounded to four decimal places). Summary of Answers: \(\widehat{p} = 0.6691\) \(\frac{\alpha}{2} = 0.05\) Margin of Error: \( E = 0.0443 \) We are \( 90 \% \) confident that the proportion of voters who feel that the economy is the most significant issue is between **0.6248 and 0.7134**.