To construct a 95% confidence interval for the proportion of adult Seattle residents who don't believe they can contract an STI, we can follow these steps:
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Calculate the sample proportion \(\hat{p}\): \[ \hat{p} = \frac{x}{n} = \frac{750}{1000} = 0.75 \]
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Determine the standard error (SE) of the proportion: \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.75(1 - 0.75)}{1000}} = \sqrt{\frac{0.75 \cdot 0.25}{1000}} = \sqrt{\frac{0.1875}{1000}} \approx \sqrt{0.0001875} \approx 0.01365 \]
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Find the z-score for a 95% confidence level:
- The z-score given is 1.96.
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Calculate the margin of error (ME): \[ ME = z \cdot SE = 1.96 \cdot 0.01365 \approx 0.0268 \]
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Construct the confidence interval: \[ \text{Lower bound} = \hat{p} - ME = 0.75 - 0.0268 \approx 0.7232 \] \[ \text{Upper bound} = \hat{p} + ME = 0.75 + 0.0268 \approx 0.7768 \]
Thus, the 95% confidence interval for the proportion of adult Seattle residents who don't believe they can contract an STI is approximately:
\[ (0.7232, 0.7768) \]
Considering the options provided, the correct confidence interval is: (.723, .777).
2 answers