To calculate a 95% confidence interval for the proportion of people who prefer seafood, we can use the formula for a confidence interval for a population proportion.
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Determine the Sample Proportion (\( \hat{p} \)):
- The sample size (\( n \)) is 500.
- The number of people who preferred seafood is 160. \[ \hat{p} = \frac{x}{n} = \frac{160}{500} = 0.32 \]
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Calculate the Standard Error (SE): The standard error for the proportion is given by the formula: \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \] Substituting the values: \[ SE = \sqrt{\frac{0.32(1 - 0.32)}{500}} = \sqrt{\frac{0.32 \times 0.68}{500}} = \sqrt{\frac{0.2176}{500}} \approx \sqrt{0.0004352} \approx 0.02085 \]
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Determine the Z-score for a 95% Confidence Level: The Z-score for a 95% confidence level is approximately 1.96.
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Calculate the Margin of Error (ME): The margin of error is calculated as: \[ ME = Z \times SE \] Substituting the Z-score and standard error: \[ ME = 1.96 \times 0.02085 \approx 0.0408 \]
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Calculate the Confidence Interval: Finally, we find the confidence interval using: \[ \hat{p} - ME \quad \text{to} \quad \hat{p} + ME \] Substituting in our values: \[ 0.32 - 0.0408 \quad \text{to} \quad 0.32 + 0.0408 \] This gives us: \[ 0.2792 \quad \text{to} \quad 0.3608 \]
Thus, the 95% confidence interval for the actual proportion of people who prefer seafood at this cafeteria is approximately (0.279, 0.361).
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