To determine the margin of error for a 99% confidence interval of the population proportion of people planning to vote for a particular candidate, we can follow these steps:
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Calculate the sample proportion \((\hat{p})\): \[ \hat{p} = \frac{x}{n} = \frac{230}{850} \approx 0.2706 \]
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Calculate the standard error (SE) of the sample proportion: The standard error of the sample proportion is given by: \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.2706(1 - 0.2706)}{850}} \approx \sqrt{\frac{0.2706 \times 0.7294}{850}} \approx \sqrt{\frac{0.1976}{850}} \approx \sqrt{0.000232} \approx 0.0152 \]
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Find the critical value: For a 99% confidence level, the critical value \(z^\) can be found using a z-table or standard normal distribution. The critical value for a 99% confidence interval is approximately: \[ z^ \approx 2.576 \]
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Calculate the margin of error (ME): The margin of error is calculated as: \[ ME = z^* \times SE \approx 2.576 \times 0.0152 \approx 0.0392 \]
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Round the margin of error: Rounding 0.0392 to the nearest hundredth gives: \[ ME \approx 0.04 \]
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Calculate the confidence interval: The lower and upper bounds of the confidence interval for the population proportion can be calculated as: \[ \hat{p} - ME \quad \text{and} \quad \hat{p} + ME \] So, \[ 0.2706 - 0.0392 \approx 0.2314 \quad \text{and} \quad 0.2706 + 0.0392 \approx 0.3098 \]
Thus, we can conclude:
- Margin of error = 0.04.
- With 99% confidence, the population proportion of people in the neighborhood who are planning on voting for a particular candidate falls between 0.2314 (or 23.14%) and 0.3098 (or 30.98%).
Interpretation:
This means that we are 99% confident that the true proportion of all residents in the neighborhood who plan to vote for the candidate lies between approximately 23.14% and 30.98%.