To estimate the true proportion of regular voters with 90% confidence, we first need to calculate the proportion of regular voters in the sample and then use the formula for the confidence interval for a proportion.
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Calculate the sample proportion (\( p \)): \[ p = \frac{x}{n} = \frac{110}{250} = 0.44 \]
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Calculate the standard error (SE): \[ SE = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.44(1 - 0.44)}{250}} = \sqrt{\frac{0.44 \cdot 0.56}{250}} = \sqrt{\frac{0.2464}{250}} = \sqrt{0.0009856} \approx 0.0314 \]
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Find the critical value for a 90% confidence interval: For a 90% confidence level, the critical value (z*) can be found from the standard normal distribution table or by using a calculator. The z-value for a 90% confidence level is approximately 1.645.
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Calculate the margin of error (ME): \[ ME = z^* \times SE = 1.645 \times 0.0314 \approx 0.0516 \]
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Construct the confidence interval: \[ \text{Lower limit} = p - ME = 0.44 - 0.0516 \approx 0.3884 \] \[ \text{Upper limit} = p + ME = 0.44 + 0.0516 \approx 0.4916 \]
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Round to the nearest thousandth: \[ \text{Lower limit} \approx 0.388 \quad \text{Upper limit} \approx 0.492 \]
Thus, the 90% confidence interval estimate for the true proportion of regular voters is:
\[ (0.388, 0.492) \]
Rounding these values to the required format, the final answer is:
\[ (39, 49) \]
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