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Answer :
Final answer:
Therefore, we estimate with 90% confidence that the true proportion of regular voters in the college town is between 38.8% and 49.2%.
Explanation:
To estimate the true proportion of regular voters in the college town with 90% confidence, we can use a confidence interval. First, we calculate the sample proportion by dividing the number of regular voters in the sample by the sample size: 110/250 = 0.44. Next, we calculate the standard deviation using the formula sqrt((p-hat * (1-p-hat))/n), where p-hat is the sample proportion and n is the sample size: √((0.44 * (1-0.44))/250) = 0.0314. Using a Z-table or calculator, we find the Z-score for a 90% confidence level is 1.645.
Now, we can calculate the margin of error by multiplying the Z-score by the standard deviation: 1.645 * 0.0314 = 0.0516. To find the lower and upper bounds of the confidence interval, we subtract and add the margin of error to the sample proportion: 0.44 - 0.0516 = 0.3884 (lower bound), 0.44 + 0.0516 = 0.4916 (upper bound).
Therefore, we estimate with 90% confidence that the true proportion of regular voters in the college town is between 38.8% and 49.2%. This means that if we were to repeat the survey multiple times, 90% of the resulting confidence intervals would contain the true proportion of regular voters.
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