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Answer: [14.4 miles; 21.6 miles]
**Calculation Steps:** 1. **Standard Error (SE)**: SE = σ/√n = 14/√100 = 14/10 = 1.4 miles 2. **Z-score for 99% Confidence Interval**: For a 99% confidence interval, the z-score (reliability factor) is 2.58 (this corresponds to α/2 = 0.005 in each tail). 3. **Margin of Error (ME)**: ME = z × SE = 2.58 × 1.4 = 3.612 miles 4. **Confidence Interval**: - Lower limit = Sample mean - ME = 18 - 3.612 = 14.388 ≈ 14.4 miles - Upper limit = Sample mean + ME = 18 + 3.612 = 21.612 ≈ 21.6 miles **Why other options are incorrect:** - **A [15.26; 20.74]**: Uses incorrect z-score (likely 1.96 for 95% CI) - **B [16.6; 19.4]**: Uses incorrect z-score (likely 1.645 for 90% CI) - **D [12.8; 23.6]**: Uses incorrect standard error calculation (likely forgot to divide by √n) **Key Concepts:** - Confidence intervals estimate population parameters from sample statistics - For normally distributed populations with known standard deviation, use z-distribution - Standard error decreases with larger sample sizes (√n in denominator) - Higher confidence levels require wider intervals (larger z-scores)
Author: Nikitesh Somanthe
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A sample of 100 students is currently renting rooms in the mean distance of 18 miles from a small U.S. College. Assuming that the population is normally distributed and the standard deviation of the sample is 14 miles, what is the 99% confidence interval for the population mean?
A
[15.26 miles; 20.74 miles]
B
[16.6 miles; 19.4 miles]
C
[14.4 miles; 21.6 miles]
D
[12.8 miles; 23.6 miles]