Answer-first summary for fast verification
Answer: [−3.466%, 11.466%]
## Explanation This is a confidence interval problem for the population mean when the population variance is unknown. Here are the key steps: **Given:** - Sample size (n) = 30 - Sample mean (x̄) = 4% - Sample standard deviation (s) = 20% - Standard error (S<sub>x̄</sub>) = 3.651% - Confidence level = 95% **Key Concepts:** 1. **Degrees of freedom**: Since we're estimating the population standard deviation from the sample, we use n - 1 = 30 - 1 = 29 degrees of freedom 2. **Critical t-value**: For a 95% confidence interval with 29 degrees of freedom and two-tailed test, we use t<sub>29,2.5</sub> = 2.045 3. **Confidence interval formula**: $$CI = \bar{x} \pm t_{\alpha/2, df} \times S_{\bar{x}}$$ **Calculation:** - Lower bound = 4% - 2.045 × 3.651% = 4% - 7.466% = -3.466% - Upper bound = 4% + 2.045 × 3.651% = 4% + 7.466% = 11.466% Therefore, the 95% confidence interval is **[-3.466%, 11.466%]**. **Why other options are incorrect:** - **Option B**: Uses incorrect critical value or calculation - **Option C**: Uses z-value instead of t-value (likely using 1.96 instead of 2.045) - **Option D**: Uses incorrect degrees of freedom or critical value
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For a sample of the past 30 monthly stock returns for McCreaery, Inc., the mean return is 4% and the sample standard deviation is 20%. The population variance is unknown but the standard error of the sample mean is estimated to be: . What is the 95% confidence interval for the mean monthly return?
A
[−3.466%, 11.466%]
B
[−3.453%, 11.453%]
C
[−2.201%, 10.201%]
D
[−2.194%, 10.194%]
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