Answer: [-3.467%, 11.467%]

## Explanation To calculate the 95% confidence interval for the mean monthly return when the population variance is unknown, we use the t-distribution. **Given:** - Sample mean (x̄) = 4% - Sample standard deviation (s) = 20% - Sample size (n) = 30 - Standard error (SE) = 20%/√30 = 3.65% - Degrees of freedom = n - 1 = 29 **Formula for confidence interval:** \[ \text{CI} = \bar{x} \pm t_{\alpha/2, df} \times SE \] For a 95% confidence interval: - α = 0.05 - α/2 = 0.025 - We need t-value for 29 degrees of freedom at 2.5% significance level From the t-table provided: t₂₉,₂.₅% = 2.045 **Calculation:** \[ \text{Margin of error} = 2.045 \times 3.65\% = 7.467\% \] \[ \text{Lower bound} = 4\% - 7.467\% = -3.467\% \] \[ \text{Upper bound} = 4\% + 7.467\% = 11.467\% \] Therefore, the 95% confidence interval is **[-3.467%, 11.467%]**, which corresponds to option A. **Why not other options:** - Option B uses the wrong t-value or calculation - Option C uses t-value for 30 degrees of freedom (2.042) instead of 29 - Option D has inconsistent bounds

Author: LeetQuiz .