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