The correct answer is A: [-3.466%, 11.466%]. The explanation for this is based on the principles of constructing a confidence interval for the mean when the population variance is unknown. Since the sample size is 30, the degrees of freedom for the t-distribution are 30 - 1 = 29. A t-reliability factor is used in place of the z-score because the population variance is not known. For a 95% confidence interval and a two-tailed test, the critical t-value from the t-table for 29 degrees of freedom at the 2.5th percentile is 2.045. The formula for the confidence interval is:

$\text{CI} = \bar{x} \pm t_{\text{critical}} \times \frac{S_x}{\sqrt{n}}$

where:

• $\bar{x}$ is the sample mean (4% in this case),
• $t_{\text{critical}}$ is the critical t-value (2.045),
• $S_x$ is the sample standard deviation (20%),
• $n$ is the sample size (30).

Plugging in the values, we get:

$\text{CI} = 4\% \pm 2.045 \times \frac{20\%}{\sqrt{30}}$

Calculating the standard error of the mean ($S_x / \sqrt{n}$) gives us 3.651%. Multiplying this by the critical t-value and then adding and subtracting from the sample mean gives us the confidence interval:

$\text{CI} = [4\% - 2.045 \times 3.651\%, 4\% + 2.045 \times 3.651\%]$ $\text{CI} = [-3.466\%, 11.466\%]$

This interval represents the range within which we can be 95% confident that the true mean monthly return of McCreary, Inc. lies.