1. **Problem Statement:** Calculate the margin of error for a 90% confidence interval for a population mean $\mu$ when $\sigma=12$ and sample size $n=9$. Also, identify what else is needed to obtain the confidence interval. 2. **Formula for Margin of Error (ME):** $$ ME = z^* \times \frac{\sigma}{\sqrt{n}} $$ where $z^*$ is the critical z-value for the confidence level. 3. **Find $z^*$ for 90% confidence:** The 90% confidence level corresponds to $\alpha=0.10$, so $\frac{\alpha}{2}=0.05$. From z-tables, $z^* = 1.645$. 4. **Calculate ME:** $$ ME = 1.645 \times \frac{12}{\sqrt{9}} = 1.645 \times \frac{12}{3} = 1.645 \times 4 = 6.58 $$ 5. **What else is needed?** To obtain the confidence interval, you also need the sample mean $\bar{x}$. **Final answer:** The margin of error is $6.58$. To construct the confidence interval, you need the sample mean $\bar{x}$.