1. **State the problem:** We are given a sample size $n=10$, sample mean $\bar{x}=12.0$, sample standard deviation $s=15$, and a confidence level $E.L=95\%$. We need to find the confidence interval for the population mean. 2. **Formula used:** For a small sample size ($n<30$) and unknown population standard deviation, we use the t-distribution confidence interval formula: $$\bar{x} \pm t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}$$ where $t_{\alpha/2, n-1}$ is the t-critical value for $\alpha=1-0.95=0.05$ and degrees of freedom $n-1=9$. 3. **Find the t-critical value:** From t-tables or calculator, $t_{0.025,9} \approx 2.262$. 4. **Calculate the margin of error (ME):** $$ME = t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}} = 2.262 \times \frac{15}{\sqrt{10}}$$ Calculate $\sqrt{10} \approx 3.162$: $$ME = 2.262 \times \frac{15}{3.162} = 2.262 \times 4.743 = 10.73$$ 5. **Calculate the confidence interval:** Lower limit: $$12.0 - 10.73 = 1.27$$ Upper limit: $$12.0 + 10.73 = 22.73$$ 6. **Interpretation:** We are 95% confident that the true population mean lies between $1.27$ and $22.73$.