1. **Problem statement:** We have a sample from a normal distribution with unknown mean $\mu$ and unknown variance $\sigma^2$. The observed values are 3.1, 3.5, 2.6, 3.4, 3.8, 3.0, 2.9, and 2.2. We want to find the shortest confidence intervals for $\mu$ at confidence levels 0.90, 0.95, and 0.99. 2. **Formula and explanation:** When variance is unknown, the confidence interval for the mean $\mu$ is given by: $$\bar{x} \pm t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}}$$ where: - $\bar{x}$ is the sample mean - $s$ is the sample standard deviation - $n$ is the sample size - $t_{\alpha/2, n-1}$ is the critical value from the Student's t-distribution with $n-1$ degrees of freedom - $\alpha = 1 - \text{confidence level}$ 3. **Calculate sample mean $\bar{x}$:** $$\bar{x} = \frac{3.1 + 3.5 + 2.6 + 3.4 + 3.8 + 3.0 + 2.9 + 2.2}{8} = \frac{24.5}{8} = 3.0625$$ 4. **Calculate sample standard deviation $s$:** First, calculate squared deviations: $$\sum (x_i - \bar{x})^2 = (3.1-3.0625)^2 + (3.5-3.0625)^2 + (2.6-3.0625)^2 + (3.4-3.0625)^2 + (3.8-3.0625)^2 + (3.0-3.0625)^2 + (2.9-3.0625)^2 + (2.2-3.0625)^2$$ $$= 0.0014 + 0.1914 + 0.2139 + 0.1139 + 0.5439 + 0.0039 + 0.0264 + 0.7456 = 1.8404$$ Sample variance: $$s^2 = \frac{1.8404}{8-1} = \frac{1.8404}{7} = 0.2629$$ Sample standard deviation: $$s = \sqrt{0.2629} = 0.5128$$ 5. **Degrees of freedom:** $$df = n - 1 = 7$$ 6. **Find critical t-values for each confidence level:** - For 90% confidence ($\alpha=0.10$): $t_{0.05,7} = 1.895$ (from t-tables) - For 95% confidence ($\alpha=0.05$): $t_{0.025,7} = 2.365$ - For 99% confidence ($\alpha=0.01$): $t_{0.005,7} = 3.499$ 7. **Calculate margin of error for each confidence level:** $$ME = t_{\alpha/2,7} \cdot \frac{s}{\sqrt{8}}$$ Calculate $\frac{s}{\sqrt{8}}$: $$\frac{0.5128}{\sqrt{8}} = \frac{0.5128}{2.8284} = 0.1813$$ - For 90%: $$ME = 1.895 \times 0.1813 = 0.3435$$ - For 95%: $$ME = 2.365 \times 0.1813 = 0.4287$$ - For 99%: $$ME = 3.499 \times 0.1813 = 0.6345$$ 8. **Construct confidence intervals:** - 90% CI: $$3.0625 \pm 0.3435 = (2.719, 3.406)$$ - 95% CI: $$3.0625 \pm 0.4287 = (2.634, 3.491)$$ - 99% CI: $$3.0625 \pm 0.6345 = (2.428, 3.697)$$ **Final answers:** - 90% confidence interval for $\mu$: $(2.72, 3.41)$ - 95% confidence interval for $\mu$: $(2.63, 3.49)$ - 99% confidence interval for $\mu$: $(2.43, 3.70)$