1. **Problem:** Given the sample data 10, 8, 12, 15, 13, 11, 6, 5 from a normal population, find: a) The point estimate of the population mean. b) The point estimate of the population standard deviation. c) The margin of error for the population mean with 95% confidence. d) The 95% confidence interval for the population mean. 2. **Formulas and rules:** - Sample mean: $\bar{x} = \frac{\sum x_i}{n}$ - Sample standard deviation: $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$ - Margin of error (ME) for mean with confidence level $1-\alpha$: $ME = t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}$ - Confidence interval: $\bar{x} \pm ME$ - Use $t$-distribution because population standard deviation is unknown and sample size is small. 3. **Calculations:** - Sample size $n = 8$ - Sum of data: $10 + 8 + 12 + 15 + 13 + 11 + 6 + 5 = 80$ - Sample mean: $$\bar{x} = \frac{80}{8} = 10$$ - Calculate squared deviations: $(10-10)^2=0$, $(8-10)^2=4$, $(12-10)^2=4$, $(15-10)^2=25$, $(13-10)^2=9$, $(11-10)^2=1$, $(6-10)^2=16$, $(5-10)^2=25$ - Sum of squared deviations: $0 + 4 + 4 + 25 + 9 + 1 + 16 + 25 = 84$ - Sample standard deviation: $$s = \sqrt{\frac{84}{8-1}} = \sqrt{12} \approx 3.4641$$ - Degrees of freedom $df = 7$ - For 95% confidence, $t_{0.025,7} \approx 2.365$ (from t-table) - Margin of error: $$ME = 2.365 \times \frac{3.4641}{\sqrt{8}} = 2.365 \times 1.2247 \approx 2.895$$ - Confidence interval: $$10 \pm 2.895 = (7.105, 12.895)$$ 4. **Interpretation:** - The best estimate of the population mean is 10. - The estimated standard deviation of the population is approximately 3.4641. - With 95% confidence, the true population mean lies between 7.105 and 12.895.