1. **State the problem:** We are given a sample of net worth values (in millions) and asked to find the confidence interval estimate for the population mean $\mu$. 2. **Data given:** The sample values are: 246, 208, 190, 169, 156, 151, 146, 146, 146, 141. 3. **Formula for confidence interval:** $$\bar{x} \pm t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}$$ where: - $\bar{x}$ is the sample mean - $t_{\alpha/2, n-1}$ is the t-score for confidence level with $n-1$ degrees of freedom - $s$ is the sample standard deviation - $n$ is the sample size 4. **Calculate the sample mean $\bar{x}$:** $$\bar{x} = \frac{246 + 208 + 190 + 169 + 156 + 151 + 146 + 146 + 146 + 141}{10} = \frac{1659}{10} = 165.9$$ 5. **Calculate the sample standard deviation $s$:** First, find each squared deviation: $$(246 - 165.9)^2 = 6408.81$$ $$(208 - 165.9)^2 = 1776.41$$ $$(190 - 165.9)^2 = 577.21$$ $$(169 - 165.9)^2 = 9.61$$ $$(156 - 165.9)^2 = 98.01$$ $$(151 - 165.9)^2 = 220.81$$ $$(146 - 165.9)^2 = 396.01$$ $$(146 - 165.9)^2 = 396.01$$ $$(146 - 165.9)^2 = 396.01$$ $$(141 - 165.9)^2 = 621.21$$ Sum of squared deviations: $$6408.81 + 1776.41 + 577.21 + 9.61 + 98.01 + 220.81 + 396.01 + 396.01 + 396.01 + 621.21 = 10599.1$$ Sample variance: $$s^2 = \frac{10599.1}{10 - 1} = \frac{10599.1}{9} = 1177.68$$ Sample standard deviation: $$s = \sqrt{1177.68} \approx 34.31$$ 6. **Determine the t-score for 95% confidence and 9 degrees of freedom:** From t-distribution tables, $t_{0.025,9} \approx 2.262$ 7. **Calculate the margin of error:** $$E = t \times \frac{s}{\sqrt{n}} = 2.262 \times \frac{34.31}{\sqrt{10}} = 2.262 \times 10.85 = 24.54$$ 8. **Calculate the confidence interval:** $$\bar{x} - E = 165.9 - 24.54 = 141.4$$ $$\bar{x} + E = 165.9 + 24.54 = 190.4$$ 9. **Final answer:** The 95% confidence interval estimate for the population mean $\mu$ is: $$141.4 \text{ million} < \mu < 190.4 \text{ million}$$ Rounded to one decimal place as requested.