1. **State the problem:** We have a sample of growing stock volumes (in m^3/ha) from randomly arranged sample plots: -803, -682, -347, -808, -664, -472, -613, -199, -403, -564, -267, -396. We want to estimate the population mean volume and the 95% confidence interval for this mean. 2. **Formula and explanation:** - The sample mean $\bar{x}$ estimates the population mean $\mu$. - The 95% confidence interval for the mean when the population standard deviation is unknown is given by: $$\bar{x} \pm t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}$$ where: - $n$ is the sample size, - $s$ is the sample standard deviation, - $t_{\alpha/2, n-1}$ is the t-value from the t-distribution with $n-1$ degrees of freedom for 95% confidence (two-tailed). 3. **Calculate sample mean $\bar{x}$:** $$\bar{x} = \frac{\sum x_i}{n} = \frac{-803 -682 -347 -808 -664 -472 -613 -199 -403 -564 -267 -396}{12}$$ Calculate the sum: $$-803 -682 -347 -808 -664 -472 -613 -199 -403 -564 -267 -396 = -6218$$ So: $$\bar{x} = \frac{-6218}{12} = -518.2$$ 4. **Calculate sample standard deviation $s$:** First, calculate each squared deviation $(x_i - \bar{x})^2$, sum them, then divide by $n-1=11$, and take the square root. Squared deviations: $(-803 + 518.2)^2 = (-284.8)^2 = 81100.0$ $(-682 + 518.2)^2 = (-163.8)^2 = 26826.4$ $(-347 + 518.2)^2 = 171.2^2 = 29310.4$ $(-808 + 518.2)^2 = (-289.8)^2 = 84088.0$ $(-664 + 518.2)^2 = (-145.8)^2 = 21258.6$ $(-472 + 518.2)^2 = 46.2^2 = 2134.4$ $(-613 + 518.2)^2 = (-94.8)^2 = 8987.0$ $(-199 + 518.2)^2 = 319.2^2 = 101888.6$ $(-403 + 518.2)^2 = 115.2^2 = 13272.0$ $(-564 + 518.2)^2 = (-45.8)^2 = 2097.6$ $(-267 + 518.2)^2 = 251.2^2 = 63104.1$ $(-396 + 518.2)^2 = 122.2^2 = 14932.8$ Sum of squared deviations: $$81100 + 26826.4 + 29310.4 + 84088 + 21258.6 + 2134.4 + 8987 + 101888.6 + 13272 + 2097.6 + 63104.1 + 14932.8 = 448997.9$$ Sample variance: $$s^2 = \frac{448997.9}{11} = 40817.1$$ Sample standard deviation: $$s = \sqrt{40817.1} = 202.0$$ 5. **Find the t-value for 95% confidence and 11 degrees of freedom:** From t-tables or calculator, $t_{0.025, 11} \approx 2.201$. 6. **Calculate margin of error:** $$ME = t \times \frac{s}{\sqrt{n}} = 2.201 \times \frac{202.0}{\sqrt{12}} = 2.201 \times 58.3 = 128.3$$ 7. **Calculate confidence interval:** Lower limit: $$-518.2 - 128.3 = -646.5$$ Upper limit: $$-518.2 + 128.3 = -389.9$$ **Final answers:** - Estimated population mean = $-518.2$ m^3/ha - 95% confidence interval = $[-646.5, -389.9]$ m^3/ha All values rounded to one decimal place as requested.