1. **State the problem:** We have a sample of newborn lengths in inches: 19.5, 17.7, 20, 19.6, 18.3, 20.5, 18.7, 20.6, 20.9, 22. We assume lengths are normally distributed and want to find the 99% confidence interval for the mean length. 2. **Formula and rules:** The confidence interval for the mean when the population standard deviation is unknown and sample size is small uses the t-distribution: $$\bar{x} \pm t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}$$ where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $t_{\alpha/2, n-1}$ is the t-critical value for confidence level $1-\alpha$ and $n-1$ degrees of freedom. 3. **Calculate sample mean $\bar{x}$:** $$\bar{x} = \frac{19.5 + 17.7 + 20 + 19.6 + 18.3 + 20.5 + 18.7 + 20.6 + 20.9 + 22}{10} = \frac{197.8}{10} = 19.78$$ 4. **Calculate sample standard deviation $s$:** First, find squared deviations: $$(19.5 - 19.78)^2 = 0.0784$$ $$(17.7 - 19.78)^2 = 4.3264$$ $$(20 - 19.78)^2 = 0.0484$$ $$(19.6 - 19.78)^2 = 0.0324$$ $$(18.3 - 19.78)^2 = 2.1904$$ $$(20.5 - 19.78)^2 = 0.5184$$ $$(18.7 - 19.78)^2 = 1.1664$$ $$(20.6 - 19.78)^2 = 0.6724$$ $$(20.9 - 19.78)^2 = 1.2544$$ $$(22 - 19.78)^2 = 4.9284$$ Sum of squared deviations: $$0.0784 + 4.3264 + 0.0484 + 0.0324 + 2.1904 + 0.5184 + 1.1664 + 0.6724 + 1.2544 + 4.9284 = 15.216$$ Sample variance: $$s^2 = \frac{15.216}{10 - 1} = \frac{15.216}{9} = 1.6907$$ Sample standard deviation: $$s = \sqrt{1.6907} = 1.3003$$ 5. **Find t-critical value for 99% confidence and 9 degrees of freedom:** From t-tables or calculator, $t_{0.005,9} = 3.2498$ 6. **Calculate margin of error:** $$ME = t \times \frac{s}{\sqrt{n}} = 3.2498 \times \frac{1.3003}{\sqrt{10}} = 3.2498 \times 0.4111 = 1.3353$$ 7. **Construct confidence interval:** $$\left( \bar{x} - ME, \bar{x} + ME \right) = (19.78 - 1.3353, 19.78 + 1.3353) = (18.4447, 21.1153)$$ **Final answer:** The 99% confidence interval for the mean length is approximately $$(18.4447, 21.1153)$$ inches.