1. **State the problem:** We are given 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 the 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 - $t_{\alpha/2, n-1}$ is the t-score 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-score for 99% confidence and 9 degrees of freedom:** From t-tables or calculator, $t_{0.005,9} = 3.2498$ 6. **Calculate margin of error:** $$ME = t \times \frac{s}{\sqrt{n}} = 3.2498 \times \frac{1.3003}{\sqrt{10}} = 3.2498 \times 0.4111 = 1.3357$$ 7. **Construct confidence interval:** $$\bar{x} - ME = 19.78 - 1.3357 = 18.4443$$ $$\bar{x} + ME = 19.78 + 1.3357 = 21.1157$$ **Final 99% confidence interval:** $$\boxed{(18.4443, 21.1157)}$$ This means we are 99% confident that the true mean length of newborns lies between 18.4443 and 21.1157 inches.