1. **Problem Statement:** We are given a sample of incubation periods (in years) for 14 HIV infected individuals: 12, 9.5, 10.5, 6.3, 5.2, 13.1, 13.5, 12.5, 10.7, 7.2, 8.1, 14.9, 7.9, 6.5. We need to: - i. Determine the sample mean. - ii. Determine the sample standard deviation. - iii. Determine the coefficient of variation. - iv. Construct a 95% confidence interval for the true average incubation period and comment on the assumption that the average is 8 years. 2. **Formulas and Important Rules:** - Sample mean: $\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ - Sample standard deviation: $s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}$ - Coefficient of variation (CV): $CV = \frac{s}{\bar{x}} \times 100\%$ - 95% confidence interval for mean (when population std dev unknown): $$\bar{x} \pm t_{\alpha/2, n-1} \times \frac{s}{\sqrt{n}}$$ where $t_{\alpha/2, n-1}$ is the t-critical value for $n-1$ degrees of freedom. 3. **Calculations:** - Sample size: $n=14$ - Data: $12, 9.5, 10.5, 6.3, 5.2, 13.1, 13.5, 12.5, 10.7, 7.2, 8.1, 14.9, 7.9, 6.5$ **i. Sample mean:** $$\bar{x} = \frac{12 + 9.5 + 10.5 + 6.3 + 5.2 + 13.1 + 13.5 + 12.5 + 10.7 + 7.2 + 8.1 + 14.9 + 7.9 + 6.5}{14}$$ Calculate sum: $$= 12 + 9.5 + 10.5 + 6.3 + 5.2 + 13.1 + 13.5 + 12.5 + 10.7 + 7.2 + 8.1 + 14.9 + 7.9 + 6.5 = 137.9$$ So, $$\bar{x} = \frac{137.9}{14} = 9.85$$ **ii. Sample standard deviation:** Calculate each squared deviation $(x_i - \bar{x})^2$: - $(12 - 9.85)^2 = 4.6225$ - $(9.5 - 9.85)^2 = 0.1225$ - $(10.5 - 9.85)^2 = 0.4225$ - $(6.3 - 9.85)^2 = 12.6025$ - $(5.2 - 9.85)^2 = 21.4225$ - $(13.1 - 9.85)^2 = 10.5625$ - $(13.5 - 9.85)^2 = 12.8225$ - $(12.5 - 9.85)^2 = 7.0225$ - $(10.7 - 9.85)^2 = 0.7225$ - $(7.2 - 9.85)^2 = 7.0225$ - $(8.1 - 9.85)^2 = 3.0625$ - $(14.9 - 9.85)^2 = 25.5025$ - $(7.9 - 9.85)^2 = 3.8025$ - $(6.5 - 9.85)^2 = 11.2225$ Sum of squared deviations: $$= 4.6225 + 0.1225 + 0.4225 + 12.6025 + 21.4225 + 10.5625 + 12.8225 + 7.0225 + 0.7225 + 7.0225 + 3.0625 + 25.5025 + 3.8025 + 11.2225 = 121.715$$ Sample variance: $$s^2 = \frac{121.715}{14 - 1} = \frac{121.715}{13} = 9.3627$$ Sample standard deviation: $$s = \sqrt{9.3627} = 3.06$$ **iii. Coefficient of variation:** $$CV = \frac{3.06}{9.85} \times 100\% = 31.07\%$$ **iv. 95% Confidence Interval:** - Degrees of freedom: $df = 13$ - From t-tables, $t_{0.025, 13} \approx 2.160$ - Standard error: $$SE = \frac{s}{\sqrt{n}} = \frac{3.06}{\sqrt{14}} = \frac{3.06}{3.7417} = 0.818$$ Confidence interval: $$9.85 \pm 2.160 \times 0.818 = 9.85 \pm 1.767$$ So, $$\text{Lower limit} = 9.85 - 1.767 = 8.08$$ $$\text{Upper limit} = 9.85 + 1.767 = 11.62$$ **Comment:** The 95% confidence interval for the average incubation period is approximately $[8.08, 11.62]$ years. Since 8 years is just below the lower bound, the data suggests the average incubation period is likely greater than 8 years, so the assumption that it is exactly 8 years is questionable based on this sample.