1. **Problem statement:** We have a sample of 9 subjects with ACL tears. The mean laxity is 17.4 mm and the standard deviation is 4.3 mm. We need to find: (a) The estimated standard error of the mean (SEM). (b) The 99% confidence interval (CI) for the population mean. (c) The precision of the estimate. (d) The assumptions for the validity of the CI. 2. **Formulas and rules:** - Standard error of the mean: $$SEM = \frac{s}{\sqrt{n}}$$ where $s$ is sample standard deviation and $n$ is sample size. - Confidence interval for mean (using t-distribution since $n<30$): $$\bar{x} \pm t_{\alpha/2, n-1} \times SEM$$ where $\bar{x}$ is sample mean, $t_{\alpha/2, n-1}$ is the t critical value for confidence level and degrees of freedom $n-1$. - Precision is half the width of the confidence interval. 3. **Calculations:** (a) Calculate SEM: $$SEM = \frac{4.3}{\sqrt{9}} = \frac{4.3}{3} = 1.4333$$ mm (b) Find $t$ critical value for 99% CI with $df=8$: From t-tables, $t_{0.005,8} \approx 3.355$ Calculate margin of error: $$ME = 3.355 \times 1.4333 = 4.805$$ mm Construct CI: $$17.4 \pm 4.805 = (17.4 - 4.805, 17.4 + 4.805) = (12.595, 22.205)$$ mm (c) Precision is half the width of CI: $$Precision = 4.805$$ mm (d) Assumptions: - The sample is a simple random sample from the population. - The population distribution of laxity is approximately normal or the sample size is large enough for the Central Limit Theorem (here $n=9$ so normality is important). - The observations are independent. **Final answers:** (a) SEM = 1.4333 mm (b) 99% CI = (12.595 mm, 22.205 mm) (c) Precision = 4.805 mm (d) Assumptions as stated above.