1. **State the problem:** We want to calculate the 90% confidence interval for the mean diameter of items produced by the machine based on a sample of 7 items with diameters: 8, 11, 13, 10, 11, 7, 9. 2. **Formula and explanation:** 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 - $t_{\alpha/2, n-1}$ is the t-score for confidence level $1-\alpha$ and degrees of freedom $n-1$ - $s$ is the sample standard deviation - $n$ is the sample size 3. **Calculate the sample mean $\bar{x}$:** $$\bar{x} = \frac{8 + 11 + 13 + 10 + 11 + 7 + 9}{7} = \frac{69}{7} = 9.857$$ 4. **Calculate the sample standard deviation $s$:** First, find squared deviations: $$(8-9.857)^2 = 3.448, (11-9.857)^2 = 1.306, (13-9.857)^2 = 9.877, (10-9.857)^2 = 0.020, (11-9.857)^2 = 1.306, (7-9.857)^2 = 8.163, (9-9.857)^2 = 0.734$$ Sum of squared deviations: $$3.448 + 1.306 + 9.877 + 0.020 + 1.306 + 8.163 + 0.734 = 24.854$$ Sample variance: $$s^2 = \frac{24.854}{7-1} = \frac{24.854}{6} = 4.142$$ Sample standard deviation: $$s = \sqrt{4.142} = 2.036$$ 5. **Find the t-score for 90% confidence and 6 degrees of freedom:** From t-distribution tables or calculator, $$t_{0.05,6} = 1.943$$ 6. **Calculate the margin of error:** $$E = t \times \frac{s}{\sqrt{n}} = 1.943 \times \frac{2.036}{\sqrt{7}} = 1.943 \times 0.770 = 1.495$$ 7. **Calculate the confidence interval:** $$\left( \bar{x} - E, \bar{x} + E \right) = (9.857 - 1.495, 9.857 + 1.495) = (8.362, 11.352)$$ **Final answer:** The 90% confidence interval for the mean diameter is approximately **(8.36, 11.35)**.