1. **Problem statement:** We have 5 measurements of a ball bearing diameter: 8.1, 9.1, 8.9, 8.9, 9.1 mm. The true diameter is 9.0 mm. (a) Calculate the unbiased estimates of the mean and variance of the measurement error. 2. **Calculate measurement errors:** The measurement error for each measurement is $\text{error}_i = \text{measurement}_i - 9.0$. Errors: $8.1 - 9.0 = -0.9$, $9.1 - 9.0 = 0.1$, $8.9 - 9.0 = -0.1$, $8.9 - 9.0 = -0.1$, $9.1 - 9.0 = 0.1$. 3. **Calculate the sample mean of errors:** $$\bar{e} = \frac{-0.9 + 0.1 - 0.1 - 0.1 + 0.1}{5} = \frac{-0.9}{5} = -0.18$$ 4. **Calculate the unbiased sample variance of errors:** Use formula: $$s^2 = \frac{1}{n-1} \sum_{i=1}^n (e_i - \bar{e})^2$$ Calculate each squared deviation: $(-0.9 + 0.18)^2 = (-0.72)^2 = 0.5184$ $(0.1 + 0.18)^2 = (0.28)^2 = 0.0784$ $(-0.1 + 0.18)^2 = (0.08)^2 = 0.0064$ $(-0.1 + 0.18)^2 = 0.0064$ $(0.1 + 0.18)^2 = 0.0784$ Sum: $0.5184 + 0.0784 + 0.0064 + 0.0064 + 0.0784 = 0.688$ Divide by $n-1=4$: $$s^2 = \frac{0.688}{4} = 0.172$$ 5. **Answer for (a):** Unbiased estimate of mean error: $-0.18$ mm Unbiased estimate of variance of error: $0.172$ mm$^2$ 6. **(b) Find 90% confidence interval for mean error assuming normality:** Sample size $n=5$, sample mean $\bar{e} = -0.18$, sample standard deviation $s = \sqrt{0.172} \approx 0.414$ Degrees of freedom $df = n-1 = 4$. From t-distribution table, $t_{0.05,4} \approx 2.132$ (two-tailed 90% CI). Margin of error: $$ME = t_{0.05,4} \times \frac{s}{\sqrt{n}} = 2.132 \times \frac{0.414}{\sqrt{5}} = 2.132 \times 0.185 = 0.394$$ Confidence interval: $$\bar{e} \pm ME = -0.18 \pm 0.394 = (-0.574, 0.214)$$ 7. **Answer for (b):** The 90% confidence interval for the mean measurement error is from $-0.574$ mm to $0.214$ mm.