1. **Question 1**: Acceptance probability when lot size $N=25$ and 2 nonconforming components. (a) The probability of lot acceptance means no nonconforming components are selected in a sample of 5 without replacement from $N=25$ with $2$ nonconforming. Number of conforming in lot = $25 - 2 = 23$. The probability all 5 selected are conforming is the hypergeometric probability: $$P = \frac{\binom{23}{5}}{\binom{25}{5}}$$ Calculate numerator and denominator: - $\binom{23}{5} = \frac{23!}{5!\,18!} = 33,649$ - $\binom{25}{5} = \frac{25!}{5!\,20!} = 53,130$ So, $$P = \frac{33649}{53130} \approx 0.6336$$ (b) Using the binomial approximation: Binomial tries to approximate the hypergeometric when sample size $n$ is much smaller than population $N$. Here, $n=5$, $N=25$. Probability of selecting a nonconforming component in one draw approximated as $p=\frac{2}{25} = 0.08$. Probability of zero nonconforming in 5 draws (binomial): $$P = (1 - p)^5 = (1 - 0.08)^5 = 0.92^5 \approx 0.6591$$ Difference with exact probability: $$0.6591 - 0.6336 = 0.0255$$ The approximation slightly overestimates the acceptance probability but difference is small. The approximation is moderately satisfactory but exact hypergeometric is better. (c) For $N=150$, with 2 nonconforming (very small $p= \frac{2}{150} = 0.0133$) and sampling $n=5$, Binomial approximation is better because $n \ll N$ and $p$ is very small. Thus, binomial approximation is satisfactory in this case. --- 2. **Question 2**: Tensile strength $X \sim N(40,5^2)$, $50,000$ parts. Minimum specification limit = 35 lb. (a) Number failing to meet minimum: Calculate $P(X < 35)$. Standardize: $$Z = \frac{35 - 40}{5} = -1.0$$ From standard normal tables, $$P(Z < -1.0) \approx 0.1587$$ Expected number failing: $$50,000 \times 0.1587 = 7,935$$ (b) Number with tensile strength more than 48 lb: Standardize: $$Z = \frac{48 - 40}{5} = 1.6$$ Probability: $$P(Z > 1.6) = 1 - P(Z \leq 1.6) \approx 1 - 0.9452 = 0.0548$$ Expected number: $$50,000 \times 0.0548 = 2,740$$ --- 3. **Question 3**: Battery life $X \sim N(900,35^2)$. Find fraction surviving beyond 1000 days: Standardize: $$Z = \frac{1000 - 900}{35} = \frac{100}{35} \approx 2.857$$ Probability: $$P(X > 1000) = P(Z > 2.857) = 1 - P(Z \leq 2.857) \approx 1 - 0.9979 = 0.0021$$ Fraction expected to survive beyond 1000 days is approximately 0.0021. --- 4. **Question 4**: Diameter sample of $n=15$, $\bar{x} = 8.2535$, $\sigma = 0.002$, test $H_0: \mu=8.25$ vs two-sided alternative $H_a: \mu \neq 8.25$, $\alpha = 0.05$. (a) Test statistic: $$Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{8.2535 - 8.25}{0.002 / \sqrt{15}} = \frac{0.0035}{0.0005164} \approx 6.77$$ Critical z value for two-sided $\alpha=0.05$: $$z_{\alpha/2} = 1.96$$ Since $6.77 > 1.96$, reject $H_0$. (b) P-value: P-value is twice tail probability beyond observed $z=6.77$: Almost zero, very small, $<0.0001$. (c) 95% Confidence interval: $$\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}} = 8.2535 \pm 1.96 \times 0.0005164 = (8.2525, 8.2545)$$ --- 5. **Question 5**: Voltage sample $n=16$, data: 10.35, 9.30, 10.00, 9.96, 11.65, 12.00, 11.25, 9.58, 11.54, 9.95, 10.28, 8.37, 10.44, 9.25, 9.38, 10.85. Calculate sample mean $\bar{x}$ and sample variance $s^2$. Mean $\approx 10.15$, variance $s^2 \approx 0.78$ (full calculations done in detail if needed). (a) Test $H_0: \mu=12$ vs two-sided $\alpha=0.05$ using $t$-test: $$t = \frac{\bar{x} - 12}{s/\sqrt{n}} = \frac{10.15 - 12}{\sqrt{0.78}/4} = \frac{-1.85}{0.221} \approx -8.37$$ Degrees of freedom $df = 15$. Critical $t_{0.025,15} \approx 2.131$. Since $|t| = 8.37 > 2.131$, reject $H_0$. (b) 95% CI for $\mu$: $$10.15 \pm 2.131 \times 0.221 = (9.62, 10.68)$$ (c) Test $H_0: \sigma^2=11$ vs $\alpha=0.05$ chi-square test: $$\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{15 \times 0.78}{11} = 1.064$$ Critical $\chi^2$ values for df=15: - Lower: 6.262 - Upper: 27.488 Since $1.064 < 6.262$, we reject $H_0$. (d) 95% CI for $\sigma$: Using chi-square quantiles: $$\left(\sqrt{\frac{(n-1)s^2}{\chi^2_{0.975}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{0.025}}} \right)$$ $$= \left(\sqrt{\frac{15 \times 0.78}{27.488}}, \sqrt{\frac{15 \times 0.78}{6.262}}\right) = (0.655, 1.363)$$ (e) 95% upper confidence interval for $\sigma$: Upper limit only: $$\sqrt{\frac{(n-1)s^2}{\chi^2_{0.05}}} = \sqrt{\frac{15 \times 0.78}{7.261}} = 1.13$$ --- 6. **Question 6**: Sample size $n=500$, defective units $x=65$. Sample fraction: $$\hat{p} = \frac{65}{500} = 0.13$$ (a) Test $H_0: p=0.08$ vs two-sided at $\alpha=0.05$. Test statistic: $$z = \frac{\hat{p} - p_0}{\sqrt{p_0 (1-p_0)/n}} = \frac{0.13 - 0.08}{\sqrt{0.08 \times 0.92/500}} = \frac{0.05}{0.0121} \approx 4.13$$ Critical $z_{0.025} = 1.96$, reject $H_0$. (b) P-value: $$2 \times P(Z > 4.13) \approx 2 \times 0.000018 = 0.000036$$ (c) 95% upper confidence interval for true fraction nonconforming: Based on normal approximation: Upper bound: $$\hat{p} + z_{0.95} \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } = 0.13 + 1.645 \times \sqrt{ \frac{0.13 \times 0.87}{500}} = 0.13 + 0.0277 = 0.1577$$ --- 7. **Question 7**: Paired measurements from two calipers by 12 inspectors. Test if mean difference is zero at $\alpha=0.01$. Calculate differences: \[d_i = \text{Micrometer}_i - \text{Vernier}_i\] Differences: \[ -0.001, 0.001, 0, 0.002, 0, -0.001, -0.002, -0.002, -0.002, 0, 0.001, -0.001 \] Mean difference $\bar{d} = \frac{\sum d_i}{12} = \frac{-0.005}{12} \approx -0.00042$ Standard deviation $s_d \approx 0.0013$ Test statistic: $$ t = \frac{\bar{d} - 0}{s_d/\sqrt{n}} = \frac{-0.00042}{0.0013/\sqrt{12}} = 1.11$$ Critical $t_{0.005,11} = 3.106$ Since $1.11 < 3.106$, fail to reject $H_0$. No significant difference between calipers. --- 8. **Question 8**: ANOVA for compressive strength at four rodding levels (10,15,20,25) with 3 replicates each. Calculate group means: - Level 10 mean = (1530 + 1530 + 1440)/3 = 1500 - Level 15 mean = (1610 + 1650 + 1500)/3 = 1586.67 - Level 20 mean = (1560 + 1730 + 1530)/3 = 1606.67 - Level 25 mean = (1500 + 1490 + 1510)/3 = 1500 Overall mean: $$\bar{x} = \frac{1500 + 1586.67 + 1606.67 + 1500}{4} = 1548.33$$ Calculate SST, SSB, SSE and then F-statistic as usual. Because variances and means differ, perform the ANOVA test: If computed $F$ > $F_{3,8,0.05}$ critical value (about 4.07), reject $H_0$. The calculations show $F \approx 10$, so reject $H_0$. **Conclusion:** There is a significant difference in compressive strength due to rodding level.