1. **Question 1: Acceptance probability for lot size $N=25$ with 2 nonconforming** (a) Problem: Probability of lot acceptance means selecting 0 nonconforming components in a sample of 5 from 25 total with 2 nonconforming. Calculate hypergeometric probability: Number of conforming = $25 - 2 = 23$ $$P = \frac{\binom{23}{5}}{\binom{25}{5}}$$ Calculate numerator and denominator: $$\binom{23}{5} = \frac{23!}{5!18!} = 33649$$ $$\binom{25}{5} = \frac{25!}{5!20!} = 53130$$ Thus, $$P = \frac{33649}{53130} \approx 0.6336$$ (b) Binomial approximation: Approximate probability $p = \frac{2}{25} = 0.08$ $$P = (1-p)^5 = (1 - 0.08)^5 = 0.92^5 \approx 0.6591$$ Difference with exact: $$0.6591 - 0.6336 = 0.0255$$ The binomial overestimates slightly but is a reasonable approximation. (c) For $N=150$ and 2 nonconforming: $$p=\frac{2}{150}=0.0133$$ Sample size $n=5 \ll N$, so binomial approximation is good here. 2. **Question 2: Tensile strength $X \sim N(40,5^2)$, $50,000$ parts, minimum spec $=35$ lb** (a) Calculate number failing ($X<35$): Standardize: $$Z = \frac{35 - 40}{5} = -1.0$$ From normal table, $P(Z < -1.0) \approx 0.1587$ Expected failures: $$50,000 \times 0.1587 = 7,935$$ (b) Number with $X>48$: $$Z = \frac{48 - 40}{5} = 1.6$$ $$P(Z > 1.6) = 1 - 0.9452 = 0.0548$$ Expected number: $$50,000 \times 0.0548 = 2,740$$ 3. **Question 3: Battery life $X \sim N(900,35^2)$, fraction surviving beyond 1000 days** $$Z = \frac{1000 - 900}{35} = 2.857$$ $$P(X > 1000) = P(Z > 2.857) = 1 - 0.9979 = 0.0021$$ Fraction surviving beyond 1000 days is approximately $0.0021$. 4. **Question 4: Diameter sample $n=15$, $\bar{x}=8.2535$, $\sigma=0.002$, test $H_0: \mu=8.25$ vs $H_a: \mu\neq8.25$, $\alpha=0.05$** (a) Test statistic: $$Z = \frac{8.2535 - 8.25}{0.002 / \sqrt{15}} = \frac{0.0035}{0.0005164} \approx 6.77$$ Critical value $z_{0.025} = 1.96$ Since $6.77 > 1.96$, reject $H_0$. (b) P-value~$ < 0.0001$, very strong evidence against $H_0$. (c) 95% CI: $$8.2535 \pm 1.96 \times 0.0005164 = (8.2525, 8.2545)$$ 5. **Question 5: Voltage sample $n=16$, data given; sample mean $\approx 10.15$, variance $\approx 0.78$** (a) Test $H_0: \mu=12$ vs two-sided, $\alpha=0.05$ using t-test: $$t = \frac{10.15 - 12}{\sqrt{0.78} / 4} = \frac{-1.85}{0.221} \approx -8.37$$ Degrees freedom $df=15$, critical $t_{0.025,15} = 2.131$ $|t|=8.37 > 2.131$, reject $H_0$. (b) 95% CI for $\mu$: $$10.15 \pm 2.131 \times 0.221 = (9.62, 10.68)$$ (c) Test $H_0: \sigma^2=11$, $\chi^2$ test: $$\chi^2 = \frac{(n-1)s^2}{\sigma_0^2} = \frac{15 \times 0.78}{11} = 1.064$$ Critical values for $df=15$: Lower $6.262$, Upper $27.488$ Since $1.064 < 6.262$, reject $H_0$. (d) 95% CI for $\sigma$: $$ \left( \sqrt{\frac{15 \times 0.78}{27.488}}, \sqrt{\frac{15 \times 0.78}{6.262}} \right) = (0.655, 1.363) $$ (e) 95% upper CI for $\sigma$: $$ \sqrt{\frac{15 \times 0.78}{7.261}} = 1.13 $$ 6. **Question 6: Sample size $n=500$, defective $x=65$, $\hat{p} = 0.13$** (a) Test $H_0: p=0.08$ vs two-sided, $\alpha=0.05$: $$z = \frac{0.13 - 0.08}{\sqrt{0.08 \times 0.92 / 500}} = \frac{0.05}{0.0121} \approx 4.13$$ 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 bound: $$0.13 + 1.645 \times \sqrt{\frac{0.13 \times 0.87}{500}} = 0.13 + 0.0277 = 0.1577$$ 7. **Question 7: Paired caliper measurements by 12 inspectors, test mean difference = 0 at $\alpha=0.01$** Mean difference: $$\bar{d} = \frac{-0.005}{12} = -0.00042$$ Std dev $s_d = 0.0013$ Test statistic: $$t = \frac{-0.00042}{0.0013 / \sqrt{12}} = 1.11$$ Critical $t_{0.005,11} = 3.106$ Since $1.11 < 3.106$, fail to reject $H_0$. No significant difference. 8. **Question 8: ANOVA for compressive strength at levels 10,15,20,25 (3 replicates each)** Group means: Level 10: 1500 Level 15: 1586.67 Level 20: 1606.67 Level 25: 1500 Overall mean: $$\bar{x} = \frac{1500 + 1586.67 + 1606.67 + 1500}{4} = 1548.33$$ F-statistic calculated to be approximately 10. Critical $F_{3,8,0.05} \approx 4.07$. Since $10 > 4.07$, reject $H_0$. Conclusion: Significant difference in compressive strength due to rodding level.