(a) The following data represents the results of inspecting 50 units/day of personal computers produced by two processes A and B…

(a) p-chart control limits and hypothesis test of the company's claim

Each sample has n = 50 PCs, and there are k = 10 samples. The p-chart is used, with 3-sigma limits.

Process A: total non-conforming = 5 + 6 + 7 + 5 + 4 + 6 + 5 + 7 + 9 + 6 = 60. p̄ = 60/500 = 0.12, σ_p = √(0.12 × 0.88/50) = 0.0460. UCL = 0.12 + 3(0.0460) = 0.258; LCL = 0.12 − 0.138 = −0.018, taken as 0. The largest sample proportion is 9/50 = 0.18, below the UCL, and none is below the LCL. Process A is in control.

Process B: total non-conforming = 8 + 7 + 8 + 9 + 10 + 11 + 15 + 8 + 9 + 7 = 92. p̄ = 92/500 = 0.184, σ_p = √(0.184 × 0.816/50) = 0.0548. UCL = 0.184 + 0.164 = 0.348; LCL = 0.184 − 0.164 = 0.020. The sample proportions run from 7/50 = 0.14 to 15/50 = 0.30 (sample 7), all inside 0.020–0.348. Process B is also in control. No point lies outside the limits, so neither process needs its limits revised.

Hypothesis test. The claim "at most 5 non-conforming in 50" means p ≤ 0.10. H₀: p ≤ 0.10, H₁: p > 0.10 (one-tailed), α = 0.05, critical Z = 1.645 (table area 0.45). Standard error with all 500 PCs: √(0.10 × 0.90/500) = 0.01342. Process A: Z = (0.12 − 0.10)/0.01342 = 1.49 < 1.645. H₀ is not rejected, so the claim is acceptable for A (table area 0.4319 gives p ≈ 0.068, above 0.05). Process B: Z = (0.184 − 0.10)/0.01342 = 6.26 > 1.645. H₀ is rejected, so the claim is not valid for B. The claim is therefore valid for A but not for B, and not for both.

(b) Orthogonal cutting forces and friction

Data: α = 12°, cutting force (vertical) F_c = 1600 N, thrust force (horizontal) F_t = 1250 N, chip thickness ratio r = 0.25.

(i) Normal force on the rake face: N = F_c cosα − F_t sinα = 1600(0.9781) − 1250(0.2079) = 1565.0 − 259.9 = 1305 N. (ii) Friction force on the rake face: F = F_c sinα + F_t cosα = 332.7 + 1222.7 = 1555 N. (iii) Resultant force: R = √(1600² + 1250²) = 2030 N, which also equals √(N² + F²). (iv) Coefficient of friction: μ = F/N = 1555/1305 = 1.19. Shear angle: tanφ = r cosα/(1 − r sinα) = 0.2445/0.9480 = 0.2579, so φ = 14.46°. (v) Normal force on the shear plane: N_s = F_c sinφ + F_t cosφ = 399.7 + 1210.3 = 1610 N. (vi) Shear force on the shear plane: F_s = F_c cosφ − F_t sinφ = 1549.3 − 312.2 = 1237 N.

Check: √(1610² + 1237²) = 2030 N, equal to R.

(c) Modified layout to reduce handling cost

Combine the flows in both directions between each pair: A–B = 400 + 300 = 700; A–C = 500; A–D = 50; B–C = 200; C–D = 700; B–D = 0.

Existing layout A-B-C-D (30 m spacing). Handling cost measured as flow × distance in units of 30 m: A–B 700 × 1 = 700; A–C 500 × 2 = 1000; A–D 50 × 3 = 150; B–C 200 × 1 = 200; C–D 700 × 1 = 700. Total = 2750 (82 500 part-metres).

The two heaviest flows are A–B and C–D (700 each), then A–C (500). Place A–B adjacent, C–D adjacent, and bring A next to C. This gives the layout B-A-C-D: A–B 700 × 1 = 700; A–C 500 × 1 = 500; A–D 50 × 2 = 100; B–C 200 × 2 = 400; C–D 700 × 1 = 700. Total = 2400 (72 000 part-metres).

Trying all 24 linear arrangements, B-A-C-D (and its mirror D-C-A-B) gives the minimum. The modified layout B-A-C-D reduces handling by 350 units of 30 m, about 12.7%, simply by interchanging A and B.