(a) The following table gives the production in three shifts and the corresponding number of defective goods that turned out in three weeks. Test at 5% level of significance whether weeks and shifts are independent.

Table :

| Shift | 1st week | 2nd

Question

(a)

The following table gives the production in three shifts and the corresponding number of defective goods that turned out in three weeks. Test at 5% level of significance whether weeks and shifts are independent.

Table :

Shift	1st week	2nd week	3rd week	Total
I	20	20	20	60
II	5	10	15	30
III	15	20	25	60
Total	40	50	60	150

15 marks

Table :

| Shift | 1st week | 2nd week | 3rd week | Total |
|-------|---------|---------|---------|-------|
| I | 20 | 20 | 20 | 60 |
| II | 5 | 10 | 15 | 30 |
| III | 15 | 20 | 25 | 60 |
| Total | 40 | 50 | 60 | 150 |

(15 marks)

(b) A random sample of 10 bags of urea are found to have the following weights (kg) : 40, 45, 49, 50, 55, 44, 52, 48, 50, 45

Test at 5% level of significance whether the average packing weight can be taken as 50 kg. (15 marks)

(c) A firm produces three products A, B and C which pass through three departments : Milling, Lathe and Grinder.

Each unit of Product A requires 8, 4 and 2 hours respectively in the three departments.

Each unit of Product B requires 2, 3 and 0 hours respectively in the three departments.

Each unit of Product C requires 3, 0 and 1 hours respectively in the three departments.

The available capacity on the machines which might limit output is given below.

Table : Capacity of the Machines for Production

| Machine Type | Available time (in machine hours per week) |
|---|---|
| Milling Machine | 250 |
| Lathe Machine | 150 |
| Grinder Machine | 50 |

The unit profit would be : ₹ 20, ₹ 6 and ₹ 8 for products A, B and C respectively.

(i) Formulate the problem in LPP.

(ii) How much of each product should the firm produce in order to maximize profit ? (8+12=20 marks)

UPSC Answer Check · Accepted Answer

(a) Chi-square test of independence of weeks and shifts

H₀: weeks and shifts are independent (the pattern of defectives across shifts is the same in every week). H₁: they are not independent. Level of significance α = 0.05.

Expected frequency = (row total × column total) ÷ grand total, with grand total 150.

Shift I (row total 60): 60 × 40 ÷ 150 = 16, 60 × 50 ÷ 150 = 20, 60 × 60 ÷ 150 = 24. Shift II (row total 30): 8, 10, 12. Shift III (row total 60): 16, 20, 24.

χ² = Σ (O − E)² ÷ E. Shift I: (20 − 16)² ÷ 16 = 1.000, (20 − 20)² ÷ 20 = 0, (20 − 24)² ÷ 24 = 0.667. Shift II: (5 − 8)² ÷ 8 = 1.125, 0, (15 − 12)² ÷ 12 = 0.750. Shift III: (15 − 16)² ÷ 16 = 0.0625, 0, (25 − 24)² ÷ 24 = 0.0417. Total χ² = 3.646.

Degrees of freedom = (3 − 1)(3 − 1) = 4. Table value at 5% is 9.488.

Since 3.646 is less than 9.488, H₀ is not rejected. At the 5% level, weeks and shifts are independent. All expected frequencies exceed 5, so the test is valid.

(b) t-test for the mean weight of urea bags

H₀: μ = 50 kg. H₁: μ ≠ 50 kg. Two-tailed test at α = 0.05, with n = 10 and σ unknown, so the t-test is used (assuming roughly normal weights).

Sample mean = 478 ÷ 10 = 47.8 kg.

Σ(x − x̄)² = 171.6, so s² = 171.6 ÷ 9 = 19.07 and s = 4.367 kg.

Standard error = s ÷ √n = 4.367 ÷ 3.162 = 1.381 kg.

t = (47.8 − 50) ÷ 1.381 = −1.593, with 9 degrees of freedom.

The table value t(0.025, 9) = 2.262. Since |−1.593| is less than 2.262, H₀ is not rejected. The average packing weight can be taken as 50 kg at the 5% level.

(c)(i) Formulation of the linear programme

Let x₁, x₂ and x₃ be the weekly units of products A, B and C.

Maximise Z = 20x₁ + 6x₂ + 8x₃ (₹ per week), subject to:

8x₁ + 2x₂ + 3x₃ ≤ 250 (milling hours), 4x₁ + 3x₂ ≤ 150 (lathe hours), 2x₁ + x₃ ≤ 50 (grinder hours), and x₁, x₂, x₃ ≥ 0.

B needs no grinder time and C needs no lathe time, so those coefficients are zero. The model assumes linearity, divisibility and certainty of the data.

(c)(ii) Optimal production plan

Add slack variables s₁, s₂, s₃ and solve by the simplex method. The starting solution has all slack, Z = 0.

Iteration 1: x₁ enters, as it has the largest profit coefficient (20). Ratios are 250 ÷ 8 = 31.25, 150 ÷ 4 = 37.5 and 50 ÷ 2 = 25, so the grinder slack s₃ leaves. This gives x₁ = 25 and Z = 500.

Iteration 2: x₂ now improves Z at 6 per unit. Ratios are 50 ÷ 2 = 25 for milling slack and 50 ÷ 3 = 16.67 for lathe slack, so s₂ leaves. This gives x₂ = 16.67 and Z = 600.

Iteration 3: x₃ improves Z at 2 per unit. Both x₁ (25 ÷ 0.5) and s₁ (16.67 ÷ 0.333) tie at 50, so x₁ leaves. This gives x₃ = 50, x₂ = 50 and Z = 700, with the final objective row Z = 700 − 4x₁ − 2s₂ − 8s₃. All coefficients are non-positive, so the solution is optimal.

Optimal plan: produce 0 units of A, 50 units of B and 50 units of C per week. Check: milling 2(50) + 3(50) = 250, lathe 3(50) = 150 and grinder 50, so all three capacities are fully used. Profit = 6(50) + 8(50) = ₹700 per week.

Check by duality: the shadow prices are 0 for milling, ₹2 per hour for lathe and ₹8 per hour for grinder, and 150 × 2 + 50 × 8 = ₹700, equal to the primal profit. Product A earns ₹20 against a resource cost of 4 × 2 + 2 × 8 = ₹24 per unit, so making A would lower profit.

(a) The following table gives the production in three shifts and the corresponding number of defective goods that turned out in…

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(a) Chi-square test of independence of weeks and shifts

(b) t-test for the mean weight of urea bags

(c)(i) Formulation of the linear programme

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More from Management 2023 Paper II

पाली	प्रथम सप्ताह	द्वितीय सप्ताह	तृतीय सप्ताह	योग
प्रथम	20	20	20	60
द्वितीय	5	10	15	30
तृतीय	15	20	25	60
योग	40	50	60	150

मशीन का प्रकार	उपलब्ध समय (मशीन घंटे प्रति सप्ताह)
मिलिंग मशीन	250
लेथ मशीन	150
ग्राइंडर मशीन	50