(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) 
H₀: Weeks and shifts are independent. H₁: They are not independent. Take α = 0.05.
Using the chi-square test of independence, Eᵢⱼ = (row total × column total)/grand total.
Expected frequencies:
Row I: 16, 20, 24
Row II: 8, 10, 12
Row III: 16, 20, 24
χ² = Σ(O−E)²/E
= (20−16)²/16 + (20−20)²/20 + (20−24)²/24 + (5−8)²/8 + (10−10)²/10 + (15−12)²/12 + (15−16)²/16 + (20−20)²/20 + (25−24)²/24
= 1 + 0 + 2/3 + 9/8 + 0 + 3/4 + 1/16 + 0 + 1/24
= 175/48 = 3.646.
Degrees of freedom = (3−1)(3−1) = 4.
Critical value χ²₀.₀₅,₄ = 9.488.
Since 3.646 < 9.488, H₀ is not rejected.
Conclusion: weeks and shifts are independent at the 5% level. All expected frequencies exceed 5, so the test is valid.

(b) 
H₀: µ = 50 kg. H₁: µ ≠ 50 kg. Use a two-tailed t-test at α = 0.05.
n = 10.
x̄ = Σx/n = 478/10 = 47.8 kg.
Σ(x−x̄)² = 171.6.
s² = Σ(x−x̄)²/(n−1) = 171.6/9 = 286/15 kg².
s = √(286/15) kg = 4.366 kg.
Standard error = s/√n = √(143/75) kg = 1.381 kg.
t = (x̄ − µ)/(s/√n) = (47.8 − 50)/√(143/75) = −11√3/√143 ≈ −1.593.
Degrees of freedom = n − 1 = 9.
From the t-table, t₀.₀₂₅,₉ = 2.262.
Since |−1.593| < 2.262, H₀ is not rejected.
Conclusion: the average packing weight can be taken as 50 kg at the 5% level. This assumes a random sample and approximate normality.

(c) 
 (i) Let x₁, x₂, x₃ be the weekly units produced of products A, B and C respectively.
Maximize Z = 20x₁ + 6x₂ + 8x₃ ₹
subject to:
8x₁ + 2x₂ + 3x₃ ≤ 250 (Milling)
4x₁ + 3x₂ ≤ 150 (Lathe)
2x₁ + x₃ ≤ 50 (Grinder)
x₁, x₂, x₃ ≥ 0.
The LPP assumes linearity, certainty, divisibility and non-negativity.

(ii) Using the simplex method, add slacks s₁, s₂, s₃:
8x₁ + 2x₂ + 3x₃ + s₁ = 250
4x₁ + 3x₂ + s₂ = 150
2x₁ + x₃ + s₃ = 50
Z − 20x₁ − 6x₂ − 8x₃ = 0.
Initial solution: x₁ = x₂ = x₃ = 0, s₁ = 250, s₂ = 150, s₃ = 50, Z = 0.
Enter x₁. Ratios: s₁: 250/8 = 31.25, s₂: 150/4 = 37.5, s₃: 50/2 = 25. Hence s₃ leaves.
After pivot:
x₁ = 25 − 0.5x₃ − 0.5s₃
s₁ = 50 − 2x₂ + x₃ + 4s₃
s₂ = 50 − 3x₂ + 2x₃ + 2s₃
Z = 500 + 6x₂ − 2x₃ − 10s₃.
Enter x₂. Ratios: s₁: 50/2 = 25, s₂: 50/3 = 16.67. Hence s₂ leaves.
After pivot:
x₂ = 50/3 − s₂/3 + 2x₃/3 + 2s₃/3
x₁ = 25 − 0.5x₃ − 0.5s₃
s₁ = 50/3 + 2s₂/3 − x₃/3 + 8s₃/3
Z = 600 − 2s₂ + 2x₃ − 6s₃.
Enter x₃. Ratios: x₁: 25/0.5 = 50, s₁: (50/3)/(1/3) = 50. Tie; let x₁ leave.
After pivot:
x₃ = 50 − 2x₁ − s₃
x₂ = 50 − 4x₁/3 − s₂/3
s₁ = 2x₁/3 + 2s₂/3 + 3s₃
Z = 700 − 4x₁ − 2s₂ − 8s₃.
All nonbasic coefficients in Z are ≤ 0, so optimality is reached. Nonbasic x₁ = s₂ = s₃ = 0.
Thus x₁ = 0, x₂ = 50, x₃ = 50.
Check: Milling = 8(0) + 2(50) + 3(50) = 250 ≤ 250; Lathe = 4(0) + 3(50) = 150 ≤ 150; Grinder = 2(0) + 50 = 50 ≤ 50.
Maximum profit Z = 20(0) + 6(50) + 8(50) = ₹700.
Final answer: produce 0 units of A, 50 units of B and 50 units of C per week; maximum weekly profit = ₹700.

(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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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