---
title: "ANOVA: productivity of four operators"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:anova-four-operators"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-four-operators
site: BCT Engineering
---

# ANOVA: productivity of four operators

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2071 Shrawan, 6 marks
- Video: [What is ANOVA (Analysis of Variance) in Statistics ? | Explained with Examples (ANOVA F - test)](https://www.youtube.com/watch?v=x_gHAly3mJo) — Digital E-Learning
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-four-operators
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

Four trained operators works on production of new product. The productivity of the operators are recorded as below:
| Operator 1 | 10 | 12 | 14 | 16 |
|---|---|---|---|---|
| Operator 2 | 12 | 11 | 13 | 16 |
| Operator 3 | 14 | 15 | 12 | 11 |
| Operator 4 | 16 | 10 | 17 | 17 |
Using ANOVA, test whether the difference in average productivity due to the difference in operators are significant. Use α = 5%

## Short answer

One-way ANOVA: F = 0.615 with df (3,12). F < critical 3.490, so fail to reject H0; the difference in average productivity due to operators is not significant.

## Answer

One-way ANOVA is used because the operators are one factor with four levels.

1. H0: μ₁ = μ₂ = μ₃ = μ₄ (average productivity of the four operators is the same)
   H1: at least one operator's mean differs (F-test, right-tailed)
2. Level of significance: α = 0.05
3. Computations:

```text
Operator 1   n=4  total=52  mean=13.0000
Operator 2   n=4  total=52  mean=13.0000
Operator 3   n=4  total=52  mean=13.0000
Operator 4   n=4  total=60  mean=15.0000

Grand total T = 216,  N = 16
Correction factor CF = T²/N = 216²/16 = 2916.0000
Total SS  = Σx² − CF = 3006.0000 − 2916.0000 = 90.0000
SS between = Σ(Ti²/ni) − CF = 2928.0000 − 2916.0000 = 12.0000
SS within  = Total SS − SS between = 78.0000
```

4. ANOVA table:

| Source | SS | d.f. | MS | F |
|---|---|---|---|---|
| Between operators | 12.0000 | 3 | 4.0000 | 0.615 |
| Within (error) | 78.0000 | 12 | 6.5000 | |
| Total | 90.0000 | 15 | | |

5. Critical value: F(0.05; 3, 12) = 3.490. Reject H0 if F > 3.490.
6. Decision: the calculated value falls in the acceptance region (p-value = 0.6181), so **fail to reject H0**.

**Conclusion:** the difference in average productivity due to operators is not significant.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is H0?
   A. mu1 = mu2 = mu3 = mu4
   B. Not all means are equal
   C. mu1 > mu4
   D. All variances are zero
   Answer: A. H0 states the average productivity of all four operators is the same. (Key point: Hypotheses)

2. What are the d.f.?
   A. (4, 16)
   B. (4, 12)
   C. (3, 15)
   D. (3, 12)
   Answer: D. Between: 4-1 = 3; within: 16-4 = 12. (Key point: Degrees of freedom)

3. How is SS within found?
   A. Total SS - SS between = 90 - 12 = 78
   B. Total SS + SS between
   C. CF - Total SS
   D. SS between / 3
   Answer: A. SS within = 78 and MS within = 78/12 = 6.5. (Key point: Within SS)

4. What is the decision?
   A. F = 6.5 < 3.490
   B. F = 0.615 > 3.490, reject
   C. F = 4 > 3.490, reject
   D. F = 0.615 < 3.490, fail to reject H0
   Answer: D. F = 4.0/6.5 = 0.615, below the critical value. (Key point: Decision)

## Close variants set in other exams

- [ANOVA: productivity of three operators](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aanova-three-operators) — Probability and Statistics, chapter 6, asked 2×

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