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

# ANOVA: productivity of three operators

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2071 Chaitra, 5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/anova-three-operators
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

Three trained operators work 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 |
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: all three operator totals equal 52, so SS between = 0 and F = 0. F < critical 4.256; the difference in average productivity due to operators is not significant.

## Answer

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

1. H0: μ₁ = μ₂ = μ₃ (average productivity of the three 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

Grand total T = 156,  N = 12
Correction factor CF = T²/N = 156²/12 = 2028.0000
Total SS  = Σx² − CF = 2072.0000 − 2028.0000 = 44.0000
SS between = Σ(Ti²/ni) − CF = 2028.0000 − 2028.0000 = 0.0000
SS within  = Total SS − SS between = 44.0000
```

4. ANOVA table:

| Source | SS | d.f. | MS | F |
|---|---|---|---|---|
| Between operators | 0.0000 | 2 | 0.0000 | 0.000 |
| Within (error) | 44.0000 | 9 | 4.8889 | |
| Total | 44.0000 | 11 | | |

5. Critical value: F(0.05; 2, 9) = 4.256. Reject H0 if F > 4.256.
6. Decision: the calculated value falls in the acceptance region (p-value = 1.0000), 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. Why is SS between zero?
   A. Data are paired
   B. N is too small
   C. All operator means are 13, equal to the grand mean
   D. SS within is zero
   Answer: C. Each operator total is 52 (mean 13), so Ti^2/ni sum equals CF. (Key point: Between SS)

2. What are the d.f. for between and within?
   A. 2 and 9
   B. 2 and 12
   C. 3 and 8
   D. 3 and 9
   Answer: A. Between: 3-1 = 2; within: 12-3 = 9. (Key point: Degrees of freedom)

3. What is F and the critical value?
   A. F = 0; 3.490
   B. F = 4.89; 4.256
   C. F = 1; 4.256
   D. F = 0; F(0.05; 2, 9) = 4.256
   Answer: D. MS between = 0 so F = 0, which is below 4.256. (Key point: F value)

4. What is the conclusion?
   A. SS within must be wrong
   B. Operators have no significant difference in average productivity
   C. Operator 1 is best
   D. Operators differ significantly
   Answer: B. F falls in the acceptance region, so fail to reject H0. (Key point: Conclusion)

## Close variants set in other exams

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

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