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- 2071 Chaitra · 5 marks
ANOVA: productivity of three operators
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.
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% |
One-way ANOVA is used because the operators are one factor with three levels.
- 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)
- Level of significance: α = 0.05
- Computations:
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
- 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 |
- Critical value: F(0.05; 2, 9) = 4.256. Reject H0 if F > 4.256.
- 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.
✓ Answer checked by Claude Sonnet 5.5
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