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- 2071 Shrawan · 6 marks
ANOVA: productivity of four operators
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.
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% |
One-way ANOVA is used because the operators are one factor with four levels.
- 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)
- 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
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
- 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 |
- Critical value: F(0.05; 3, 12) = 3.490. Reject H0 if F > 3.490.
- 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.
✓ Answer checked by Claude Sonnet 5.5
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