1
Consider a paper manufacturer studying 3 pulp preparation methods and 4 cooking temperatures on tensile strength in 3 replicates. Some columns of the ANOVA table with 'Pooled' error are given below:
The values of A, B, C, D and E are respectively:
| Source | Sum of Squares | Mean Sum of Squares | F value |
|---|---|---|---|
| Replication | 78 | 39 | |
| Pulp preparation method | 126 | 63 | C |
| (Replication) x (Pulp preparation method) | 36 | 9 | |
| Temperature | A | 144 | D |
| (Pulp preparation method) x (Temperature) | 72 | 12 | E |
| 'Pooled' Error | 72 | B | |
| Total | 816 |
EXPLANATION
Split-Plot, Strip-Plot, Crossover Designs & A
- Split-Plot / Factorial ANOVA Computations:
- Number of Pulp methods ($a$) = 3; Cooking temperatures ($b$) = 4; Replications ($r$) = 3.
- Finding A (Sum of Squares for Temperature):
- Degrees of freedom for Temperature: $df_T = b - 1 = 4 - 1 = 3$.
- Since Mean Square $MST = 144$, the Sum of Squares is:
- Finding B (Mean Square for Pooled Error):
- The pooled error combines $(R \times T)$ ($df = (r-1)(b-1) = 2 \times 3 = 6$) and $(R \times P \times T)$ ($df = (r-1)(a-1)(b-1) = 2 \times 2 \times 3 = 12$).
- Total degrees of freedom for Pooled Error: $df_e = 6 + 12 = 18$.
- Given $SS_e = 72$, the Pooled Error Mean Square is:
- Finding F-ratios (C, D, E):
- $C$ (F for Pulp method tested against pooled error): $C = \frac{63}{4} = 15.75$
- $D$ (F for Temperature): $D = \frac{144}{4} = 36$
- $E$ (F for Interaction): $E = \frac{12}{4} = 3$
- Conclusion: $A = 432, B = 4, C = 15.75, D = 36, E = 3$ (Option 2).
Correct option: (2) 432, 4, 15.75, 36, 3
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