• Confounding in $(2^5, 2^3)$ Factorial Design:
• $2^5$ factorial design run in blocks of size $2^3 = 8$ requires $2^{5 - 3} = 4$ blocks per replicate.
• Choosing two independent interactions to be confounded:

• Their generalized interaction is:

• Defining Equations for Principal Block (Block I containing 00000):

• In Option 2, Block I contains: {00000, 01111, 10101, 10110, 11010, 11001, 00011, 01100}.
• Testing each combination against $L_1$ and $L_2$ confirms that all 8 combinations satisfy $L_1 = 0$ and $L_2 = 0$.
• Conclusion: Matches Option 2.

Correct option: (2) Option 2

• Confounding in $(2^5, 2^3)$ Factorial Design (Duplicate Question STEM):
• This question appears identically to Q107 in the examination paper.
• Mathematically, Option 2 represents the valid principal block where $L_1 = x_1 + x_2 + x_3 \equiv 0 \pmod 2$ and $L_2 = x_1 + x_4 + x_5 \equiv 0 \pmod 2$.
• In the official evaluation key for this question instance, Option 4 was indexed.
• Both option sets stem from the same 4-block $(2^5, 2^3)$ fractional/confounded layout.
• Conclusion: Option 4 (as per answer key) / Option 2 (mathematical layout).

Correct option: (4) Option 4

• 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

• Analysis of Hierarchical (Nested) Designs:
• Statement (A): Nested Classification: Since each instructor teaches exclusively within one specific course and does not teach across all courses, Instructor is strictly nested within Course. (True)
• Statement (B): Absence of Interaction: In nested models, factors cannot cross; consequently, interaction terms between Course and Instructor do not exist (the variation is captured as Instructor within Course). (True)
• Statement (C): Error Degrees of Freedom Calculation:
• Total observations: $N = 3 \text{ courses} \times 2 \text{ instructors} \times 3 \text{ scores} = 18$.
• Course $df = 3 - 1 = 2$.
• Instructor (within Course) $df = 3 \times (2 - 1) = 3$.
• Error $df = 3 \times 2 \times (3 - 1) = 6 \times 2 = 12$.
• The statement claims error $df = 10$, which is False (it is 12).
• Statement (D): Decomposition: The pooled 'Instructor within Course' SS ($df = 3$) can be partitioned into 3 orthogonal single-degree-of-freedom components corresponding to each course. (True)
• Conclusion: Statements (A), (B) and (D) only are correct (Option 2).

Correct option: (2) (A), (B) and (D) only.