Paper: Design of Experiments and Sample Survey Module: Factorial Design-V

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Paper: Design of Experiments and Sample Survey Module: Factorial Design-V

Module No: DoE-19

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Paper co-ordinator:

Dr. Bikas Kumar Sinha,

Retired Professor, Indian Statistical Institute, Kolkata

Content writer: Dr. Santu Ghosh,

Lecturer, Department of Environmental Health Engineering, Sri Ramachandra University, Chennai

Content reviewer: Dr. Sugata SenRoy,

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of a dialogue between the Instructor [Professor Bikram Kanti Sahay - BKS] and his students [Ms. Sagarika Ghosh - SG and Mr. Subhra Sankar Gupta - SSG]. Further to this, while covering factorial designs, I have ventured into MCQ style all over, since the students are already expected to have introductory concepts from undergraduate studies. I have continued with that style for response surface methodology part as well. I fondly hope this instructional discourse and my efforts on two of my favorite topics in Statistics will be appreciated and found useful.

In the video recordings, I will impersonate as BKS.Mr. Samopriya Basu[MSc (Statistics), Calcutta University] andMs. Moumita Chatterjee[University of Calcutta, Kolkata] will impersonate as the students [SSG and SG] respectively.

Professor Bikas k Sinha Retired Professor of Statistics Indian Statistical Institute Kolkata

July 10, 2015

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List of Factorial Plans and Defining Set

Plan #1 : [(1) (a) (b) (c)] Defining Set (I) : I = ABC Plan #2 : [(1) (ab) (ac) (bc)] Defining Set (II) : A = BC Plan #3 : [(1) (a) (bc) (abc)] Defining Set (III) : B = AC Plan #4 : [(1) (a) (b) (ab)] Defining Set (IV) : C = AB Plan #5 : [(1) (c) (ab) (abc)] Defining Set (V) : I = ABCD Plan #6 : [(1) (a) (bc) (c)] Defining Set (VI) : I =AB Plan #7 : [(1) (a) (b) (c) (d) (abc) (abd) (acd)] Defining Set (VII) : I = CD Plan #8 : [(1) (ab) (cd) (abcd)] Defining Set (VIII): I = BCD

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Consider 2³ factorial expt. in 2 blocks with the Interaction Effect ABC confounded. Then the key block is given by

(a) Plan #1;

(b) Plan #2;

(c) Plan #3;

(d) None of the above.

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Consider 2³ factorial expt. in 2 blocks with the 2-factor interaction effect AB confounded. Then the key block is given by

(a) Plan #1;

(b) Plan #3;

(c) Plan #5;

(d) Plan #7.

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Consider 2³ factorial expt. with the key block given by Plan# 3. Then the confounded effect/interaction is given by

(a) AB;

(b) BC;

(c) AC;

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Consider 2³ factorial expt. with the key block given by Plan #5. Then the confounded effect / interaction is given by

(a) AB;

(b) BC;

(c) ABC;

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Consider 2⁴ factorial expt. with the key block given by Plan # 7. Then the confounded effect / interaction is given by

(a) ABD;

(b) ACD;

(c) BCD;

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Consider 2⁴ factorial expt. in 4 blocks and with the key block given by Plan #8. Then the effects / interactions confounded are

(a) A, BC, ABC;

(b) B, AC, ABC;

(c) AB, CD, ABCD;

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Consider 2⁴ factorial expt. conduced in 4 blocks, confounding the effects / interactions : AB, CD and ABCD. Then the key block is given by

(a) Plan #3;

(b) Plan #5;

(c) Plan #8;

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Consider 2⁴ factorial expt. in 4 blocks. Then the effects that may be confounded are

(a) AB, AD, BD;

(b) AB, CD, AD (c) A, BC, ABD;

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In a 2 factorial expt., we plan to confound the interactions AC, BD and ABCD. Then we need to start with two sets of blocks arising out of the Defining Set given by

(a) I=AC=BD;

(b) I=ABC=D;

(c) I=AB=CD,

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In order to give a plan for a confounded 2 factorial expt in 4 blocks of equal size, Defining set of the form I =ABC =ABD is formed. Then the confounded interactions are likely to be

(a) ABC, ABD and ABCD;

(b) ABC, ABD and CD;

(c) AB, CD and ABCD;

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Q-11

In Q10, the key block is (a) [(1),(a),(b),(ab)];

(b) [(1),(ab),(cd),(abcd)];

(c) [(1),(ad),(bc),(abcd)];

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obtained from the Defining Set (I) given on the top. This means that the Key Block will be formed of

(a) One of the components in each of Defining Sets (I) to (IV);

(b) Only the LHS component in each of the Defining Sets (I) to (IV);

(c) The level combinations obtained after expanding

(a−1)(b−1)(c−1) = (abc)+(a)+(b)+(c)−(1)−(ab)−(ac)−(bc) and using ALL FOUR with ”+” sign;

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(b) Only the RHS component in each of the Defining Sets (I) to (IV);

(a−1)(b−1)(c−1) = (abc)+(a)+(b)+(c)−(1)−(ab)−(ac)−(bc) and using ALL FOUR with ”-” sign;

(d) none of the above.

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(b) BOTh the components in ANY TWO of the Defining Sets (I) to (IV);

(a−1)(b−1)(c−1) = (abc)+(a)+(b)+(c)−(1)−(ab)−(ac)−(bc) and using ALL FOUR with ”+” sign;

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(b) BOTH the component in each of the Defining Sets (I) and (II);

(a−1)(b−1)(c−1) = (abc)+(a)+(b)+(c)−(1)−(ab)−(ac)−(bc) and using ALL FOUR with ”-” sign;

(d) none of the above.

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Here is the end.

SG

Sir, what should I say ? I am simply spell-bound ! I had least idea about MCQs in factorial designs. We studied and learnt basic techniques and solved long questions . . . we never realized that there is so much of beauty in this topic . . . it is simply amazing . . . definitely helps in thorough understanding and eliminating all our doubts.

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