ASQ Exam CQE Topic 10 Question 95 Discussion

Actual exam question for ASQ's CQE exam

Question #: 95
Topic #: 10

[All CQE Questions]

The number of runs required by a full-factorial design with three factors each at two levels is equal to

D12

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Suggested Answer: C

by Berry at Aug 23, 2024, 01:45 PM

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Carmela

2 months ago

Wait, isn't a full-factorial design the one where you test every possible combination? That sounds like a lot of work - I'll stick to my 2^k designs, thank you very much!

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Veta

19 days ago

That's good to know, thanks for the information!

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Kattie

23 days ago

The number of runs for a full-factorial design with three factors at two levels is 8.

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Roslyn

1 months ago

I agree, 2^k designs are much simpler and easier to manage.

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Chanel

1 months ago

It's true, full-factorial designs can be quite intensive with all the combinations.

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Crissy

2 months ago

Trick question! The real answer is to order a pizza and take a break from studying. Factorial designs are the worst...

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Donte

2 months ago

D) 12 has got to be the right answer. I remember learning this in my statistics class. Glad I brushed up on my notes!

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An

1 months ago

I'm pretty sure it's 12 as well. Good thing we remembered that from our statistics class.

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Gertude

1 months ago

Yes, I agree. It's definitely 12 runs for a full-factorial design with three factors at two levels.

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Delpha

2 months ago

I think you're right, D) 12 seems to be the correct answer.

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Stephania

3 months ago

Hmm, I'm not so sure. I was always a bit shaky on factorial designs. Let me double-check the formula...

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Truman

1 months ago

C) 9

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Giuseppe

1 months ago

B) 8

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Ayesha

2 months ago

C) 9

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Luke

2 months ago

A) 6

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Meghann

2 months ago

B) 8

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Diane

2 months ago

A) 6

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Tony

3 months ago

I think the answer is B. 8 runs are required for a full-factorial design with three factors at two levels. It's basic experimental design stuff!

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Freida

3 months ago

I'm not sure, but I think the answer might be D) 12 because each factor at two levels means 2^3 = 8 runs, and then we need to include all possible combinations, which would be 8 + 4 = 12.

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Rosalind

3 months ago

I agree with Rochell, because each factor at two levels means 2^3 = 8 runs, but we also need to include the full factorial, so it's 8 + 1 = 9.

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Rochell

4 months ago

I think the answer is C) 9.

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