Deal of The Day! Hurry Up, Grab the Special Discount - Save 25% - Ends In 00:00:00 Coupon code: SAVE25
Welcome to Pass4Success

- Free Preparation Discussions
Mail Us support@pass4success.com
Location US
Mob_nav_barsMENU

ASQ CQE Exam - Topic 10 Question 95 Discussion

The number of runs required by a full-factorial design with three factors each at two levels is equal to
C) 9
A) 6
B) 8
D) 12

ASQ CQE Exam - 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

Show Suggested Answer Hide Answer
Suggested Answer: C

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

Limited Time Offer

25%

Off

Get Premium CQE Questions as Interactive Web-Based Practice Test or PDF

Contribute your Thoughts:

0/2000 characters
Submit Cancel
Renato
7 months ago
Full-factorial design = 2^3 = 8 runs.
upvoted 0 times
...
Cordelia
8 months ago
Are we sure about that? Seems off.
upvoted 0 times
...
Delisa
8 months ago
Wait, I thought it was 12?
upvoted 0 times
...
Paz
8 months ago
Nope, it's actually 8.
upvoted 0 times
...
Carlee
8 months ago
It's 8 runs for 3 factors at 2 levels!
upvoted 0 times
...
Veronika
9 months ago
I’m not confident, but I feel like it could be 12? I remember something about needing to account for interactions, but I can't quite remember how that fits in.
upvoted 0 times
...
Jimmie
9 months ago
If I recall correctly, for three factors at two levels, it should be 2^3, which equals 8. I hope that's right!
upvoted 0 times
...
Abel
9 months ago
I remember practicing a similar question, and I think the answer was 8. But I'm not entirely sure if I got the factors mixed up.
upvoted 0 times
...
Gerald
9 months ago
I think a full-factorial design with three factors at two levels means we multiply 2 by itself three times, right? So that should be 2^3?
upvoted 0 times
...
Elly
9 months ago
Okay, I remember learning about full-factorial designs in class. The number of runs is equal to the number of possible combinations, which is 2^k where k is the number of factors. So for 3 factors, that's 2^3 = 8 runs. I'm confident that's the right answer.
upvoted 0 times
...
Javier
9 months ago
Hmm, I'm a bit unsure on this. Let me think it through - there are 3 factors, each with 2 levels, so that's 2 x 2 x 2 = 8 possible combinations. But I'm not sure if that's the same as the number of runs required. I'll have to double-check my notes.
upvoted 0 times
...
Frank
9 months ago
I think I know this one - a full-factorial design with 3 factors at 2 levels means there are 2^3 = 8 possible combinations, so the number of runs required is 8.
upvoted 0 times
...
Elroy
9 months ago
Wait, I'm confused. Isn't a full-factorial design supposed to have all possible combinations of the factor levels? If there are 3 factors at 2 levels each, wouldn't that be 2 x 2 x 2 = 12 runs? I'll have to re-read the question carefully.
upvoted 0 times
...
Trinidad
9 months ago
I’m not entirely sure, but I feel like it might be the Access Point Identifier? I remember something about identifiers in class.
upvoted 0 times
...
Carmela
1 year 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!
upvoted 0 times
Veta
1 year ago
That's good to know, thanks for the information!
upvoted 0 times
...
Kattie
1 year ago
The number of runs for a full-factorial design with three factors at two levels is 8.
upvoted 0 times
...
Roslyn
1 year ago
I agree, 2^k designs are much simpler and easier to manage.
upvoted 0 times
...
Chanel
1 year ago
It's true, full-factorial designs can be quite intensive with all the combinations.
upvoted 0 times
...
...
Crissy
1 year ago
Trick question! The real answer is to order a pizza and take a break from studying. Factorial designs are the worst...
upvoted 0 times
...
Donte
1 year 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!
upvoted 0 times
An
1 year ago
I'm pretty sure it's 12 as well. Good thing we remembered that from our statistics class.
upvoted 0 times
...
Gertude
1 year ago
Yes, I agree. It's definitely 12 runs for a full-factorial design with three factors at two levels.
upvoted 0 times
...
Delpha
1 year ago
I think you're right, D) 12 seems to be the correct answer.
upvoted 0 times
...
...
Stephania
1 year ago
Hmm, I'm not so sure. I was always a bit shaky on factorial designs. Let me double-check the formula...
upvoted 0 times
Truman
1 year ago
C) 9
upvoted 0 times
...
Giuseppe
1 year ago
B) 8
upvoted 0 times
...
Ayesha
1 year ago
C) 9
upvoted 0 times
...
Luke
1 year ago
A) 6
upvoted 0 times
...
Meghann
1 year ago
B) 8
upvoted 0 times
...
Diane
1 year ago
A) 6
upvoted 0 times
...
...
Tony
1 year 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!
upvoted 0 times
...
Freida
1 year 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.
upvoted 0 times
...
Rosalind
1 year 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.
upvoted 0 times
...
Rochell
1 year ago
I think the answer is C) 9.
upvoted 0 times
...

Save Cancel