Calculate no. of choice situations (labeled, full factorial)

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greenvanilla
Posts: 16
Joined: Sun May 05, 2019 5:40 pm

Calculate no. of choice situations (labeled, full factorial)

Post by greenvanilla »

Dear all,

I am reading the Ngene handbook and have a problem understanding the calculation on page 95, referring to Table 7.1.
It is said that "the full factorial design has 2^1 x 3^8 x 4^2 = 209,952 choice situations".
I don't understand how exactly this is calculated, particularly the 3^8 part.
I thought that you would multiply the number of levels, but I really don't get where there are 8 times 3 levels.
Can anybody help me?

Thanks a lot in advance!
Michiel Bliemer
Posts: 2057
Joined: Tue Mar 31, 2009 4:13 pm

Re: Calculate no. of choice situations (labeled, full factor

Post by Michiel Bliemer »

Yes you can simply multiply the levels, so in that example that would be (looking at Table 7.1):
3*3*4*3*3*4*3*2*3 = 2^1 * 3^6 * 4^2 = 23,328.

So it seems that the calculation in the manual is wrong, it should be 3^6 instead of 3^8. We will correct this in the manual.

Michiel
greenvanilla
Posts: 16
Joined: Sun May 05, 2019 5:40 pm

Re: Calculate no. of choice situations (labeled, full factor

Post by greenvanilla »

Thank you very much for your helpful reply, Prof. Bliemer!
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