ABX P Value

2004-09-22 12:14:48

I definately remember reading a discussion on the correct calculation of the ABX p-value (probability you were guessing) a while back but a couple of searches and a read through the FAQ were unhelpful. From my limited stats knowledge (half a semester on probability) I came up with (for p correct from n trials) nCk*(0.5)^n*(0.5)^(n-k) which simplifies to nCk/2^n. This doesn't seem correct to me.

What is the correct equation for working this out?

ABX P Value

Reply #1 – 2004-09-22 21:02:34

I don't know if there's an easier mathematical equation for it, but this one works:

n=number of trials
k=number correct
q=temp variable

sum( q=0 to n-k : ([span style='font-size:8pt;line-height:100%']n[/span]C[span style='font-size:8pt;line-height:100%']q[/span]) / (2^n))
For n=5 and k=4, this gives 1/32 + 5/32 = 3/16 ~ 0.1875

[edit: deleted quote]

ABX P Value

Reply #2 – 2004-09-22 22:05:50

http://www.hydrogenaudio.org/show.php/showtopic/14679

ABX P Value

Reply #3 – 2004-09-23 07:40:44

Thanks Omion and kjoonlee, that's just what I was looking for.