Re: Tests for significance applied to signals

From: Ray Koopman (koopman_at_sfu.ca)
Date: 01/24/05 

Date: 24 Jan 2005 10:27:28 -0800

Bala Vamsi Tatavarthy wrote:
>>
>> If you are willing to say that the results for the 5000 periods
>> are independent and identically distributed then put the data into
>> a 2 x 2 table and do the usual chi-square test of independence:
>> n(n11*n00-n10*n01)^2/((n11+n10)(n01+n00)(n11+n01)(n10+n00)).
>>
>
> I am sorry but I lost you here. What would be n11, n00 , n10 and n01.
> Lets say I have M time samples in which events occur m1 times in
first
> signal and m2 times in the second signal, and m12 times
simulatneously
> in both signals. How do I show non-independence here using
chi-square
> test. (I am a first-timer here).
> The generation of events for a signal at all time instants, are due
to
> i.i.d
>
> Thanx
> Vamsi

n11,n10,n01,n00 are the joint counts inside the 2 x 2 table.

n = M
n11 = m12
n10 = m1 - m12
n01 = m2 - m12
n00 = M - m1 - m2 + m12

m12 = n11
m1 = n11 + n10
m2 = n11 + n01
M = n = n11 + n10 + n01 + n00
chisquare = M*(M*m12-m1*m2)^2/(m1*(M-m1)*m2*(M-m2))

Relevant Pages