Re: Tests for significance applied to signals

From: Ray Koopman (koopman_at_sfu.ca)
Date: 01/26/05 

Date: 25 Jan 2005 17:31:39 -0800

Bala Vamsi Tatavarthy wrote:
> Hi,
>
> Please read my problem statement and tell whether my method and
> conclusions are correct , and also please clarify on whether I have
to
> use Bonferroni's correction and if I have to then how do I use it.
>
> I have six systems, each system receives a series of volatges at
> regular intervals(all of them simultaneously), and when the voltage
> crosses a threshold (on positive or negative side) an output impulse
is
> emitted. The threshold vary for each system.
>
> So, now I want to understand whether the impulse generation or
siomply
> events is not an independent process. So , I took two series of
events
> and found the common occurences (ie events occuring at the same
> period). And then I permformed the chi-square test for independence.
>
>
> The null hypothesis being that : The two system's event generation is
> independent.
> Based on two-tailed Chi-Square values I found that for all C(6,2)
pairs
> under alpha = 0.05, it is false . This stands true for another
> threshold value too.
>
> >From 6 similar systems (with different voltage generators and time
> periods)
> I find that for alpha=0.05 , 4 falses with one threshold value and 5
> falses with another.
>
>
> Ok, based on the first 6 systems, can I make a claim that my the
event
> generation of the 6 systems is not independent. And when do I have
use
> Bonferroni's correction.
>
> Please do correct me if I am wrong anywhere or making any implicit
> assumptions, all this from one day reading of Handbook on Applied
> Statistics :)
>
>
> Thanx
> Vamsi

Your basic data are in the form of an M x 6 matrix U whose elements
are 0 or 1. Let p_j be the observed proportion of 1s in column j.

Let x_i be the value obtained by reading row i of U as
a binary integer. 0 <= x_i <= 63. Then for k = 0...63, let:
f_k = the number of rows in U whose x-value = k;
u_1...u_6 = the binary expansion of k;
e_k = M * Product{(p_j)^(u_j) * (1-p_j)^(1-u_j), j = 1...6}.

f & e are the observed & expected frequencies for the chi-square.
However, you will have to pool k-values because many of the e_k
will be far too small.

I suggestion you start by pooling all the values of k whose binary
expansion has more than two 1-bits. With M = 5000 and univariate
counts on the order of 100,..,150, this will give a pooled e of 1.5
or so, so further pooling will probably be necessary.
In general, decisions about pooling should be based only on e,
without considering f.

Let n = the length of the f- and e-vectors after pooling.
Then chi-square = Sum{(f-e)^2/e}, with df = n-1 - 6 = n-7.

This should be more powerful than doing C(6,2) bivariate chi-squares,
and will need no Bonferroni correction.