Re: sampling from a circular distribution

From: Ross Clement (clemenr_at_wmin.ac.uk)
Date: 08/13/04 

Date: 12 Aug 2004 23:01:42 -0700

scalesymmetry@optusnet.com.au (David Alexander) wrote in message news:<dcd49649.0408121826.674bbacb@posting.google.com>...
> Dear sci.stat.math,
> Here is the problem:
> I have an image made up of grey-scale pixel values--1 to 180 (pi/2) in
> integer steps. These values are circular i.e. 180+1=1. The population
> (all the pixels in the image) distribution is roughly uniform, but not
> perfectly so. Now, I have made a measurement that samples about 20% of
> the image. The distribution of this sample is even less uniform. What
> I want to show is that the sample is not a random sampling of the
> population i.e. that it is systematically biased.
> Thanks,
> David.

Health warning: I'm no expert statistician but that rarely shuts me
up.

If you only have 180 possible pixel values (did I understand your
description correctly). If your full image has >> 180 pixels, then you
can easily calculate the probability distribution for the entire
image. Trivially the probability of any pixel value (e.g. 24) is p(24)
= count( pixel_value = 24 ) / count( pixels in image ). Then, since
you know the size of the sample N, you know that the expected number
of pixels of value 24 in the sample is p(24) * N. You can then use
standard chi-square test for significance since you know the observed
and expected values for each pixel value.

Note that if you have small values for various pixel value counts (N
is not >> 180), then you will probably need to either use a different
test for significance, or use Yates correction, not Pearsons (the
standard) chi-square test.

Cheers,

Ross-c

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