Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm attempting to develop a standardized method to check whether one set of locations are representative of a larger set. In this particular case, I'm attempting to look specifically at their geographical representativeness.

One method is to look at a two sample t-test for latitude and longitude * independently*, but that clearly ignores the possibility that there could be correlation between the values. Another option is to look at a categorical grouping of location (eg state, market, or any other gridding of the network), and use a chi-squared test. Neither of these strike me as optimal, however.

Is anyone familiar with a test that can check bias of a sample based on two dimensions simultaneously? Any thoughts would be greatly appreciated.

share|improve this question
This question suggests you may have different ideas of what "representative" means. The use of t-tests implies you just want the sample to have a comparable mean to the population. The reference to correlation implies you are looking at a second-order comparison between the sample and the population. A chi-squared test examines the full distribution of the sample in comparing it to the population. So, the first thing you could do is reflect on what precisely you mean by "representative," then come back here and share that with us. Could you also say why you're checking "representativeness"? – whuber Jun 8 '12 at 15:23
Fair questions-- let me clarify. The check is aimed at ensuring that a the smaller sample would be a representative proxy of the larger group if used in a test. The desire is to extrapolate findings of a test on the smaller group out for the larger group. – Greg Jun 8 '12 at 15:43
In terms of the different tests mentioned-- I realize the Chi-squared test looks at the full distribution, while the t-test only looks at the means. Given the goal, a test for checking the distributions is more valuable than simply checking the mean. – Greg Jun 8 '12 at 15:47
(+1) Thank you for the clarifications. You might get some interesting replies, because this goes to a fundamental issue about how one obtains representative samples in the first place. I have added some tags to reflect that. – whuber Jun 8 '12 at 15:54
@whuber - I'm under the impression that I've read several other questions, perhaps many, which deserve this tag. I can't say I'll hunt them down and tag them, but I think it'll get a fair amount of use going forward. – jbowman Jun 8 '12 at 16:44

Fasano and Franceschini suggested a multi-dimensional version of the Kolmogorv-Smirnov test which they show to be preferable to the $\chi^2$-test for 2- and 3-dimensional data in Monthly Notices of the Royal Astronomical Society 225:155-170. The paper is freely available here.

share|improve this answer

Actually, I had the same questoin recently. By scanning rapidly through the published literature, I came to realize that a general test has been developed by Friedman & Rafsky. Their approach is to use a minimum spanning tree, which is the smallest tree that connects the points of the cloud in $n$ dimension, and compute a statistic from it that is distributed as a Student's t. Unfortunately, I am not aware of any implementation of that test.

All I can suggest is the trick that consists in normalizing your variables in the square $(0,1)\times(0,1)$, applying the inverse erf function to get a bivariate gaussian, square them and sum them, wich should give you a sample distributed as a $\chi^2(2)$, which you can check with your favorite goodness of fit test.

Update: There is C library to test for uniformity in several dimension written by Ben Pfaff. At the section Uniformity testing library you can download the source code and the documentation. If I understood well, this is an implementation of the Smith & Jain test which is a refinement of the Friedman & Rafsky test in case the boundaries of the domain are not defined.

share|improve this answer

You can still do the chi-square test. Nothing says that the bins have to be 1 dimensional. Divide the globe into longitude by latitude segments and count the number of cases in each bin for the two samples. The same chi square test applies.

share|improve this answer
So, are you suggesting discretizing a continuous distribution and then doing a $\chi^2$ test as though this were a discrete distribution? How exactly would you choose the binning to produce the categorization? (that's a question that interests me in a completely different context -… - but it is relevant to this question) – Macro Jun 8 '12 at 16:02
@Macro That is exactly what is done with the chi-square test in 1 dimension. I wouldn't characterize it as treating it 'as though it were a discrete distribution". It just tests that the area under the curve in the bins are nearly the same as they should be if the continuous distributions are identical. As far as binning goes the same problem exists in the 1 dimensional case. You want to take enough bins to represent the shape of the distribution but not so many that you get a lot of low frequency or empty cells. – Michael Chernick Jun 8 '12 at 16:12
I think I would try to make the cells the same size as much as possible. Using the same values for delta longitude x delta latitude in each cell. – Michael Chernick Jun 8 '12 at 16:13
Re: "You want to take enough bins to represent the shape of the distribution but not so many that you get a lot of low frequency or empty cells." - are there methods for doing that or is this generally done in an ad-hoc way? – Macro Jun 8 '12 at 16:13
This is a pretty good idea, but there are some subtleties that deserve explanation. The most important perhaps is that the degrees of freedom will be just one less than the number of bins. @Macro brings up some others, but they are less of an issue when one is planning to compare a sample to a known population which is measured within standardized administrative units (such as states, counties, etc.) If such units are not available, then Macro's point gains more force: it's the MAUP. – whuber Jun 8 '12 at 16:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.