I think linear regression could handle this but I am asking myself if that method has good statistical power.

Let's think of circles with different radii and let there be points inside those circles.

          radius points
Circle A      20      8
Circle 1      10      7
Circle 2      25     12
Circle 3       8     10

How can I test the null hypothesis that the number of points in circle A is the same as in the circles 1,2 and 3 given that the area has an linear influence?

My idea

Here is a very basic idea: What do we expect for each Circle given the simplification that the area of the circle is given and not the radius?

In total we have an area of $63$ with $37$ points. Thus $0.59$ points per area. So we would expect $11.75$ points for circle A.

As I am interested in the aspect that A differs from the other circles which can be seen as an homogene area i will sum them up to an area of $43$ with $25.25$ expected points.

This leads to the Chi-Square-Test Statistic: $$ \chi^2 = \frac{(20-11.74603)^2}{11.74603} + \frac{(43-25.25397)^2}{25.25397} \sim \chi_1 $$ After calculation we obtain the p-value: $0.186$ for this example


As we have a sample of of point numbers $x=(7,12,10)$ we can calculate the variance $Var(x) = 6.3$

That's why the formula will change to: $$ \chi^2 = \frac{(20-11.74603)^2}{11.74603} + \frac{(43-25.25397)^2}{3 \cdot 6.333} \sim \chi_1 $$ And the p-value will be: $0.0164$

Is this "safe" ?


2 Answers 2


You certainly could test this with a regression model, although Poisson regression may be a more appropriate choice--it depends on what the nature of the response variable is. I'm guessing that the number of points in a circle is a count variable. That is, it can't be negative (e.g., $-8$), and can't be a fractional value (e.g., $.5$); in addition, such variables are often generated by a Poisson process, with the result that the variance is proportional to the mean. (Of course, it is possible that these might not be true, and that the number of points is sufficiently normal that linear regression would be acceptable.)

The question of statistical power is different. The power of this analysis will mostly be driven by the amount of data you have. In your case, you have only 4 data points, so your power is likely to be quite weak. I should note that power is maximized when the model used matches the structure of the underlying data generating process, so for instance, if the number of points in a circle is a function of the area of the circle, rather than its radius, then the relationship would not be linear, and the model would be more powerful if a squared term were added. However, you will have a problem in practice adding more terms to your model, as you only have 4 points, and you will quickly exhaust your degrees of freedom. This means that (counter-intuitively) you might have more power without the squared term. Your situation reminds me of an anecdote in which a researcher found his results were not significant and asked Sir Ronald Fisher what he should do, Fisher responded "go get more data".

On the other hand, you seem to be focused on the nature of your covariate. That is unlikely to be an issue. You should know that regression (whether OLS or Poisson) does not make assumptions about the nature or distribution of your covariates. For more on that issue, you should read this CV question and its answers: What if the residuals are normally distributed, but Y is not?

  • $\begingroup$ +1 Gung. It is a good answer. I mention Pearson correlation even though I realize that the number of points is a count variable. Other correlation measures could be used. I don't think it is really a regression problem though. The OP is not looking to predict the radius based on points inside nor vice versa. He just want do determine if they are positively related (tend to increase or decrease together). $\endgroup$ Aug 15, 2012 at 22:11
  • $\begingroup$ What do you think of my added idea? $\endgroup$
    – jakob-r
    Aug 15, 2012 at 23:59
  • $\begingroup$ To be honest, I don't follow your new idea or its improvement. $\endgroup$ Aug 16, 2012 at 1:43

You could call it linear regression but I would simply call it correlation. Computer the pearson correlation between radius and points inside. If the correlation is statistcally significantly different from 0 then it appears that there is some relationship. I would imagine that you would speculate that there is positive correlation, larger circles include more points.

I think this directly addresses the issue and don't see a method that would have more power.


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