I have to test for significant differences between scenarios. Data consist of the length of a segment divided by the total length of the network. They are distributed between 0 (never equal to 0) and 1, with the length of the network changing. My data contains also many 1s. An example is:

basin scenario 1 scenario 2 scenario 3 Network length
1 1.00 0.11 0.5 15
2 0.95 0.7 0.13 1500

I tried a GLM with quasi-binomial family in R with model 'data ~ scenario' The summary is :


Estimate Std. Error t value Pr(>t)
(Intercept) 2.91314 0.10747 27.107 <2e-16 ***
scenario 1 -0.05700 0.15008 -0.380 0.704
scenario 2 -0.08756 0.14910 -0.587 0.557
scenario 3 -0.10635 0.14851 -0.716 0.474

------- cut ----------

(Dispersion parameter for quasibinomial family taken to be 0.594791)

Null deviance: 3166.8 on 10539 degrees of freedom.

Residual deviance: 3165.0 on 10530 degrees of freedom

The test is never significant neither for scenarios nor for pairwise comparison (with means). Am I using the correct distribution for these data? Is it fully correct to use a GLM to test for significant differences among groups of data in a case like this?

  • 1
    $\begingroup$ What do you want to test? $\endgroup$
    – Dave
    Commented Jun 16, 2021 at 15:37

1 Answer 1


A suggestion: You could investigate (test via simulation) based on the Triangular Distribution.

Not confident that you get anything useful here (as your data appears to be somewhat sparse), but a simulation exercise could provide insights on expected accuracy as your sample size increases. Results could also be compared (and/or combined) with other approaches.

Background: Per Wikipedia, to quote:

Distribution of the absolute difference of two standard uniform variables

This distribution for a = 0, b = 1 and c = 0 is the distribution of X = |X1 − X2|, where X1, X2 are two independent random variables with standard uniform distribution.

The expected value is 1/3 and associated variance 1/18.

More research here finds the following more precise commentary:

Note that X − Y will only have a triangular distribution if b − a = d − c (that is, the joint support is a square interval), otherwise it will have a trapezoidal distribution.

Note: I am looking at the problem from the perspective that the data may consist inherently of samples taken from, say, a Uniform Distribution (hence, the non-traditional analysis). This may be representative of your data upon excluding your reported presence 'of many ones'.


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