I need to test for the differences between three groups of observations (grouped along $x$-axis), which appear to follow an exponential distribution along the $y$-axis dimension (see example fig.). Tentatively, I used the MLE for $\mu$ and their respective 95% CI for each group and deemed them significantly different if their CI are non-overlapping. However, I'm not certain this is formally appropriate, particularly given that the sample sizes are markedly different.
The figure below shows the individual observations (blue) for the 3 groups (<span class=$x$ axis = {groups a, b, c} shown with jitter); the sample size is $n$ in each case. The red line and pink shaded area depict the MLE $\mu$ and C.I., respectively, across all groups. Only the C.I. intervals between groups a & c are not overlapping (the triangle symbol is just a flag for the non-overlapping CIs, and the 'a' on upper-left is the subplot id)">

  • $\begingroup$ Welcome to our site. Please explain what you mean by the "n-size" of a distribution. Indeed, since nothing in the graphic looks remotely like it is described by an exponential distribution, would you mind explaining what you mean by this term? $\endgroup$ – whuber Oct 30 '18 at 15:44
  • $\begingroup$ n meaning sample size $\endgroup$ – Juan Vainas Oct 30 '18 at 16:03
  • $\begingroup$ Okay. Are you trying to ask about the meaningfulness of comparing group means by evaluating whether confidence intervals are overlapping? We have some answers about that, including stats.stackexchange.com/questions/18215 and stats.stackexchange.com/questions/31657. $\endgroup$ – whuber Oct 30 '18 at 17:04
  • $\begingroup$ Yes, exactly that $\endgroup$ – Juan Vainas Oct 30 '18 at 17:09

Increasing the sample size decreases the width of confidence intervals because it decreases the standard error and approaches towards original dataset. As long as we're talking about a CI for a population percentage. The standard error for a population percentage has the square root of the sample size in the denominator. Hence, increasing the sample size by a factor of 4 (i.e., multiplying it by 4) is equivalent to multiplying the standard error by 1/2. Hence, the interval will be half as wide. This also works approximately for population averages as long as the multiplier from the t-curve doesn't change much when increasing the sample size (which it won't if the original sample size is large).

Thus, given exponential distribution parameter- μ have lower C.I. for larger sample size. Here with the larger differences in sizes of datasets, you might want to play with the significance level. I would suggest using 99% C.I. for the smaller dataset and 90% C.I. for the larger dataset then see the difference in width and overlapping in groups.

  • $\begingroup$ Are you, or are you not, suggesting that one can conduct this test by comparing confidence intervals for overlap? $\endgroup$ – whuber Oct 31 '18 at 15:47
  • $\begingroup$ We are remained with choices in altering groups sample sizes n or changing the value of alpha. Keeping one constant - meaning nearly equal, we need to change other. Removing data might seem okay for large dataset but not with relatively smaller n as it might deviate from original mean. I hope this answers your question. $\endgroup$ – Gazal Patel Nov 1 '18 at 6:38
  • $\begingroup$ I'm afraid I don't see an answer to it in your reply. $\endgroup$ – whuber Nov 1 '18 at 12:19
  • $\begingroup$ okay.. so for simplification, yes, I suggest the change in alpha and check for the change in C.I. and overlapping between groups. If not, then one can also try finding the proper test size for given alpha. $\endgroup$ – Gazal Patel Nov 2 '18 at 6:23

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