I've run into an odd problem that I have a simple answer to but would like to know if there's something deeper to it.

I have a data set with 7 observations, each with 2 variables (X,Y). I then run a linear regression between the two and find a statistically significant relationship and all is well.

I then start to create prediction intervals (PIs) for the function across the range of my X independent values, range of 5 to 180.

To summarize: Observations: 7 Range of X: [5,180] Mean of X: 45

Now the odd part, when I create PIs across the range of my independent variable x I get a Mean that increases as well as a Standard Deviation (sd) that also increases the further I go from the mean of 45. Which makes sense given the function of PIs.

The odd part is the further I get from the mean of 45 for my PIs the coefficient of variation (CV) begins to decrease which doesn't make much sense as one assumes the further you get from the mean the more variability you should expect. Which is supported by the idea that the sd of the prediction interval is also increasing the further I go from the mean of 45.

So my explanation to this problem is that the further I go from the mean for my PIs the sd of the PIs does increase at a certain rate, as I should expect, but given the range of X is so spread from the mean the CV of the PI will decrease given the growth of the sd is overtaken by the increased spread out range.

Does my simple explanation make sense or is their something I'm missing? Thanks.

  • 1
    $\begingroup$ Characterizing the CV as an appropriate measure of "variability" in this setting seems a stretch. Regardless, how is this related to prediction intervals? How exactly are you defining the "CV"? $\endgroup$ – whuber Dec 13 '17 at 19:18
  • $\begingroup$ You are not asking about it, but I will flag that a sample size of 7 is pretty small for most purposes. Still, it's a role of any interval estimation to show the resulting uncertainty. That side, I don't think that the coefficient of variation of anything has any special meaning here, but the observation that CV decreases with the mean follows from the definition. If you want the lowest possible CV, look for the highest possible mean (for given SD). $\endgroup$ – Nick Cox Dec 13 '17 at 19:19

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