This is a very basic questions however I have not been able to find an answer. When you plot your regular bar plot with your standard errors of the means for comparison do you plot the standard errors and means from the raw data ?
Or do you plot the ones predicted by the model you are fitting. Thank you in advance for your help. I usually use the latter please advice.


  • $\begingroup$ I alfred. What model (on your data) do you use to estimate the mean and SE ? $\endgroup$ – Tal Galili Aug 22 '10 at 14:59
  • $\begingroup$ Can you post an example graph? $\endgroup$ – user88 Aug 22 '10 at 16:10

The list of things to say here...

As Tai's question suggests, it's hard to directly answer your question without information on the actual model. Nevertheless, it's usually good to present data reflective of the model. Typically that is the means with t-tests or ANOVAs. With something else it's probably close to that.

How does a standard error on a graph work for comparison? Are you going to be putting the N on the graph and the multiplying factor needed for comparisons?

How many data points do you have? You could probably just put up the entire data set, with a line indicating your predicted value and some kind of measure of variability around it. Perhaps even an overlayed boxplot.

The measure of variability should reflect what you want to say about the data. If you just want people to compare values then std. err. isn't a very good idea because it's dependent upon n and requires some value of multiplication greater than 2 (i.e. bars have to not overlap by some amount for significant effects). Instead, put up 0.5*LSD bars (comparison bars) (about an 84% confidence interval, or a 0.5 * 95%CI * sqrt(2)). Those bars would show significant differences at the point of bar overlap.

Or, you could be wanting to represent how well you estimated the predicted values. In that case a more convenient confidence interval (about 95%) would be best or even the std. err would be ok. If you want to reflect the estimated variability of the population you put up the standard deviation.

  • $\begingroup$ Thank you John & Tai I guess that covers any potential avenue of my poorly formulated question. regards Alfredo $\endgroup$ – Alfred Aug 22 '10 at 19:23
  • $\begingroup$ You're welcome Alfred. Please choose Johns answer so to verify you found your answer :) Cheers, Tal $\endgroup$ – Tal Galili Aug 22 '10 at 23:12

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