The best option is if you have a proper error model so that you can use this to calculate confidence intervals. If not, I suppose that the errors of logarithm of your data will be relatively symmetric, so that the normal distribution (t-distribution) can be used to calculate (approximate) confidence intervals. In the latter case I'd show the data (and the intervals) directly on the log scale. However, if you feel urged to show it on the original scale, you need to anti-log the estimate and the upper and lower limits of the intervals.

The best option is if you have a proper error model so that you can use this to calculate confidence intervals. If not, I suppose that the errors of logarithm of your data will be relatively symmetric, so that the normal distribution (t-distribution) can be used to calculate (approximate) confidence intervals. In the latter case I'd show the data (and the intervals) directly on the log scale. However, if you feel urged to show it on the original scale, you need to anti-log the estimate and the upper and lower limits of the intervals.

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If you have enough data, you could report some quantiles (on a log scale) ?

Sure, this is not an "error bar" but, at least, it conveys some useful information about the distributions !

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As it turns out, taking the log of exponentially distributed values doesn't render them particularly symmetric. Fortunately I found this nice reference that tells how to calculate confidence intervals from discrete values. http://people.missouristate.edu/songfengzheng/Teaching/MTH541/Lecture%20notes/CI.pdf

please kindly look at the link; Analysis of Environmental Data Problem Set

Conceptual Foundations:

Fre q u e n tis t In fe re n c e : Maxim um Like lih o o d

1., http://www.umass.edu/landeco/teaching/ecodata/schedule/likelihood.problems.answers.pdf, regards

If you have lots of data points, for each of your bins try fitting the Gamma distribution to your frequency histogram. You can do this using maximum likelihood estimation. The two parameters of the Gamma, the shape and the scale, can be estimated and plotted on the Gamma (shape, scale)-plane with 95% confidence intervals for each. In your case the shape should be ~1, falling to the left of the horizontal axis, on the exponential range of the Gamma.

If you use Matlab you can try

[phat, ci] = gamfit( X , alpha )

where X is your array of values from your biological data and the default alpha is 0.05, and you can change the alpha to whatever confidence level. Then plot on the (shape, scale)-plane each point given by the estimated shape and scale values returned in the 1x2 array phat, with the ci error bars returned in the 2x2 matrix ci, one for the scale and one for the shape.

In the attached file Figure 8 gives an example of how to plot on the Gamma plane the estimated values and the 95% confidence intervals (though I used a log-log plot).

Hope that his helps you visualize your data.

Regards

Glad William found that reference. I think a point that bears noting for all here is that 95% confidence intervals and +- 2 std. dev. only equate for the normal distribution. So reviewers "should" be looking for confidence intervals in general and not standard error based bars. Also, using a log scale is valid in presenting a distribution with right skew even when the result remains unsymetric and (alas) might help with reviewer concern about wide confidence intervals.

William, there may be other better approaches, but this is what it works for me: 1) Build a Lorenz curve with your data for descending order; 2) Select the two extreme points of the tails plus 4 inner points (Xi; Li). 3) Obtain the media of your total sample as a simple average- Once you have this data of 2) and 3) try to use the steps I explain in the 3 pages text attached -or send data in a new message- so I can work it. If you have a decils table publish it here that it is also workable. There is no problem if it is real data or not and there is no need to know real units employed. OK, thanks, emilio

So you bin the data according to values of one variable, and then you see that the values of another variable are exponentially distributed, per bin, but with presumably different rate parameter lambda per bin. You can estimate each bin's lambda by 1 divided by the average and you can give a confidence interval for each estimated lambda. Its width will be inversely proportional to the square root of the number of observations in the corresponding bin. There are standard recipes for doing this in every first textbook on mathematical statistics, based on the fact that the sum of independent exponentials with the same rate parameter has a gamma distribution. (If the number of observations per bin is large you can use the delta-method and +/-2 estimated standard deviations for an approximately 95% confidence interval, as others have pointed out).

In short, my advice is: collaborate with a competent statistician. Statistics *is* rocket science. Biologists and statisticians need to work closely together. Either on their own is useless. You say your data is exponentially distributed because there are many small and few large observations. You might as well say some other data is normally distributed because the histogram is kind-of bell-shaped. Sweet dreams!

Richard: I said that the data are exponentially distributed because they are, in fact, exponentially distributed, both in theory (chemical bond lifetimes, in this case) and as shown in a log survival probability plot. In saying that there are many small and few large observations I was clarifying my problem for the respondents, not stating how I came to that conclusion.

Richard, interesting comment (and very true too)

However, I thought that talking to each other in this very forum with minimum arrogance and maximum openness was precisely a form of collaboration in it of itself. In fact many interesting suggestions have come up thus far and they are from a good mixture of backgrounds. Personally, this has been one of the questions in Research Gate that I have enjoyed the most.

I am sorry if I find it hard to combine minimum arrogance with maximum openness. If one wants to be both brief and clear, maximum openness could well come across as arrogance.

Elizabeth, I don't think that Richard's answer was intentionally arrogant (I don't even find it arrogant at all). I generally that all participants honestly try to share knowledge and experience, that they really want to help, are open-minded, willing to learn, aware of being possibly wrong in some own convictions, and acknowleding the value of such a forum to provide and get ideas and new aspects, and develop the understanding. (This presumption might get invalidated by some persons, though, but this will take some empirical evidences ;) )

I have to admit that I am writing this post because I can imagine that others might interpret some of my posts as being rude or arrogant, what is clearly not intended. I also think like Richard that brevity, clearity, openness -and a lack of proficiency in the English language- might make answers look arrogant or impolite for some readers.

There you go then. It's all good.

Thank you, everyone, for your answers. I certainly gained a broader perspective on the problem and its several possible approaches. In the end the approach we'll use is to calculate the confidence intervals thus:

P{(2*sum(x1..xn))/chisq(2n,1-alpha/2) < mu < (2*sum(x1..xn))/chisq(2n, alpha/2)} = 1-alpha

since it seems to account for the distribution of the data in a "small" sample rather than relying solely on knowledge the mean. I posted this approach before (I found it in the attached link), but thought I'd repeat it here since the discussion seems to be winding down.

This was a particularly interesting discussion in light of the fact that most folks in my field use the standard error of the mean on everything with little consideration given to the nature of the data. I prefer to do things correctly. Thank you all for helping me do so!

http://people.missouristate.edu/songfengzheng/Teaching/MTH541/Lecture%20notes/CI.pdf