How should I compute an "average" EC50 for several experiments? How can I express its uncertainty or standard deviation?

If you used the one-site competition or sigmoid dose-response curve-fit models, you presumably used X values (concentrations) that were equally spaced on a log scale (e.g., 1, 3, 10, 30, 100, 300...). The midpoint that Prism computes is the logEC50, which will have a distribution closer to a Gaussian distribution than will the EC50. Therefore, you should average the logEC50 values, not the EC50 values. Since the distribution of EC50s is usually not symmetrical, the mean of the EC50s is not a good way to pool the data.

Since some people have trouble thinking about logarithms, it is conventional to report the antilog of the mean of the logEC50 values. This is the same as the geometric mean of the EC50 values, and it is expressed in the same concentration units as the EC50.

An alternative is to report the median, rather than the mean. If you have an odd number of values, the median of the EC50 will be equivalent to the mean of the logEC50. If you have an even number of values, then the median involves taking the mean of the middle two values, so the result will be a bit inconsistent in the EC50 and logEC50 scales, but the difference is not likely to be huge.

What about quantifying the scatter? You can compute the SD or SEM of the logarithms of the EC50, and report the mean logarithm of the EC50 along with the SD or SEM of the logEC50 values. This makes sense, but some scientists are confused by logs.

If you take the antilog of the Standard Error of the logEC50, you get a puzzling number. Is not a value you can add to and subtract from the mean EC50. Instead, you can divide and multiply by that number. So if you take the antilog of the se of the logec50, the result is multiplied and divided by the geometric mean to give a sense of error. It is a multiply/divide factor, not a plus/minus standard error. This is conceptually sound, but not commonlh done.  Here is a more conventional approach. Forget the SE of the logEC50s. Or rather, use the SE to compute the 95% confidence interval of the logec50. Then antilog both limits to get the confidence interval of the EC50 -- which will not be symmetrical around the ec50 itself.

Here is an example. Five repeated experiments give logEC50 values:

-7.59
-7.45
-7.66
-7.48
-7.54

The mean of the logEC50 values is -7.544, and the  standard error of the mean is  of 0.03776. The antilog of the mean (10^-7.544) is 2.86E-08 or 28.6 nM. This is the geometric mean of the EC50 values.

The antilog of the standard error (10^0.03776) equals 1.09. The standard error (0.03776) is a number that you can add or subtract from the mean to get a sense of its uncertainty. The antilog of the standard error of a set of logs (1.09) is a number that you can multiply the mean by or divide into the mean. Although quite logical, this is way of presenting uncertainty is not commonly used.

Standard errors really are not useful except as a way to compute a confidence interval. The 95% confidence interval for the mean of the logEC50 values ranges from -7.649 to -7.439. This range is centered on the mean logEC50 (-7.544). Taking the antilog of both ends of the confidence interval, the 95% confidence interval for the EC50 ranges from  2.2439E-08 to 3.6392E-08 or  22.4 to 36.4 nM. Note that this interval is NOT centered on the geometric mean of the EC50 values (28.6 nM). 

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Edited 2/11 to correct the SEM value, which was wrong.