Say I want to plot summary data with a point and a 95% confidence interval around that point. What should my point really be? Mean, mode, or median?

I know that mean = median for any symmetrical distribution, and the most common distributions used in data analysis (t distribution and normal distribution) have the convenient property of mean = mode = median, but what about the log-normal distribution:

$\frac{1}{{x\sqrt {2\pi {\sigma ^2}} }}\exp \left( { - \frac{{{{\left( {\ln x - \mu } \right)}^2}}}{{{\sigma ^2}}}} \right)$

transformed back onto a linear scale? This is a common situation in biology (at least it would be if biologists were better with their statistics).

Here are my arguments for all three moments. I cannot decide which is the strongest argument, so I am asking here:

  1. Median The lower bound of the 95% confidence interval is the value which divides the lower 2.5% of the distribution from the upper 97.5%. Likewise, the upper bound divides the lower 97.5% from the upper 2.5%. Thus, the point that divides the upper 50% from the lower 50%, the median, should be the point between the upper and lower 95% CI bounds. In the case of the log-normal distribution, this is equal to $e^{\mu}$.

  2. Mode If I am estimating a true value via regression, I want to show the value that is the most likely of any value. That would be the peak of the distribution- the mode. In the case of the log-normal distribution, this is equal to $e^{\mu-\sigma^2}$, which is the lowest of the three moments measures of central tendency.

  3. Mean Along the same logic used for mode, but now I am not just looking for the most right answer, but the answer that will minimize how wrong I'll be if I don't get it right. In the case of the log-normal distribution, this is equal to $e^{\mu+\frac{1}{2}\sigma^2}$, which is the highest of the three moments measures of central tendency.

    • Which do you think is the strongest case?
    • Is the answer the same if the value being plotted came from a regression model vs. summarizing raw data (note: I know that summaries of raw data are just single-parameter regressions, but not many biologists make that distinction).
    • at the heart of the issue: what are we really trying to show with the point in the center of a confidence limit?

EDIT 01: @user603's answer has some good discussion and a request for more details. Here's some context that made me ask the question in the first place:

Let's say I am doing regression on a dose-inhibition experiment with different drugs added to cultured cells. My model for the regression is:

$ M = F+\frac{C-F}{1+\exp{\left( H(\ln{dose} - I_{\ln} \right) }}$

where $M$ is the measurement, $F$ is the lower asymptote (signal floor), $C$ is the uninhibited response (ceiling), and $I_{\ln}$ is the natural log of the half-inhibitory concentration for a particular drug. I do my regression under the assumption that $I_{\ln}$ is log-normal, so I get a regressed value (and confidence limits) for $I_{\ln}$. I want to report my findings in a nice chart showing the half-inhibitory concentration for the drugs on a linear scale.

The goal of this or any scientific reporting of measure is to give our best guess at what is True in Reality, which we can approximate by experimentation and repeated measures. So I guess my question just became much more philosophical: what are summary statistics in science really trying to show? Our best estimate for a value (mode)? The estimate at which we have a 50% chance of over/undershooting (median)? The estimate by which we have the lowest probable deviation from the Truth (mean)? I often see it taught that confidence limits are are based on standard error (of the mean), but it is starting to seem to me that it should really be standard error (of the median) which just so happens to be (of the mean) when we're dealing with normal- and t-distributed uncertainty. So far, I am inclined to agree with @user603.

Follow up question: If indeed I did want to show mean $\pm$ confidence limits, or mode $\pm$ confidence limits, how would those be calculated? Can you have standard error of the mode?

  • 5
    $\begingroup$ The median and the mode are not moments. In terms of your question, you get to decide, depending on what you are interested in. There is far more machinery available for inferences for the mean than for the median and in turn than for the mode, but (with all sorts of reservations too small to be written in the margin) bootstrapping makes confidence intervals easier even when there is not much machinery for confidence intervals. The central position of the mean (all puns should be considered intentional) is part history, part a consequence of the central limit theorem. $\endgroup$
    – Nick Cox
    Commented May 20, 2013 at 16:08
  • $\begingroup$ Re the update: You would do well to review the definition of confidence interval, because most of this exposition appears to be developed from an implicit confusion among the distinctions between parameters, estimates, and confidence limits. $\endgroup$
    – whuber
    Commented May 20, 2013 at 19:24
  • 1
    $\begingroup$ @whuber: I realized that a few minutes after I submitted the edit. I think it clicked. $\endgroup$
    – DocBuckets
    Commented May 20, 2013 at 19:38

1 Answer 1



Note these advantages:

  1. the median and it's C.I. (see below) are equivariant to monotone transformation of your data: $$\mathrm{med}(g(x))=g(\mathrm{med}(x))$$ for any function $g$ monotone on the domain of $x$ (i.e. $\log()$ if $x>0$).
  2. It's robust in the sense that it's minimally changed when you replace any fraction $\varepsilon<1/2$ of your observations by arbitrary points (the min maxbias property of the median).
  3. the median is interpretable without reference to an underlying distribution for your data --and so are its confidence intervals --see below.
  4. The 95% confidence intervals for the median are the smallest observations with rank $j$ and $k$ where: $$j=\lceil n/2-1.96\sqrt{n/4}\rceil$$ $$k=\lceil n/2+1.96\sqrt{n/4}\rceil$$ for fat tailed and/or asymmetric distributions this yields C.I. that are much more precise than the Gaussian ones (and not that much less precise when the underlying data is narrow tailed and drawn from a symmetric distribution). These C.I. remain meaningful in many cases (bounded or discrete distributions) where that much cannot be said of the mean/sd based ones.
  • 1
    $\begingroup$ But if you're interested in the mean and the confidence limit you have obtained is for the mean, then why is the median relevant at all? I agree with @Nick Cox that the correct answer is determined by the research or decision objectives. $\endgroup$
    – whuber
    Commented May 20, 2013 at 16:32
  • $\begingroup$ @whuber: why would anyone be interested in a particular statistic? I interpret the question to mean 'among many measures of central tendency'...IMO this is very clear from the first line of the OP's question... $\endgroup$
    – user603
    Commented May 20, 2013 at 16:53
  • $\begingroup$ The example of the lognormal and the reference to biology strongly suggest the issue concerns the biological distinction between an arithmetic and geometric mean. Because those are two different things, a fully valid and objective answer ought to acknowledge that. A recommendation made solely on statistical principles mistakenly ignores the importance of understanding why a user might be using a statistical procedure in the first place. $\endgroup$
    – whuber
    Commented May 20, 2013 at 17:28
  • 3
    $\begingroup$ Doc, I believe you do not characterize confidence limits correctly. There's not enough space in a comment to cover all the problems, but one is that a CI is for a particular statistic. If you intend it to be for a mean, as confidence goes to zero it (usually) converges to the estimate of the mean on which it is based--not a median of anything. You may be confusing confidence with the Bayesian idea of posterior credibility. Note, too, that a CI applies perfectly well to asymmetric distributions--but what matters in its calculation is the sampling distribution of the statistic it covers. $\endgroup$
    – whuber
    Commented May 20, 2013 at 19:20
  • 2
    $\begingroup$ There's a folk myth that once you have asymmetry, you should forget the mean. Exaggerated! (Even perfect symmetry does not help the mean much if you have a Cauchy.) Mean and median are typically close for a Poisson and can be identical for an asymmetric binomial. The mean is well defined for an exponential. And so on. The mean is what it is, regardless of distribution, and its additive properties can be important in applications because of the link to the total. We are not talking sports teams here where you are expected to choose one and only one favourite. @user603 is right too. $\endgroup$
    – Nick Cox
    Commented May 20, 2013 at 19:29

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