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Clearly median seems to be the statistic of choice when it comes to ages.

I am not able to explain to myself why arithmetic mean would be a worse statistic. Why is it so?

Originally posted here because I did not know this site existed.

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4  
It seems like you already had a reasonable answer on the other site? –  Shane Sep 10 '10 at 20:40
    
@Shane: But perhaps different sites hold the potential to garner different answers from different points of view? –  whuber Sep 11 '10 at 14:56

12 Answers 12

up vote 32 down vote accepted

Statistics does not provide a good answer to this question, IMO. A mean is ok to use, too, and is relevant in mortality studies for example. But ages are not as easy to measure as you might think: older people, illiterate people, and people in some third-world countries tend to round their ages to a multiple of 5 or 10, for instance. The median is more resistant to such errors than the mean. Moreover, median ages are typically 20 - 40, but people can live to 100 and more (an increasing and noticeable proportion of the population of modern countries now lives beyond 100). People of such age have 1.5 to 4 times the influence on the mean than they do on the median compared to very young people. Thus, the median is a bit more up-to-date statistic concerning a country's age distribution and is a little more independent of mortality rates and life expectancy than the mean is. Finally, the median gives us a slightly better picture of what the age distribution itself looks like: when you see a median of 35, for example, you know that half the population is older than 35 and you can infer some things about birth rates, ages of parents, and so on; but if the mean is 35, you can't say as much, because that 35 could be influenced by a large population bulge at age 70, for example, or perhaps a population gap in some age range due to an old war or epidemic. Thus, for demographic, not statistical, reasons, a median appears more worthy of the role of an omnibus value for summarizing the ages of relatively large populations of people.

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I think you meant "The median is more resistant to such errors than the mean". I agree with your comments though, and I believe the US census typically reports median's for many categories in official reports (not just the age) for basically all of the same reasons. Income is maybe even a better example than age to illustrate such points. –  Andy W Sep 11 '10 at 1:54
    
Oops: thanks for catching that typo, Andy. –  whuber Sep 11 '10 at 14:54
    
You have substituted a fact—the mean is sensitive to outliers/skewed distributions—for a value statement about the preference for the median over the mean. In effect, you have argued that the mean is not to be preferred because it is not the median (much like those who say one should only use the mean on symmetric distributions, i.e. when the mean and median are equal). –  Alexis May 4 at 18:43
    
@Alexis I do not follow your criticism. Could you elaborate? After all, this answer provides far more than "a fact": it contains quite a few of them, along with an analysis of their implications. And specifically to what "value statement" do you refer? –  whuber May 5 at 14:19
    
My concern is that factual characteristics of the mean and median (e.g. the former is sensitive to outliers, viz "People of such age have 1.5 to 4 times the influence on the mean than they do on the median compared to very young people.") become translated into values about their worth, viz "the median gives us a slightly better picture of what the age distribution itself looks like". The former is a fact, the later a valuation of that fact. My concern is with the switch between the two. More: stats.stackexchange.com/questions/96371/… –  Alexis May 5 at 14:28

John gave you a good answer on the sister site.

One aspect he didn't mention explicitly is robustness: median as a measure of central location does better than the mean as it has a higher breakdown point (of 50%) whereas the mean has a very low one of 0 (see wikipedia for details).

Intuitively, it means that individual bad observations do not skew the median whereas they do for the mean.

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+1 Beat me..... –  mbq Sep 10 '10 at 20:50
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Breakdown isn't an issue for a descriptive statistic of an entire population. –  whuber Sep 10 '10 at 22:06
    
@whuber (+1) ...about the first remark. –  mbq Sep 15 '10 at 22:41

Here's my answer first posted on math.stackexchange:

Median is what many people actually have in mind when they say "mean." It's easier to interpret the median: half the population is above this age and half are below. Mean is a little more subtle.

People look for symmetry and sometimes impose symmetry when it isn't there. The age distribution in a population is far from symmetric, so the mean could be misleading. Age distributions are something like a pyramid. Lots of children, not many elderly. (Or at least that's how it is in a sort of steady state. In the US, the post-WWII baby boom generation has distorted this distribution as they age. Some people have called this "squaring the pyramid" because the boomers have made the top of the pyramid wider than it has been in the past.)

With an asymmetrical distribution, it may be better to report the median because it is a symmetrical statistic. The median is symmetrical even if the sampling distribution isn't.

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In what sense is the median a "symmetrical" statistic? It's certainly not the case that distributions tend to be symmetrically distributed about their medians (nor about their means). If you mean merely what you wrote in another comment that the "median splits the population in half" (which defines the median), your argument sounds circular: the median is good because the median is the median! –  whuber Jan 2 '12 at 15:58

Why is an axe better than a hatchet?

That's similar to your question. They just mean and do different things. If one is talking about medians then the story they are trying to convey, the model they are trying to apply to the data, is different than one with means.

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For a concrete example, consider the mean ages for the Congo (DRC) and Japan. One is devastated by civil war, the other is well developed with an ageing population. The mean isn't terribly interesting for an apples to apples comparison. On the other hand, the median can be informative as a measure of central tendency since by definition we have half above, half below. The wikipedia article on Population Pyramid might be enlightening (see the sections on youth bulge, ageing populations).

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I don't think there is a good descriptive reason for choosing median over mean for age distributions. There is one of practicality when comparing reported data.

Many countries report their population in 5-year age intervals with the top band open-ended. This causes some difficulties calculating the mean from the intervals, especially for the youngest interval (affected by infant mortality rates), the top "interval" (what is the mean of an 80+ "interval"?), and the near top intervals (the mean of each interval is usually lower than the middle).

It is far easier to estimate the median by interpolating within the median interval, often approximating by assuming a flat or trapezium age distribution in that interval (death rates in many countries are relatively low around the median age, making this a more reasonable approximation than it is for the young or old).

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You're getting good answers here, but let me just add my 2 cents. I work in pharmacometrics, which deals in things like blood volume, elimination rate, base level of drug effect, maximum drug effect, and parameters like that.

We make a distinction between variables that can take on any value plus or minus, versus values that can only be positive. An example of a variable that can take on any value, plus or minus, would be drug effect, which could be positive, zero, or negative. An example of a variable that can only realistically be positive is blood volume or drug elimination rate.

We model these things with distributions that are typically either normal or lognormal, normal for the any-valued ones, and lognormal for the only-positive ones. A lognormal number is the number E taken to the power of a normally distributed number, and that is why it can only be positive.

For a normally distributed variable, the median, mean, and mode are the same number, so it doesn't matter which you use. However, for a lognormally distributed variable, the mean is larger than both the median and the mode, so it is not really very useful. In fact, the median is where the underlying normal has its mean, so it is a much more attractive measure.

Since age (presumably) can never be negative, a lognormal distribution is probably a better description of it than normal, so median (E to the mean of the underlying normal) is more useful.

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The distribution of age is certainly not log normal. –  Rob Hyndman Sep 14 '10 at 0:51
    
I do not think you can infer age is log-normally distributed just from the fact that it is always positive. The gamma and the Weibull distributions are also always positive, so why not choosing those ones? –  nico Sep 15 '10 at 23:08
    
@Rob: @nico: I'm sure you're right. It was a poor choice of example. Typically we model pharmacometric parameters like volume and clearance. –  Mike Dunlavey Sep 18 '10 at 12:53

John's answer on math.stackexchange can be viewed as the following:

When you have a skewed distribution the median may be a better summary statistic than the mean.

Note that when he says that there are more infants than adults he essentially is suggesting that the age distribution is a skewed distribution.

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Actually I think nowadays the skew in a lot of countries is more towards seniors, not tots. –  J. M. Sep 11 '10 at 0:09
    
Perhaps, it is skewed the other way but the general point stands. For skewed distributions a median may make more sense than the mean. –  user28 Sep 11 '10 at 0:21
    
I just updated my answer on math.stackexchange to emphasize just that point. People look for symmetry and can incorrectly impose symmetry when it isn't there. When you report the median, you give an answer that is symmetric -- the median splits the population in half -- even though the distribution is not symmetric. –  John D. Cook Sep 11 '10 at 14:11
    
This answer always seems somewhat furtive to me: when distributions are not skew (i.e. they are symmetric), the mean equals the median, so saying that the median is "better" when the distribution is skew is a backdoor way of saying "only use the median." –  Alexis May 4 at 18:38

Public Health Data repositories in the United States are moving toward an AGE in years format of five year increments due to the impact of the HIPAA regulations regarding the intentional blinding and masking of data for personal privacy reasons.

Given this challenge to what was had been in the past (prior to HIPAA) a fairly scale level of measure data element based upon the difference between date of birth and date of death, we may need to reconsider AGE as a scale variable that can be parametrically described at all in public health data sets, in favor of models that describe AGE in a non-parametric fashion, as an ordinal level of measure. I know this may seem "over the top" to many factions within the biomedical informatics community, but this idea may have some merit in terms of "interpretation" as described in the comments above.

What about all of the analytical power that is available to the non-parametric approaches? Yes, it is true that every one of us almost universally will attempt to apply GLM (general linear model) techniques to a variable that presents itself to us in distributions that behave the way AGE does.

At the same time the shape of that distribution and how that shape is being determined by multiple dimension interaction effects upon multi-dimensional centroids and sub-group centroids present in the distribution, must be taken into consideration. What to do with these very complex data sets?

When a data element fails to meet the "assumptions of the model", we progressively scan across (I said across, not down; we should be equal opportunity employers of method, each tool comes from the factory with form follows function rules) the list of other possible models to find the ones that "do not fail" the assumptions tests.

In the present format in public health data sets, we really do need (as a data visualization community) to come up with a more standard model for handling AGE in five year increments (5YI). My vote for data visualization of AGE (given the new 5YI format) is to use histograms and box and whisker plots. Yes, this means the median. (No pun intended!)

Sometimes a picture really is worth a thousand words, and an abstract is a summary of a thousand words. The box and whisker plot shows the "shape" of the distribution as a meaningful symbolic representation of the histogram at nearly an iconic level of resolution. Comparing the distributions of five year age increments by showing "side by side" box and whisker plots where one can instantly visually compare patterns of 75th to 50th (median) to lower 25th ntiles, would make an elegant "universal standard" for comparing AGE across the world. For those of us that continue to enjoy the thrill of data representation through the textual mechanics of tabular display, the "stem and leaf" diagram may also be of service when employed as an animated visual graphics element in a "sparkline" approach that portrays variation of the shapes of distributions over time.

AGE has come of age. It needs to be explored further with the more powerful computational algorithms that are now available.

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This is a well written post, but it doesn't appear to have any connection to the original question. –  Andy W Jan 2 '12 at 16:53
    
I think it indirectly but appropriately addresses the apparent intent of the question, @Andy. The fault, if any, lies in the question itself, which is ambiguous because it does not specify the sense in which a mean might be "worse" than a median. A good answer therefore has to explore this and consider the purpose of summarizing an age distribution with a single statistic. Here, this leads naturally to a discussion of what an "age" might mean and how appropriately to compare age distributions. –  whuber Jan 2 '12 at 17:47

I hope the mean age would be influenced by the outliers in your data set while this is not the case for a median age. Let us take an example of a data set vaccinated patients: 1,2,3,4,4,5,6,6,6,78 years the mean would be:11.5 and median age of these patients is 4.5. this mean age has been affected by the outlier 78. median is the best while dealing with data sets of the skewed distribution.

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See my response to User28. –  Alexis May 4 at 18:40

To give a useful answer the original question requires we know the question behind the question. In other words, "Why do you want some sort of summary statistic comparing the age distribution of different countries?" The median might be the most useful for some questions. The mean might be the most useful for others. And there are probably questions where "percent above (or below) some particular age" would be the most useful statistic.

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Certainly in the case of demographic analysis, I would think that both the mean and median would be valuable, especially in combination with each other, if you are looking for outliers or areas of growth that may be mislabeled by the median alone. In communities with a large retirement community or in an area with a birth rate explosion, the median alone may not give you the whole picture, and that is where the mean, in comparison, can be very useful.

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