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While reading an article (published and peer-reviewed, referenced below) investigating proportions of different tissue types in the human brain I came upon a table presenting data on the different subgroups in the study. Data was presented as means +/- SE (which I assume to be standard error, although I could not see it specified in the text) and range.

One example is the age of one of the subgroups: n= 23 mean age = 66 SE = 2.9 range of age 60-69

This puzzled me. If the SE is 2.9, then given the sample size of 23 the SD of the age should be:

SD = sqrt(23) * 2.9 = 13.9

This would lead to a standard deviation that is quite a lot larger than the specified range for the age, which should not be possible.

I'm a beginner in the world of statistics, so my question is simply if it from this is possible to conclude that one or several of the values given must be incorrect, or if I'm missing something that would make this data to make sense.

Thank you!

Reference: Guttmann CR, Jolesz FA, Kikinis R, et al. White matter changes with normal aging. Neurology 1998;50:972-978.

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  • $\begingroup$ There is inconsistency in reporting parameters of variables in articles. Variable (age) probably does not have normal distribution in this particular case. Therefore is "range" also given - which is common in the case of non-normal distribution. I think that the "strict" rules should be applied when reporting parameters. Probably most appropriate should be reporting of mean and SD (or SEM) for normal distribution and in the case of non-normal distribution - median and range. $\endgroup$ Feb 24, 2014 at 16:35
  • $\begingroup$ @Ladislav Nado. Thank you for the comment! I agree with your suggestions regarding the reporting of variance. In this specific case, do you also find the SE to be too large, or is it statistically/mathematically possible for this SE to occur in relation to the range and number of observations? Any insight into whether or not I have understood this correctly would be appreciated! $\endgroup$
    – Mattias
    Feb 24, 2014 at 19:36
  • $\begingroup$ I finally got the point. Yes, the value 13.9 for SD is impossibly large when we know that range is from 60 to 69 (difference only 9). The assumed SE is probably SD. You have a sharp mind. $\endgroup$ Feb 24, 2014 at 20:07

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I'm looking at the paper now (figures are at the end).

I may have missed something, but so far I see nothing in the paper that states that the 2.9 is intended to be a standard error (I can't find "SE" or "standard error" in the paper, for example).

Edit: Mattias points out in comments that (unlike the html version I linked), the pdf version definitely says 'SE', which invalidates what follows. It means the article has it wrong.

You may have inferred that it is a standard error from the way the information is presented in Table.1, where for example the mean "Age" for age group 60-69 is given as 66.0$\pm$ 2.9.

However, it's not unusual for $\pm$ to be used in different ways, such as to indicate the standard deviation, or some multiple of the standard error (leaving us to always require it to be spelled out if we are to know for sure what it means).


Bounding the standard deviation and the standard error of the mean

In any case, it's a good question, and such investigation of the information in papers is important.

The largest possible value for a population standard deviation for a bounded continuous variable on $[a,b]$ is $(b-a)/2$; this happens when half the observations are at the lower limit and half at the upper limit.

So for example if we knew that the age group is $60-69$ (assuming ages are recorded only in whole years), the biggest standard deviation possible is $4.5$, and the biggest standard error of the mean would be $4.5/\sqrt{n}$.

enter image description here

Of course, if the sample variance is based on an $n-1$ denominator, then the standard deviation can slightly exceed half the range (in an easily computable way).

The simple rule of thumb - the standard deviation shouldn't be more than half the range - is one worth remembering, as long as for small samples we keep in mind it's really $s_n$ that it holds for.

However, we can bound it further. First note that $n$ was odd in the table in question, so that we can't actually place half at each end, and it's possible to compute the (slightly smaller) standard deviation that would allow. Even more importantly, we're told the mean, and that can have a greater impact, reducing the maximum standard deviation to roughly 4.15 ($s_n$) or 4.24 ($s_{n-1}$).

Note that if the age had been uniformly distributed, it would have given about the right standard deviation:

enter image description here

(It isn't actually uniform - we can tell that because the mean is higher than the center value, but it gives us some idea of the kind of spread we have.)

The "sd < $\frac{1}{2}$ range" rule of thumb remains probably the most useful one to remember, unless you're doing some very detailed investigation.

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  • $\begingroup$ Thank you for the very clear and informative answer! Regarding the mentioned paper; in the pdf-version there is a legend under the table you linked to, stating that "values are means +/- SE with ranges in parentheses", so that's what sparked my question. $\endgroup$
    – Mattias
    Feb 25, 2014 at 7:27
  • $\begingroup$ Thanks Mattias. In that case, yes, you're right, the standard error of the mean (conditional on age being in that range) can't be that large, at least not calculated in the usual way. I suspect it may be a typographical error - 2.9 is a reasonable value for the standard deviation. $\endgroup$
    – Glen_b
    Feb 25, 2014 at 11:28

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