# can I work out *the* standard deviation from a value given as “+/- 1 standard deviation”

I'm confronted by a table of values:

17.1  (16.6)
18.0  (9.5)
38.1 (22.5)
30.6 (16.7)


In the left-hand column we have some mean values. I’m told that the values in parenthesis are 'one standard deviation of the mean' Which I understand to be a range, within which 68% of values would fall.

What I want to know is THE standard deviation associated with the numbers on the left. Am I missing something? Is something perhaps wrong with the information supplied to me in the table? If not, is there a way to move from 'one standard deviation' to 'the standard deviation'. The sample size in every case is 5.

EDIT

In response to the excellent answers and insights already posted, I have added more detail in the comments. Here is the table I am looking at:

Here is a reference to the information:p184 for the data table in question (Table5), Catena, 2008, 'Surface soil hydraulic properties in four soil series under different land uses and their temporal changes' X. Zhou a,⁎, H.S. Lin a, E.A. White p:184

• Much more detail would help us here as you have two answers which are in sharp conflict. What is this variable, can you tell us where this all comes from, why have you tagged it meta-analysis? – mdewey Feb 27 at 16:42
• Hi thanks for your comments. Yes, we seem to be in the realm of guesswork slightly! I have tagged in meta analysis, as I am compiling data for said analysis. These numbers come from a hydrology paper. I suspect the author is neither a native English speaker or a statistician - the only information I have is that the numbers here (measures of hydraulic conductivity - unlikely to be normally distributed) are the mean values on the left and ('One standard deviation of the mean' in parenthesis) – francis Feb 28 at 9:03
• . I'm trying to work out whether I can include this data or not as I am not sure, as has been pointed out here, whether or not the author in fact just means 'standard deviation' or whether I am getting confused, they do infact mean 'one SD' , and further, if they did mean that, would there be some way of deriving the actual SD from their report of 1SD.... Thanks again for all your comments, its definitely helping (if not quite definitively) understand whats going on! – francis Feb 28 at 9:04
• So, you are using a published source. Full reference to journal paper, book or technical report would do no harm: at least some readers may have access to the original. – Nick Cox Feb 28 at 10:40
• once again thanks - i wasn't sure whether a paper reference would be useful or not, here it is,p184 for the data table in question (Table5), Catena, 2008, 'Surface soil hydraulic properties in four soil series under different land uses and their temporal changes' X. Zhou a,⁎, H.S. Lin a, E.A. White p:184 .......................Once again, many many thanks for the insightful comments – francis Feb 28 at 12:06

To underline the obvious, we don't have any more information than you give, as you don't give a source that we can look at ourselves, or any wider context on what variable or variables are being discussed. Is this something conveyed informally? Can't you illuminate what the context is? Homework? You're the assistant or consultant?

"one standard deviation of the mean" strikes me as awkward wording. It could mean the standard error of the sample mean.

From those values, however, my guess is that you are being given the means and in parentheses the standard deviations of various data. So, 17.1 is the mean and 16.6 is the standard deviation of the first set (subset, variable: I am just guessing).

Note that the 68% you cite is a result for the normal distribution. 68% of the values of a normal distribution lie within 1 SD of the mean. But if your data are not normally distributed, then that may not hold.

In particular, if your data are for variables that can only be positive, or can only be zero or positive, then the SDs given are roughly the same order as the means, and the implication is that the variables are positively skewed and not normally distributed. If mean $$-$$ SD is close to zero and negative values are not allowed, then you can't have a normal distribution.

I have used the words guess and guessing already. Here's a third: this is all just a guess. You don't give enough information for this analysis to be confident. (If someone is garbling or limiting what they tell you, that's not good either.)

I can't begin to guess what A means.

My answer is not at all the same as that of @asdf, which should be an extra warning.

Let me try to end confidently: I can't think that there is a difference in substance between "one standard deviation" and "the standard deviation".

• Given the relative size of the means and supposed standard deviations I suspect they are not standard errors assuming the variable is bounded at zero. – mdewey Feb 27 at 16:41
• @mdewey That's my guess too. OTOH, the sample sizes are just 5, so large SE means aren't absolutely ruled out. But some wild guesses on plausible values support SD not SE mean. – Nick Cox Feb 27 at 17:32
• I have done some calculations and posted them as an answer. If you have a moment you might like to check my working. – mdewey Feb 28 at 10:28
• @mdewey Indeed. Similar arguments, but not nearly so well developed, underlay my previous comment. Soil hydraulic conductivity is nearer my tramping grounds than yours, or so I infer, and I underline that I expect large variance and strong skewness in the data. – Nick Cox Feb 28 at 10:37

Let us assume that conductivity is non-negative and that there are only two possibilities: standard deviation (sd) and standard error of the mean (sem). Then let us look at the second value for Glenelg which is 1.6 (2.1). Since there are 5 values we know their sum is 1.6 * 5 = 8. The maximum sd for five values adding to 8 of a non-negative variable is found for the values 0, 0, 0, 0, 8 and is 3.58. So the maximum value of the sem would be $$3.58 / \sqrt(5 - 1) = 1.79$$ which is less than the value of 2.1 quoted. So we deduce that 2.1 cannot be the sem as it is too large so must be the sd.

Note however that this is based on the assumptions stated in the first sentence so we cannot be 100% sure based on the evidence we have.

If the numbers in parenthesis are one standard deviation, you can just subtract them from the left "mean" column and get the actual standard deviation, so I think I am not understanding your question.

• HI thanks, no, if that is the relationship between ' ONE standard deviation' and 'THE standard deviation', then you have totally understood me, and answered my question! I could not find this explained anywhere...(at least not with 10/15 minutes of looking online) Thanks again – francis Feb 27 at 9:02
• This answer seems to me to be mistaken. Can you explain your reasoning in detail? – mdewey Feb 27 at 16:39