I am looking at someone else’s published data where they report treatment means and then a single standard error value across treatment means.

Sanderson et al. (2018)

I want to know if certain treatments are significantly different from each other. For example, I’ve highlighted a comparison I am interested in in green.

I am thinking that I might be able to use the Tukey HSD test, but I am not sure because I am not sure how standard error is defined and used in the Tukey HSD.

I am looking at an example from the following textbook by Robert O. Kuehl: “Design of Experiments: Statistical Principles of Research Design and Analysis. 2nd Edition.”

In the example (p. 107-109), you calculate the HSD by multiplying the Studentized range statistic (q) by the standard error.

Here’s my confusion. On page 107 it says the standard error is “the standard error of a treatment mean” which, to me, evokes the idea of a unique standard error for each treatment mean calculated from that treatments individual replications.

That understanding of it doesn’t reconcile well with the idea that the HSD is going to be the same across treatments comparisons. On page 109 there’s an actual numeric example, and the standard error used is constant no matter what means are being compared. That suggests that the standard error is calculated from all the treatments, but I am not sure how.

What does the standard error used in the Tukey HSD calculation refer to and how is it calculated?

Also, is it safe to assume that the standard error reported in the table above (highlighted in blue) is the standard error you would use in a Tukey HSD calculation? There's no additional information in the text of the article further defining what is meant by SE in the table.


1 Answer 1


Here’s my confusion. On page 107 it says the standard error is “the standard error of a treatment mean” which, to me, evokes the idea of a unique standard error for each treatment mean calculated from that treatments individual replications.

Statistical analysis often involves pooling information among observations to get more reliable estimates of parameter values. In an analysis of variance like what you show in your question, you do not use calculations "from that treatment's individual replications" to get standard errors. Rather, based on an assumption that the underlying variance of observations is the same among all treatments, you use all the replications on all treatments to get an overall estimate of the variance. For each treatment you estimate the variance around that treatment's mean value, and then pool those estimates from all treatments.

If all treatments had the same number of replicates, then the standard errors of the mean values are the same for all treatments. The use of "SE" by the authors of that table suggests that the values reported are for those standard errors of the mean values, taking the number of replicates into account.

For t-tests in general when testing the difference of a value from 0, the statistic you calculate is the ratio of that value to the standard error of that value. For Tukey's range test, instead of evaluating the statistic against a t distribution with the appropriate number of degrees of freedom you evaluate it against a studentized range distribution that also takes into account the number of values that are under consideration for testing. This provides a correction for multiple comparisons.

So Tukey's test uses a t-test type of statistic that is the ratio of the difference between two mean values divided by the standard error of the difference of those two means. Assuming independence and the same individual standard errors, that would be $\sqrt 2$ times the standard error of an individual mean.* You would have to check the software you are using to see whether it expects from you the standard error of an individual mean or the standard error of a difference of two means.

That said, the Tukey test might not be the best way to accomplish what you want. It's appropriate if you have several treatments (with a common estimated standard error) that you want to compare in a way that corrects for multiple comparisons. If you have a single pre-specified comparison in mind, not developed based on looking at the data, then you don't have to correct for multiple comparisons. Note in particular that if a difference isn't significant without that correction, then it certainly won't be significant after correction. In the particular comparison you highlight, the difference of 567 in treatment means is only 1.55 standard errors and thus would not pass the standard test of statistical significance at p < 0.05.

*The Wikipedia web page I linked says it's the standard error of the sum of the values, but for uncorrelated variables that's the same as the standard error of their difference. One situation in which you do have to deal with treatments individually is if there are different numbers of replicates for the two treatments being compared, not addressed directly in the Wikipedia page.

  • $\begingroup$ That clarifies things. Thank you. If there's no need to correct for multiple comparisons, then would the approach be to use a t-test? In which case, when calculating the standard deviation from the given standard error, would I multiply the standard error by the square root of the number of replications per treatment (4) or the total number of replications across all treatments (4*15) or something else entirely? What is considered the sample size when a standard error is calculated across multiple replicates and treatments? $\endgroup$
    – Angela
    Nov 19, 2019 at 1:01
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    $\begingroup$ @Angela I edited my answer to say more about the t-test statistic. It's the ratio of a value to the standard error of the value, so you don't have to back-calculate standard deviations from the standard error. For a difference of 2 mean values each having the same individual standard error, the standard error is $\sqrt 2$ times the individual standard error. The numbers of treatments and replicates enters into the number of degrees of freedom to use for the t-distribution against which you compare the statistic. That's discussed in the t-test Wikipedia page I added as a link. $\endgroup$
    – EdM
    Nov 19, 2019 at 15:46
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    $\begingroup$ @Angela one more hint and one warning. The hint: with large numbers of degrees of freedom the t-distribution approaches a normal distribution. For a normal distribution, you need to have a value with an absolute magnitude at least 1.96 standard errors above zero for p < 0.05. For a difference between 2 means that would be about 2.8 times the standard error of an individual mean. For looking quickly at data, that's a useful rule of thumb--if a difference isn't significant based on the normal approximation, it won't be with a proper t-test either. $\endgroup$
    – EdM
    Nov 19, 2019 at 15:55
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    $\begingroup$ @Angela The warning: in my answer I was careful to say "If you have a single pre-specified comparison in mind, not developed based on looking at the data" as a situation for which a standard t-test could be OK. If you test an hypothesis based on looking at the results of a data set or if you are testing multiple such hypotheses, then a t-test isn't appropriate without some form of correction for multiple comparisons, like the Tukey test. But for ruling out possible significant differences, if a value doesn't pass significance without correction it won't with correction, either. $\endgroup$
    – EdM
    Nov 19, 2019 at 16:00
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    $\begingroup$ @Angela specifying the "highest performing" complicates things, as the identification of which cases are "highest performing" does depend on the data. That raises issues of extreme-value analyses on which I have no expertise. Multiple-comparison correction might thus be necessary. Also, there would have to be some correction for multiple comparisons if you analyzed all time points. In this particular case, it doesn't look like your null hypothesis would be ruled out in any situations even without the correction, so it would also not be ruled out with the correction. $\endgroup$
    – EdM
    Nov 19, 2019 at 18:10

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