Both the procedures are correct but the stds have different meaning. The first std.(red)=0.06 measures the dispersion of the three sample means with respect to the grand mean 0.52. The second std. (blue)=0.3 measures the dispersion of the 36 samples with respect to the grand mean 0.52. The standard uncertainty (SU) of the grand mean estimated from the two procedures can be calculated: (1) SU=0.06/sqrt(3)/c4=0.039 (c4 is the bias correction factor, c4=0.8862 at n=3), (2) SU=0.3/sqrt(36)/c4=0.050 (c4=0.993 at n=36). Please refer to the following paper for the unbiased estimator of SU.

Article Comparison of three approaches for computing measurement uncertainties

The standard uncertainty is the best representation of the dispersion of the data. It can be calculated as the standard deviation of the means if the measurements were conducted under (statistically controlled) or the repeatability conditions. Thus (assuming unbiased measurements);

SU=Std.Dev.[0.59 0.47 0.50]=0.06 (without dividing by sqrt(3), I could not understand why Hening Huang divided by sqrt(3) ).

Thus, the data can be represented as X (grand mean) +/- 0.06.

Nevertheless, if a single group of measurements was conducted, the standard uncertainty can be estimated as:

SU=Std.Dev(Data)/sqrt(n)

Thus, if it has been assumed that all the 36 observations were conducted as a single group of measurements, then uncertainty can be estimated as:

SU=0.3/sqrt(36)=0.0503.

Ideally, the estimated SU=0.3/sqrt(36) should equal Std.Dev[0.59 0.47 0.50]. However, this is valid only under the following conditions;

i. A normal distribution is assumed for the measurements' results;

ii. The measurements were conducted under repeatability (or statistically controlled) conditions;

iii. A sufficient data size is available to approach the population (Typically, as the replicated number within each group increases, the estimate Std.Dev.(x)/sqrt(n) approaches the standard deviation of the means).

It seems that the results and the measurement process do not fulfill these conditions. Although you may not be interested, using one-way ANOVA would be better for correct evaluation of uncertainty.

Amr M. Sadek Hening Huang thank you both for the answers and the explanations you have provided. Actually, the dataset I provided was randomly generated via excel. I was simply interesed in the correct representation of the dispersion of a mean of a three years measurements of different variables (such as soil temperature).

Just to clarify, the model I have used to compare different groups was a RMANOVA (GLM. SPSS. Data Box-Cox transformed to meet all the assumptions, checked with Shapiro, Levene's, and Mauchly tests).

So, practically I was interested in the correct way to report the data dispersion on the three-years mean of each group, regardless of time. Just the mean, something like:

"throughout the three year-monitor period, the mean soil temperature was 24+/- SD (or better, +/- Standard Error) and 27 +/- SD in A and B, respectively"

This is an interesting experiment. I assume that you randomly drawn 36 samples from a normal distribution with a mean (about 0.5) and a std. (about 0.3). You then divided the 36 samples into 3 groups, each having 12 samples. I also assume that you want to report the mean (i.e. the grand mean) and the associated standard uncertainty (SU) (or standard error). The two procedures you described in your question give the same grand mean=0.52. However, neither 0.30 nor 0.06 is the SU of the sample grand mean. There are two procedures to estimate the SU, corresponding to the two procedures of estimating the grand mean. Procedure 1 is simply SU=0.3/sqrt(36)/c4=0.050, where c4=0.993 at n=36. Procedure 2 is a little bit complicated because we may need to account for the heterogeneity between the 3 groups (I didn’t know your 36 sample were from the same distribution). In fact, SU=0.06/sqrt(3)/c4=0.039 (c4=0.8862 at n=3) is an unconditional SU that accounts for the heterogeneity (supposedly). The conditional SU can be estimated as 1/3*sqrt[Var(group1)+Var(group2)+Var(group3)]=1/3/sqrt(0.307^2+0.280^2+0.329^2)=0.051. The SU of the sample grand mean should be the unconditional SU or the conditional SU, whichever is greater (Huang 2019). Therefore, Procedure 2 gives SU=0.051 that is very close to SU=0.050 Procedure 1 gives. The results comply with your experiment in which the samples were drawn from the same distribution so that there is no heterogeneity between the 3 groups. You may do a meta-analysis to verify this.

Huang, H. (2019). A unified formula for uncertainty estimation in interlaboratory studies and key comparisons. Cal Lab the International Journal of Metrology, 26(4), 22-29. Available from https://www.researchgate.net/publication/338344682_A_Unified_Formula_for_Uncertainty_Estimation_in_Interlaboratory_Studies_and_Key_Comparisons