It is obvious that when we group individual values, the std deviation considering the means of all groups is lower than the std deviation of the individual values. So, to reduce variation around the mean of all values, would it pay off to group individual values whenever possible?
It is best to use individual values with a caveat.
Means are estimated with +/- half length of the confidence interval (hCI) — assuming normally distributed individual values. The smaller the hCI, the more precisely you estimate the mean. The formula for hCI is:
hCI = std deviation*t/sqrt(n*g)
; where g is the number of groups, n is the number of individual values per group, t is the t-value with n-1 degrees of freedom and alpha = 0.05, and the std deviation refers to that of individual values. Let N = n*g, the total number of individual values.
So, for a given N, as you group values — as g increases — you forcedly reduce n to keep N constant. If you don’t group values, N = n. Then, for a given N, n is greater when you don’t group individual values. The value of n affects the t-value, because of the n-1 degrees of freedom. The higher the n, the lower the t-value — though not linearly. As such, with a lower t-value and same N, the mean estimated from individual values results in a smaller hCI — i.e. more precise.
The plot below shows what happens when std deviation = 1 — the standardized hCI (hCI*) — vs. N, with the number of groups g annotated. Note that as g increases from 1 to 4, hCI* increases for a given N. However, the difference in hCI* between the group sizes is reduced as N increases. So, for a large N, it does not matter in practical terms whether you group individual values. For a small N, it matters quite a lot.
Thus, it is best to estimate the mean from individual values. If you have a large N, grouping may at best be only as good as not grouping, but never better.
The sensation that grouping is best comes from the fact that the std deviation of group means (m) is: m = std deviation/sqrt(n). But if you use group means the hCI is:
hCI = m*t/sqrt(g) = [std deviation/sqrt(n)]*[t/sqrt(g)] = std deviation*t/sqrt(n*g).
Now, the caveat. The sensation that grouping is best may also come from the fact this is quite often the case in many testing standards, process control charts, etc...
Say you have individual measurements of a set of parts and these parts represent a lot. The reasoning behind the plot above applies to the simplest case where you want to come up with the mean of a specific part from its individual measurements — you are not interested in the lot mean or in other parts — so the data has no hierarchy.
If you are interested in the lot mean, then grouping the individual measurements by part is a must because parts might not all have the same mean. In this case, if you don’t group by parts and the number of individual measurements by part is highly unequal, the lot mean you get by averaging all individual values might be biased towards those parts with more measurements. Besides, even if the number of individual measurements by part is equal, the variation within the lot will be inflated by using individual values because it will reflect not only the part-to-part variability but also the measurement variability.
If you average the averages of each part, the impact of measurement variability is reduced on the lot variability. The higher the number of measurements per part, the lower this impact because the parts will be more precisely estimated — smaller hCI as per g = 1 in the plot above — so less biased by the measurement. A few individual measurements per part can reduce their hCI many folds.
