Why the Number of Observations Matters
When subjects contribute different numbers of measurements, an average over all recorded observations does not give every subject the same total influence. A subject with many measurements contributes many terms to the pooled sum; a subject with few measurements contributes fewer. This is an arithmetic fact about how the mean is constructed, not by itself evidence that the pooled mean is wrong.
To make the distinction precise, suppose a numeric variable is measured for \(S\) subjects. Subject \(i\) contributes \(n_i\) observations \(x_{i1},\ldots,x_{in_i}\), where each subject has at least one observation. Let \(N=\sum_{i=1}^{S}n_i\) be the total number of observations and let \(\bar{x}_i\) be subject \(i\)'s mean. The measurements must be on a common scale and represent the quantity being averaged. For example, measurements in milligrams per litre can be averaged with other measurements in the same units.
The two means summarize the same collection of subjects and measurements, but they give different total weights to subjects whenever the \(n_i\) differ. The pooled mean gives every observation weight \(1/N\). Therefore, subject \(i\)'s total weight is \(n_i/N\). In the equal-subject mean, subject \(i\)'s total weight is \(1/S\), and each of that subject's observations has weight \(1/(S n_i)\) in the resulting summary.
Algebra of the Unequal Weights
The pooled sum can be regrouped by subject. Within each subject, the sum of the measurements is \(n_i\bar{x}_i\). Substituting that identity into the pooled mean gives a weighted average of the subject means:
The weights \(n_i/N\) are nonnegative and sum to one, since \(\sum_i n_i/N=N/N=1\). Thus the pooled mean is a weighted average of subject means, with weights proportional to the number of observations each subject contributes. By contrast, \(\bar{x}_{\mathrm{subject}}=\sum_i(1/S)\bar{x}_i\) assigns the same weight to every subject.
A useful expression makes the direction of the difference explicit. Let \(\bar{n}=N/S\), the average number of observations per subject, and define the finite-subject covariance between counts and subject means as \(C_{n,\bar{x}}=(1/S)\sum_{i=1}^{S}(n_i-\bar{n})(\bar{x}_i-\bar{x}_{\mathrm{subject}})\). Expanding the product gives \(C_{n,\bar{x}}=(1/S)\sum_i n_i\bar{x}_i-\bar{n}\bar{x}_{\mathrm{subject}}\). Since \(\bar{x}_{\mathrm{pool}}=(\sum_i n_i\bar{x}_i)/(S\bar{n})\), it follows that:
This identity says that the pooled mean exceeds the equal-subject mean when subjects with more observations tend, in this set of subjects, to have higher subject means; it is lower when they tend to have lower means. If counts and subject means have zero finite-subject covariance, the summaries are equal, even if counts are unequal. Equal counts are another sufficient condition for equality, because then every pooled subject weight is \(1/S\). Neither condition should be mistaken for a claim about how the measurements were generated.
Worked Example: Two Subjects With Unequal Counts
Suppose two hypothetical subjects have concentration measurements, in milligrams per litre, as follows. Subject A has \(2\) and \(4\), while Subject B has \(10\), \(12\), \(14\), and \(16\). Calculate the pooled observation mean and the equal-subject mean.
Subject A's mean is \(\bar{x}_A=(2+4)/2=3\) mg/L. Subject B's mean is \(\bar{x}_B=(10+12+14+16)/4=13\) mg/L. There are \(N=2+4=6\) observations in total.
The pooled mean is computed from all six observations: \((2+4+10+12+14+16)/6=58/6=9.667\) mg/L. Equivalently, the subject-mean representation gives \((2/6)(3)+(4/6)(13)=1+8.667=9.667\) mg/L.
The equal-subject mean is \((3+13)/2=8\) mg/L. The summaries differ because Subject B's subject mean is higher and that subject supplied two-thirds of all observations. In this hypothetical data set, the pooled mean is 1.667 mg/L higher than the equal-subject mean. The calculation alone does not establish which summary is preferable; that depends on the question the summary is intended to answer.
Worked Comparisons
Worked Example: More Observations From a Lower-Mean Subject
In a hypothetical measurement exercise, Subject C contributes one value, \(20\) units, and Subject D contributes three values, \(8\), \(10\), and \(12\) units. Compare the two summaries.
Subject C's mean is \(20\) units. Subject D's mean is \((8+10+12)/3=10\) units. The pooled mean is \((20+8+10+12)/4=50/4=12.5\) units. Using the weighted subject means gives the same result: \((1/4)(20)+(3/4)(10)=5+7.5=12.5\) units.
The equal-subject mean is \((20+10)/2=15\) units. Here the subject with the lower mean contributes more observations, so the pooled mean is below the equal-subject mean by \(2.5\) units. The subject counts are the weights: \(1/4\) for C and \(3/4\) for D in the pooled summary, versus \(1/2\) for each subject in the equal-subject summary.
Worked Example: Unequal Counts but Equal Summaries
Consider hypothetical subject means of \(5\) units for Subject E and \(11\) units for Subject F. E contributes two observations, and F contributes four. What are the pooled and equal-subject means if each subject's observations have the stated subject mean?
The pooled mean is \((2/6)(5)+(4/6)(11)=10/6+44/6=54/6=9\) units. The equal-subject mean is \((5+11)/2=8\) units, so these would not be equal. This calculation illustrates that unequal counts alone do not guarantee equality: here the higher-mean subject also has more observations.
Now consider a different hypothetical pair of subject means, \(5\) units for E and \(11\) units for F, but with E contributing four observations and F contributing two. The pooled mean is \((4/6)(5)+(2/6)(11)=20/6+22/6=7\) units, while the equal-subject mean remains \(8\) units. These examples make the role of the direction of the weighting visible.
For an exact equality with unequal counts, suppose E's mean is \(5\) units and F's mean is also \(5\) units, with E contributing two observations and F contributing four. The pooled mean is \((2/6)(5)+(4/6)(5)=5\) units, and the equal-subject mean is \((5+5)/2=5\) units. The subject means are identical, so changing the weights cannot change the result.
Choosing and Checking a Summary
The pooled and equal-subject means answer distinct descriptive questions. The pooled mean answers: what is the mean among the recorded observations, giving each recorded value equal weight? The equal-subject mean answers: what is the average of the observed subject-specific means, giving each included subject equal weight? These are not competing arithmetic rules for one uniquely defined target. The intended target determines which weighting is suitable.
A pooled mean can be appropriate when the object of interest is the collection of recorded observations and equal weighting of those observations is intended. An equal-subject mean can be useful when the intended summary is the average of the included subjects' means. The latter does not automatically represent every subject in a broader population, and the former does not automatically represent a typical subject. Both are summaries of the data as observed; wider population interpretations require assumptions and a defined target.
A practical check is to calculate the counts \(n_i\) and means \(\bar{x}_i\) separately before reporting an overall average. Then verify the pooled calculation in both forms: \(\sum_{i,j}x_{ij}/N\) and \(\sum_i(n_i/N)\bar{x}_i\). Agreement checks the arithmetic and makes the weighting transparent. Compare this result with \(\sum_i\bar{x}_i/S\), and state clearly which one is being reported.
- Counts vary: inspect the subject weights \(n_i/N\); do not describe the pooled mean as equally weighted across subjects.
- Counts do not vary: the pooled and equal-subject means coincide, because every subject supplies the same fraction of all observations.
- Means differ but the summaries coincide: this can occur if the weighting differences balance, as when count and subject mean have zero finite-subject covariance.
- Values are missing or not comparable: first clarify which observations and subjects are included and whether the measured quantity has the same meaning and units across them. The identities above describe the chosen observed values; they do not settle whether those values form an appropriate target.
Common Mistakes and Limitations
A common mistake is to calculate the pooled mean correctly and then call it an equally weighted average of subjects. The wrong reasoning is: “Every measurement is included, so every subject counts equally.” The correction is that every measurement has equal weight \(1/N\), while a subject with \(n_i\) measurements has total weight \(n_i/N\). Equal weighting of subjects instead requires taking the mean of the subject-specific means.
A second mistake is to assume that the equal-subject mean is always more representative. It gives each included subject the same weight by definition, but that feature does not prove it is the scientifically relevant summary. Conversely, unequal subject weights do not make the pooled mean mathematically biased by themselves. The weighting may be exactly what the descriptive question calls for. The essential limitation is that neither calculation can determine the intended target on its own.
These formulas are descriptive identities. They do not account for within-subject dependence when calculating uncertainty, and they do not make observations from the same subject independent. A mean's weighting and the method used to quantify its uncertainty are separate issues.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions below.
- Three subjects have \(1\), \(2\), and \(5\) observations. What fraction of the pooled mean's total weight belongs to the third subject?
- Subject means are \(4\) and \(10\) units. The first subject contributes three observations and the second contributes one. Calculate the pooled mean and the equal-subject mean.
- Give one condition under which the pooled and equal-subject means must be equal, and explain why.
- If subjects with more observations tend to have lower subject means, which summary tends to be lower in that set of subjects? Explain in terms of the weights.
Solutions. (1) There are \(N=1+2+5=8\) observations, so the third subject's pooled weight is \(5/8=0.625\), or \(62.5\%\). (2) The pooled mean is \((3/4)(4)+(1/4)(10)=3+2.5=5.5\) units. The equal-subject mean is \((4+10)/2=7\) units. The pooled mean is lower because the lower-mean subject contributes more observations. (3) If all subjects have the same number of observations, each has pooled weight \(1/S\), the same as its equal-subject weight; the means must therefore coincide. They also coincide if all subject means are identical. (4) The pooled mean tends to be lower: it assigns larger weights to subjects with more observations, and here those subjects tend to have lower means.