Two Ways to Average Repeated Concentrations
When subjects contribute different numbers of concentration measurements, the phrase “mean concentration” can refer to more than one calculation. One approach gives every recorded concentration the same weight. Another first calculates a mean for each subject and then gives every subject the same weight. These are distinct summaries when visit counts differ, and each answers a different question.
The distinction is about the quantity being averaged, not about which arithmetic is correct. A pooled mean describes the average among the observed concentration records. An equally subject-weighted mean describes the average of the subject-specific observed means in the group. Neither calculation automatically describes concentrations at a common time point or a broader target population.
In these definitions, each subject must have at least one observation for their subject-specific mean to be defined. The summaries also require a clear rule about which records are included: for example, whether the calculation uses all available records or only measurements in a specified time window. The units of both means are concentration units, such as mg/L.
A Direct Calculation From Unequal Visit Counts
Consider the following entirely hypothetical dataset. The observations are concentrations in mg/L, and subjects contribute different numbers of records.
| Subject | Observed concentrations (mg/L) | \(n_i\) | \(\bar C_i\) (mg/L) |
|---|---|---|---|
| A | 12, 8, 4 | 3 | 8 |
| B | 10, 6 | 2 | 8 |
| C | 2 | 1 | 2 |
Worked Example: Pooling Six Concentration Records
Calculate the observation-weighted pooled mean and the equally subject-weighted mean for the table.
Observation-weighted pooled mean. There are \(N=3+2+1=6\) records. Their total concentration is \(12+8+4+10+6+2=42\) mg/L, so
Equally subject-weighted mean. The subject means are 8, 8, and 2 mg/L. Giving each of the three subjects one equal share,
The pooled mean is the average concentration among these six observed records. The subject-weighted mean is the average of these three subjects’ observed means. They differ by 1 mg/L because the pooled calculation gives subject A three records, subject B two, and subject C one; the subject-weighted calculation gives each subject one-third of the total weight.
The same pooled calculation can be written using subject means and visit counts:
Thus, the pooled mean is a weighted average of the subject means, with subject \(i\)'s weight equal to their fraction of all records, \(n_i/N\). The weights are nonnegative and sum to one. In the table, those weights are \(3/6\), \(2/6\), and \(1/6\), whereas the equally subject-weighted calculation uses \(1/3\) for each subject.
Why the Summaries Differ
The difference is determined by how visit counts align with subject means. If subjects with more records tend to have higher means, the pooled mean tends to exceed the equally subject-weighted mean. If they tend to have lower means, it tends to be smaller. If all subjects contribute the same number of records, the two means are equal.
This relationship can be shown directly. Write \(\bar n=N/S\) for the average number of observations per subject and \(\bar C_{\mathrm{subject}}\) for the mean of the subject means. Because \(\sum_i(n_i-\bar n)=0\) and \(\sum_i(\bar C_i-\bar C_{\mathrm{subject}})=0\), subtracting the subject-weighted mean from the pooled mean gives
For the second equality, \(1/S=\bar n/N\), so the difference is initially \((1/N)\sum_i(n_i-\bar n)\bar C_i\). Subtracting \(\bar C_{\mathrm{subject}}\) inside each product leaves the sum unchanged because \(\sum_i(n_i-\bar n)=0\). The resulting expression shows that the difference depends on the alignment of each subject’s record count and mean, not just on the fact that counts are unequal.
Worked Example: More Records From a Higher-Mean Subject
In a second hypothetical dataset, subject A has four concentrations with mean 10 mg/L, subject B has one concentration of 6 mg/L, and subject C has one concentration of 2 mg/L. For example, A’s four values could be 16, 12, 8, and 4 mg/L. Calculate both summaries.
The total of A’s four observations is \(40\) mg/L, and B and C contribute \(6\) and \(2\) mg/L. There are \(N=4+1+1=6\) records, so
The equally subject-weighted mean is
Here the pooled mean is 2 mg/L higher. Subject A has both the most records and the highest subject mean, so giving every observation equal weight gives A more influence than an equal-subject calculation does. This arithmetic does not establish why A has more observations; it only identifies the consequence of the chosen weighting.
Worked Example: More Records From a Higher-Mean Subject Can Reverse the Comparison
Consider three hypothetical subject means: 5, 10, and 15 mg/L. Suppose their respective record counts are 1, 2, and 5. Calculate both summaries and interpret their ordering.
The equally subject-weighted mean is
For the pooled mean, the total number of records is \(N=1+2+5=8\). The weighted total is \(1(5)+2(10)+5(15)=5+20+75=100\) mg/L, giving
The pooled mean exceeds the equal-subject mean because the highest-mean subject contributes the most records. This example also shows why counts alone cannot predict the size of the difference: the subject means matter as well.
What Population Summary Does Each Mean Represent?
For a fixed set of enrolled subjects and the observations specified in the calculation, the pooled mean describes the average over the \(N\) recorded concentrations. Equivalently, imagine selecting one of those records at random, with each recorded value equally likely: the mean of that record-level distribution is the pooled mean. A subject who supplied more records has a greater chance of being represented in that draw.
The equally subject-weighted mean describes the average of the \(S\) subject-specific observed means. Equivalently, select one of the included subjects with equal probability, then consider that subject’s mean over their included records. This gives each subject equal weight, but it does not give each measurement time equal weight or ensure that every subject was measured at the same times.
So the question “which summary should be reported?” depends on the intended target. If the target is the average of all available records, the pooled mean matches that arithmetic target. If the target is the average subject-specific observed mean among the included subjects, the equal-subject calculation matches it. The label “population mean” should not be used without specifying the population and sampling target: neither calculation, by itself, guarantees representativeness of a wider patient population.
| Summary | Weight assigned | Direct interpretation for the dataset |
|---|---|---|
| Observation-weighted pooled mean | Each record has equal weight; subject \(i\)'s total weight is \(n_i/N\). | Mean of all included concentration records. |
| Equally subject-weighted mean | Each included subject has weight \(1/S\). | Mean of the included subjects’ record-level means. |
| Mean at a specified time | Defined by selecting or summarizing records at that time. | Mean concentration at the specified time, if the data support that comparison. |
A concentration mean across records collected at different times is not automatically a mean concentration at any particular time. If subjects are measured at different points on a concentration-time profile, the pooled records combine both subjects and times. The equal-subject mean still averages each subject’s available records, but subjects may have different sets of times. To make a time-specific comparison, the records and target time must be defined explicitly rather than inferred from either overall mean.
Checks, Pitfalls, and Limits
A practical check is to calculate a short subject-level table before reporting an overall mean. For each subject, record \(n_i\), the sum of concentrations, and \(\bar C_i\). Confirm that the counts add to \(N\), that the pooled total agrees with \(\sum_i n_i\bar C_i\), and that the equal-subject mean uses exactly one subject mean per included subject. These checks can expose a missing record, an accidental omission of a subject, or an unintended weighting choice.
Also inspect what the records represent. If the pooled mean is intended to describe all observations, unequal record counts are part of its definition. If the goal is an equal-subject summary, unequal counts should not inadvertently be allowed to change subject weights. But switching to equal subject weights does not correct every design or interpretation problem: the subjects may have measurements at different times, and the subjects included may not represent the population of interest. The arithmetic alone cannot decide whether additional weighting, a restricted time window, or a different analysis is appropriate.
- Mistake: treating the two means as interchangeable. They are equal when every subject has the same number of records, but generally differ when counts are unequal and subject means vary. State which weighting was used.
- Mistake: dividing the total by the number of subjects. For the pooled mean, divide by the number of concentration records \(N\). For the equally subject-weighted mean, first calculate each \(\bar C_i\), then divide their sum by \(S\).
- Mistake: interpreting an overall record mean as a concentration at a common time. Check the times attached to the records. An across-record average does not become time-specific just because every record is a concentration measurement.
- Limitation: the summaries do not explain unequal counts. They show how observations were weighted, not why some subjects contributed more records or whether the observed subjects represent a broader population.
The key decision is to name the target summary before calculating. “Average observed record” and “average subject-specific observed mean” are precise descriptions; an unqualified “average concentration” can conceal an important change in weighting.
Check Your Understanding
Answer each question, then compare with the solutions below.
- Two subjects have concentrations 4 and 8 mg/L, and 10 mg/L and 14 mg/L, respectively. Calculate the pooled and equally subject-weighted means.
- In a dataset, three subjects have means of 3, 6, and 9 mg/L and contribute 1, 1, and 4 records, respectively. Calculate both means and say which is larger.
- When must the pooled and equally subject-weighted means be equal?
- A pooled mean combines records from different sampling times. Can it automatically be interpreted as the mean concentration at one specified time? Explain.
Solutions. (1) Each subject has two records, so the pooled mean is \((4+8+10+14)/4=36/4=9\) mg/L. The subject means are 6 and 12 mg/L, giving \((6+12)/2=9\) mg/L. (2) The equal-subject mean is \((3+6+9)/3=6\) mg/L. The pooled mean is \((1(3)+1(6)+4(9))/6=45/6=7.5\) mg/L, which is larger because the highest-mean subject contributes the most records. (3) They are equal if all subjects contribute the same number of records; they can also coincide in particular datasets when the record counts and subject means happen to produce no net weighting difference. (4) No. The pooled mean averages the included records, which may represent several times. A time-specific interpretation requires defining and using observations for that time.