Why Pooled Proportions Are Weighted Averages

Why a Combined Proportion Depends on Group Sizes

A two-way table can show an outcome separately for several groups and also show the total across those groups. The overall outcome proportion is not generally found by treating every subgroup rate equally. It depends on how many observations are in each subgroup: a rate calculated from a large subgroup represents more observations in the combined total than a rate calculated from a small subgroup.

This relationship is useful whenever a table is stratified by a second variable. It explains how subgroup proportions combine into a marginal proportion for all observations. The calculation is a direct consequence of counting: add the outcome counts, then divide by the total number of observations. Rewriting that calculation in terms of subgroup proportions reveals the weights.

Definition: Suppose observations are divided into \(k\) non-overlapping groups, and the outcome of interest has two categories. For group \(j\), let \(n_j\) be its number of observations and \(x_j\) its number with the outcome of interest. Its observed subgroup proportion is \(\hat{p}_j=x_j/n_j\). The pooled proportion, \(\hat{p}\), is the proportion with the outcome among all observations in all groups.

Deriving the Weighted-Average Formula

Let \(N=\sum_{j=1}^{k}n_j\) be the total number of observations. Since each observation belongs to exactly one group, the total number with the outcome is \(\sum_{j=1}^{k}x_j\). Therefore, the pooled proportion is the total outcome count divided by the total number of observations.

\hat{p} = \frac{\sum_{j=1}^{k}x_j}{\sum_{j=1}^{k}n_j}

For each group, \(x_j=n_j\hat{p}_j\), because multiplying the group size by its observed proportion gives its outcome count. Substituting these counts into the pooled-proportion formula gives the weighted-average form.

\hat{p} = \frac{\sum_{j=1}^{k}n_j\hat{p}_j}{N} = \sum_{j=1}^{k}\left(\frac{n_j}{N}\right)\hat{p}_j

The weight for group \(j\) is \(n_j/N\), that group’s share of all observations. The weights are nonnegative and sum to one, because \(\sum_{j=1}^{k}(n_j/N)=N/N=1\). Thus, each subgroup proportion contributes according to the fraction of the combined data that it represents.

Theorem: The pooled proportion equals the group-size-weighted average of the subgroup proportions. The weight attached to each subgroup proportion is that subgroup’s size divided by the total size. This formula requires a consistent definition of the outcome and a complete partition of the observations into the groups being combined.

This is a descriptive identity, not an inferential procedure. It does not require a probability model or a significance test. For it to represent the intended overall rate, each observation must be counted once, the groups must not overlap, and the outcome category must mean the same thing in every group. The denominators must also refer to the observations eligible for that outcome in the table.

Find each subgroup size and outcome count.
Use the table’s counts, checking that each subgroup denominator includes the observations relevant to the outcome proportion.

Calculate the subgroup proportions.
For each group, divide its outcome count by its group size: \(\hat{p}_j=x_j/n_j\).

Combine using counts or weights.
Either add outcome counts and divide by the total size, or multiply each subgroup proportion by its share of the total and add.

Worked Calculations

Worked Example: Two Groups of Unequal Size

Suppose a hypothetical table records whether participants completed a short task. In Group A, 10 of 50 participants completed it. In Group B, 48 of 60 completed it. Find the pooled completion proportion and compare it with the ordinary average of the two subgroup proportions.

First calculate each subgroup proportion:

\hat{p}_A=\frac{10}{50}=0.20, \qquad \hat{p}_B=\frac{48}{60}=0.80

There are \(50+60=110\) participants and \(10+48=58\) completions. The pooled proportion is therefore:

\hat{p}=\frac{58}{110}=0.5273

The weights are \(50/110\) for Group A and \(60/110\) for Group B. Applying them to the subgroup rates gives the same result:

\hat{p} = \left(\frac{50}{110}\right)(0.20) + \left(\frac{60}{110}\right)(0.80) = \frac{10+48}{110} = 0.5273

Thus, about 52.7% of all 110 participants completed the task. The ordinary average of the subgroup proportions is \((0.20+0.80)/2=0.50\), or 50.0%, which is not the pooled proportion here. Group B’s rate receives more weight because Group B contains more participants.

Worked Example: Equal Group Sizes

In another hypothetical setting, 10 observations are recorded in each of two groups. Two observations in Group A and eight in Group B have the outcome. Calculate the pooled proportion and show why the ordinary average works in this case.

The subgroup proportions are \(\hat{p}_A=2/10=0.20\) and \(\hat{p}_B=8/10=0.80\). Together, there are \(2+8=10\) observations with the outcome among \(10+10=20\) observations, so:

\hat{p}=\frac{10}{20}=0.50

Each group has weight \(10/20=0.50\). Consequently, the weighted average is also the ordinary average:

\hat{p} = (0.50)(0.20)+(0.50)(0.80) = 0.10+0.40 = 0.50 = \frac{0.20+0.80}{2}

When all subgroup sizes are equal, all weights are equal to \(1/k\). The weighted average then reduces to the ordinary arithmetic average of the subgroup proportions. The ordinary average is therefore valid for the pooled proportion in this example precisely because the group sizes match.

Worked Example: Three Groups with Different Sizes

Consider hypothetical records from three service centers. The outcome is whether a request was resolved on the first contact. Center A resolved 60 of 100 requests, Center B resolved 18 of 30, and Center C resolved 9 of 10. Find the pooled proportion and compare it with the unweighted average of the three rates.

The subgroup proportions are:

\hat{p}_A=\frac{60}{100}=0.60,\qquad \hat{p}_B=\frac{18}{30}=0.60,\qquad \hat{p}_C=\frac{9}{10}=0.90

Across all centers, \(60+18+9=87\) of \(100+30+10=140\) requests were resolved on first contact. Thus:

\hat{p}=\frac{87}{140}=0.6214

The corresponding weighted calculation makes each center’s contribution explicit:

\hat{p} = \left(\frac{100}{140}\right)(0.60) + \left(\frac{30}{140}\right)(0.60) + \left(\frac{10}{140}\right)(0.90) = \frac{60+18+9}{140} = 0.6214

The unweighted average of the three rates is \((0.60+0.60+0.90)/3=0.70\), or 70.0%. That number gives each center equal influence despite their very different request counts. It is the average of the three center rates, but it is not the proportion resolved on first contact among all 140 requests. The pooled rate is about 62.1%.

When Does the Ordinary Average Give the Pooled Proportion?

The ordinary average of \(k\) subgroup proportions is \((\hat{p}_1+\cdots+\hat{p}_k)/k\). The pooled proportion instead assigns weight \(n_j/N\) to \(\hat{p}_j\). If every subgroup has the same size, then \(n_j/N=1/k\) for every group, so the two expressions are identical.

With unequal sizes, the two calculations can still happen to produce the same number, for example if all subgroup proportions are equal. But equality is not guaranteed. In general, use the weighted calculation for the proportion among all observations. Use the unweighted average when the quantity of interest is explicitly the average of subgroup rates, giving each group equal importance regardless of its size.

Why the Distinction Matters

A pooled proportion answers a question about the combined observations: “What fraction of all observations had this outcome?” A simple average of subgroup proportions answers a different question: “What is the average of the subgroup rates when each group counts equally?” Both quantities can be meaningful, but they describe different targets. Naming the target before calculating helps prevent a denominator or weighting error.

The subgroup sizes determine the weights, not whether a subgroup’s rate is high or low. A large subgroup has a large weight because it contributes many observations to the total. A small subgroup has a small weight. Since the weights sum to one, the pooled proportion lies between the smallest and largest subgroup proportions (including the endpoints). If a calculated pooled value falls outside that range, recheck the arithmetic, counts, or denominators.

This calculation also clarifies why a pooled rate can differ from rates shown within groups: combining the observations changes how much each group contributes. The weighted-average identity tells exactly how that happens, without treating the overall rate as an equal-weight summary of groups.

Common Mistake

Wrong reasoning: “There are three groups, so the overall proportion is the sum of their three rates divided by three.” That gives each group one-third of the influence, even if one group contains most of the observations and another contains very few.

Correction: For the proportion among all observations, add the outcome counts and divide by the total count of observations, or weight each subgroup proportion by its share of the total. The ordinary average gives the same pooled result when all group sizes are equal; with unequal sizes, it is generally a different quantity. It remains appropriate if the intended target is specifically the average of group rates with equal importance assigned to each group.

Key takeaway: The pooled proportion is \(\hat{p}=\sum_{j=1}^{k}(n_j/N)\hat{p}_j\). Each subgroup rate is weighted by its fraction of the total observations. Equal subgroup sizes make this a simple average; unequal sizes generally do not.

Check Your Understanding

For each question, distinguish a proportion among all observations from an equal-weight average of group rates.

Group A has 12 outcomes among 30 observations, and Group B has 28 outcomes among 70. Find the pooled proportion and the ordinary average of the subgroup proportions. Why are they equal?
Two groups have sizes 20 and 80, with subgroup proportions 0.30 and 0.50. Write the weighted-average calculation for the pooled proportion.
Explain why a subgroup containing 200 observations usually receives more weight in a pooled proportion than a subgroup containing 20 observations.
A report gives subgroup rates of 0.40, 0.50, and 0.70 for groups of unequal sizes. What additional information is needed to calculate the pooled proportion?
State a situation in which the unweighted average of subgroup proportions is the appropriate summary, even when subgroup sizes differ.