Unequal Trial Totals and Precision

Why Unequal Trial Totals Matter

A group’s observed proportion is not equally stable at every sample size. If one clinic observes 10 patients and another observes 1,000, the proportions can differ noticeably just by chance even when the clinics have the same underlying success probability. The denominator—the number of trials—determines how much sampling variability to expect, provided the probability model is otherwise the same.

This distinction matters whenever grouped binomial data are summarized by proportions. A proportion based on many trials generally gives a more precise estimate than one based on few trials. But trial total is not the whole story: the probability itself also affects variance, and pooling groups requires a substantive assumption that their underlying probabilities are equal.

Definition: For group $i$, let $n_i$ be the number of trials, $Y_i$ the number of successes, and $p_i$ the probability of success on each trial. Under the binomial model, $Y_i\sim\operatorname{Bin}(n_i,p_i)$, and the observed proportion is $\hat p_i=Y_i/n_i$. Its variance measures its sampling variability; its standard error is the square root of that variance.

The binomial model here assumes that, within a group, the trials are independent and each has the same success probability. For comparisons across groups, we will also state when we assume that the groups are independent. These assumptions are what allow trial totals to translate into the variance calculations below.

Variance of an Observed Proportion

Write the success indicator on trial $j$ in group $i$ as $X_{ij}$, where $X_{ij}=1$ for a success and $X_{ij}=0$ otherwise. Then $Y_i=\sum_{j=1}^{n_i}X_{ij}$ and $\hat p_i=n_i^{-1}\sum_{j=1}^{n_i}X_{ij}$. Each indicator has expectation $p_i$ and variance $p_i(1-p_i)$.

Result: If the $n_i$ trials in group $i$ are independent Bernoulli trials with common probability $p_i$, then $$ E[\hat p_i]=p_i,\qquad \operatorname{Var}(\hat p_i)=\frac{p_i(1-p_i)}{n_i},\qquad \operatorname{SE}(\hat p_i)=\sqrt{\frac{p_i(1-p_i)}{n_i}}. $$

Derivation. By linearity of expectation, $$ E[\hat p_i]=\frac{1}{n_i}\sum_{j=1}^{n_i}E[X_{ij}] =\frac{1}{n_i}(n_i p_i)=p_i. $$ Independence makes the covariances between distinct indicators zero, so the variance of their sum is the sum of their variances: $$ \operatorname{Var}(\hat p_i) =\frac{1}{n_i^2}\operatorname{Var}\left(\sum_{j=1}^{n_i}X_{ij}\right) =\frac{1}{n_i^2}\sum_{j=1}^{n_i}p_i(1-p_i) =\frac{p_i(1-p_i)}{n_i}. $$ Taking the nonnegative square root gives the standard error. $\square$

For a fixed $p_i$, doubling $n_i$ halves the variance but does not halve the standard error: it multiplies the standard error by $1/\sqrt{2}$. More generally, multiplying the trial total by $c$ divides the variance by $c$ and the standard error by $\sqrt{c}$. Thus, obtaining four times as many trials reduces the standard error by a factor of two, not four.

Worked Example: Four Times the Trials

Suppose two independent groups each have success probability $p=0.20$, but one has $n_1=25$ trials and the other has $n_2=100$. Find the variance and standard error of each observed proportion.

For the first group, $$ \operatorname{Var}(\hat p_1)=\frac{0.20(1-0.20)}{25} =\frac{0.16}{25}=0.0064, \qquad \operatorname{SE}(\hat p_1)=\sqrt{0.0064}=0.08. $$

For the second group, $$ \operatorname{Var}(\hat p_2)=\frac{0.20(1-0.20)}{100} =\frac{0.16}{100}=0.0016, \qquad \operatorname{SE}(\hat p_2)=\sqrt{0.0016}=0.04. $$

The second group has four times as many trials, one quarter of the variance, and half the standard error. The calculation concerns sampling variability under the specified model; it does not guarantee that the second group’s observed proportion will be closer to $0.20$ in any particular data set.

Comparing Precision Across Groups

For an estimate whose variance is $V$, a useful variance-based measure of precision is $1/V$: smaller variance means greater precision. For a binomial proportion this measure is $$ \frac{1}{\operatorname{Var}(\hat p_i)} =\frac{n_i}{p_i(1-p_i)} $$ when $0<p_i<1$. This is a comparison based on the variance of the proportion, not a claim that trial total alone determines precision in every comparison.

If groups have the same probability, the factor $p_i(1-p_i)$ is common, so their variance-based precision is proportional to their trial totals. If their probabilities differ, the factor $p_i(1-p_i)$ differs too. In particular, the binomial variance factor $p(1-p)$ is largest at $p=0.5$ and smaller near 0 or 1. Therefore a group with more trials need not have lower proportion variance than a smaller group if their probabilities are sufficiently different.

Worked Example: Comparing Two Group Proportions

A study observes 10 successes in $n_1=40$ trials in one group and 40 successes in $n_2=160$ trials in another. Both observed proportions are $0.25$. Use $0.25$ as the probability value for the variance calculation, and compare their standard errors and variance-based precisions.

For each group, the variance factor is $0.25(1-0.25)=0.1875$. The first group has $$ \operatorname{Var}(\hat p_1)=\frac{0.1875}{40}=0.0046875, \qquad \operatorname{SE}(\hat p_1)=\sqrt{0.0046875}\approx0.0685. $$

The second group has $$ \operatorname{Var}(\hat p_2)=\frac{0.1875}{160}=0.001171875, \qquad \operatorname{SE}(\hat p_2)=\sqrt{0.001171875}\approx0.0342. $$

The second group’s trial total is four times larger, so its variance is one quarter as large and its standard error is half as large. The inverse-variance precision values are approximately $1/0.0046875=213.3$ and $1/0.001171875=853.3$, again a four-to-one ratio. The observed proportion is being used as a plug-in value for the unknown probability; these are estimated standard errors, not known exact sampling variances unless $p=0.25$ is in fact the group’s probability.

Pooling When Groups Share a Probability

Suppose $m$ independent groups all have the same success probability $p$, although their trial totals $n_1,\ldots,n_m$ may differ. Let $N=\sum_{i=1}^{m}n_i$. The pooled proportion is the total number of successes divided by the total number of trials. Equivalently, it is a weighted average of group proportions, with each group weighted in proportion to its trial total.

Theorem: Assume groups are independent and, within each group, trials are independent Bernoulli trials with the same probability $p$, where $0<p<1$. Among all linear unbiased estimators of $p$ formed as $\sum_{i=1}^{m}a_i\hat p_i$, with fixed coefficients satisfying $\sum_{i=1}^{m}a_i=1$, the estimator $$ \hat p_{\mathrm{pool}}=\sum_{i=1}^{m}\frac{n_i}{N}\hat p_i =\frac{\sum_{i=1}^{m}Y_i}{N} $$ has the smallest variance. Its variance is $p(1-p)/N$.

Proof. Each $\hat p_i$ is unbiased for $p$, so $\sum_i a_i\hat p_i$ is unbiased when $\sum_i a_i=1$. Because the groups are independent, the variance of this weighted sum is $$ \operatorname{Var}\left(\sum_{i=1}^{m}a_i\hat p_i\right) =\sum_{i=1}^{m}a_i^2\operatorname{Var}(\hat p_i) =p(1-p)\sum_{i=1}^{m}\frac{a_i^2}{n_i}. $$ By the Cauchy–Schwarz inequality, $$ 1=\left(\sum_{i=1}^{m}a_i\right)^2 =\left(\sum_{i=1}^{m}\frac{a_i}{\sqrt{n_i}}\sqrt{n_i}\right)^2 \leq \left(\sum_{i=1}^{m}\frac{a_i^2}{n_i}\right) \left(\sum_{i=1}^{m}n_i\right). $$ Thus $\sum_i a_i^2/n_i\geq1/N$. Equality holds when $a_i/\sqrt{n_i}$ is proportional to $\sqrt{n_i}$, which means $a_i=c n_i$; the constraint $\sum_i a_i=1$ gives $c=1/N$. Therefore the minimum-variance coefficients are $a_i=n_i/N$, and substituting them gives $$ \operatorname{Var}(\hat p_{\mathrm{pool}}) =p(1-p)\sum_{i=1}^{m}\frac{(n_i/N)^2}{n_i} =p(1-p)\frac{\sum_i n_i}{N^2} =\frac{p(1-p)}{N}. $$ Finally, since $\hat p_i=Y_i/n_i$, the weighted expression is $\sum_i(n_i/N)(Y_i/n_i)=\sum_iY_i/N$. $\square$

This result gives a precise sense in which groups with larger denominators contribute more to a pooled estimate: their proportions receive weights proportional to their trial totals. A group with twice as many trials receives twice the weight under the shared-probability assumption. This is not a rule for combining arbitrary group proportions; the common probability assumption is essential to the derivation.

Worked Example: Unequal Groups and the Pooled Proportion

Suppose group 1 has $n_1=10$ trials and $Y_1=1$ success, while group 2 has $n_2=90$ trials and $Y_2=27$ successes. If it is appropriate to regard both groups as having a common success probability, calculate the pooled proportion and compare it with the unweighted average of the two observed proportions.

The group proportions are $\hat p_1=1/10=0.10$ and $\hat p_2=27/90=0.30$. Their unweighted average is $$ \frac{0.10+0.30}{2}=0.20. $$ But the trial totals are unequal. Here $N=10+90=100$, and the trial-count weights are $10/100=0.10$ and $90/100=0.90$. Thus $$ \hat p_{\mathrm{pool}} =0.10(0.10)+0.90(0.30) =0.01+0.27=0.28. $$ Directly counting successes gives the same check: $(1+27)/(10+90)=28/100=0.28$.

For a variance calculation using $p=0.28$ as a plug-in value, the pooled standard error is $$ \sqrt{\frac{0.28(1-0.28)}{100}} =\sqrt{0.002016}\approx0.0449. $$ The unweighted average answers a different question: it gives equal weight to each group, regardless of how many trials each proportion summarizes. It is not the pooled proportion, and it does not have the variance-minimizing property proved above under a common probability.

Limits of the Denominator Rule

A larger denominator improves precision when the probability and sampling assumptions are held fixed. It does not make different groups interchangeable. If group probabilities differ, weighting by trial totals estimates an overall trial-weighted proportion, not a single common group probability. Whether that summary is useful depends on the question being asked.

The variance formula also relies on independent Bernoulli trials within a group. If outcomes are dependent, or if success probabilities vary within the group, then the binomial variance need not apply. In those settings, treating the denominator as if it alone determined precision can substantially understate uncertainty. The standard error formula should therefore be read together with its model assumptions, not as a universal property of every observed proportion.

Common Mistakes

Dividing the standard error by $n_i$. The variance has $n_i$ in its denominator, but the standard error has $\sqrt{n_i}$ in its denominator.
Assuming larger $n_i$ always means smaller variance. That conclusion holds when comparing groups with the same probability. In general, the variance also depends on $p_i(1-p_i)$.
Giving every group proportion equal weight when pooling trials. When a common probability is assumed, the pooled proportion weights groups by their trial totals, not by giving each group equal influence.
Using the pooled proportion without a common-probability interpretation. If group probabilities differ, pooling still produces a numerical overall proportion, but it does not estimate one shared group probability.
Treating a plug-in standard error as exact. Replacing an unknown $p_i$ with an observed proportion gives an estimated standard error; it is not the exact variance unless that value equals the true probability.

Key takeaway: Under independent binomial sampling, a group’s proportion has variance $p_i(1-p_i)/n_i$. At a common probability, variance-based precision grows in proportion to the trial total, and the minimum-variance linear unbiased pooled estimate weights each group proportion by its share of all trials.

Check Your Understanding

Use the stated binomial assumptions when answering.

For a group with $n=64$ and $p=0.25$, calculate the variance and standard error of its observed proportion.
At the same success probability, one group has 25 trials and another has 100. What is the ratio of their variances, and what is the ratio of their standard errors?
Why might a group with more trials still have a larger proportion variance than a smaller group?
Two independent groups share probability $p$, with trial totals 30 and 70. What weights do their proportions receive in the pooled estimator?
Explain why trial-count weighting does not, by itself, justify treating groups with different probabilities as if they shared one probability.