Pooled And Unpooled Standard Errors For Proportions

Why Two Standard Errors Appear

When comparing two proportions, the observed difference is calculated in the same way for both estimation and testing: subtract one sample proportion from the other. The standard error, however, depends on the question. For a confidence interval, the samples estimate two proportions that may differ, so each sample proportion supplies its own estimate of variability. For the usual large-sample test of equal proportions, the null hypothesis says both samples share one underlying proportion; the variability estimate is therefore pooled.

Using the pooled standard error in a confidence interval, or the unpooled standard error in the usual pooled equality test, changes the calculation because it changes the assumptions used to estimate variability. Keep the purpose of the calculation in view before choosing a formula.

Model and notation: Suppose two independent samples have fixed sizes \(n_1\) and \(n_2\). In sample \(j\), each observation is an independent Bernoulli outcome with success probability \(p_j\), where \(j=1,2\). Let \(X_j\) be the number of successes and \(\hat p_j=X_j/n_j\) the sample proportion. The estimated difference is \(\hat p_1-\hat p_2\), corresponding to the population difference \(p_1-p_2\). The formulas below use a large-sample normal approximation; they are not exact small-sample procedures.

The Unpooled Standard Error for an Interval

When estimating \(p_1-p_2\), do not assume the two population proportions are equal. The sample proportion in group 1 estimates \(p_1\), and the sample proportion in group 2 estimates \(p_2\). The resulting Wald standard error estimates the sampling variability of the difference using a separate estimated variance for each sample.

Unpooled standard error: For a large-sample Wald confidence interval for \(p_1-p_2\), use

SE_{\text{unpooled}} = \sqrt{ \frac{\hat p_1(1-\hat p_1)}{n_1} + \frac{\hat p_2(1-\hat p_2)}{n_2} }.

A two-sided Wald interval with confidence level \(1-\alpha\) is

(\hat p_1-\hat p_2) \ \pm\ z_{1-\alpha/2}SE_{\text{unpooled}},

where \(z_{1-\alpha/2}\) is the standard normal quantile with cumulative probability \(1-\alpha/2\).

This interval is an approximation. It relies on independent observations and sufficiently large samples for the normal approximation to be reasonable. If the data are sparse or a sample proportion is near zero or one, the Wald method may perform poorly; the displayed formula should not be treated as a universally suitable interval method.

The Pooled Standard Error for an Equality Test

Now consider the null hypothesis that the two population proportions are equal: \(H_0:p_1=p_2\). Under this null, both groups share a common success probability, say \(p\). The two observed proportions still differ due to sampling variation, but their variance is calculated using the same underlying \(p\). Estimate that common probability by combining the successes and sample sizes.

Pooled estimate and standard error: Under \(H_0:p_1=p_2\), the pooled estimate is

\hat p_{\text{pool}} = \frac{X_1+X_2}{n_1+n_2} = \frac{n_1\hat p_1+n_2\hat p_2}{n_1+n_2}.

The usual pooled large-sample standard error for testing equality is

SE_{\text{pooled}} = \sqrt{ \hat p_{\text{pool}}(1-\hat p_{\text{pool}}) \left(\frac{1}{n_1}+\frac{1}{n_2}\right) }.

The corresponding standardized statistic is

z = \frac{\hat p_1-\hat p_2}{SE_{\text{pooled}}}.

Pooling is justified here by the equality assumption in the null hypothesis. It is not a general instruction to combine the two samples whenever proportions are compared. In particular, the pooled standard error is not the standard error used in the Wald confidence interval above. Other testing procedures can use different constructions; this tutorial’s pooled formula is for the usual large-sample test of equality.

Why the Formulas Differ

For a Bernoulli observation with success probability \(p_j\), the variance is \(p_j(1-p_j)\). Therefore, the variance of the sample proportion \(\hat p_j\) is \(p_j(1-p_j)/n_j\). Independent samples have zero covariance, so the variance of their difference is the sum of those two variances. Estimating each \(p_j\) separately gives the unpooled standard error.

\begin{aligned} \operatorname{Var}(\hat p_1-\hat p_2) &= \operatorname{Var}(\hat p_1)+\operatorname{Var}(\hat p_2)\\ &= \frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}. \end{aligned}

Under the null hypothesis \(p_1=p_2=p\), this expression becomes \(p(1-p)(1/n_1+1/n_2)\). Replacing the unknown common \(p\) by the combined sample estimate gives the pooled formula. The pooled estimate is the combined successes divided by the combined number of observations because, under the common-probability model, the likelihood for the two independent Bernoulli samples depends on their total successes and total sample size. Maximizing that likelihood gives \((X_1+X_2)/(n_1+n_2)\).

For an interval.
Estimate each group’s variance separately, using \(\hat p_1\) and \(\hat p_2\). This allows \(p_1\) and \(p_2\) to be different.

For an equality test.
Impose the null condition \(p_1=p_2\), combine the observed successes to estimate the common probability, and calculate variability under that condition.

Check the purpose.
Choose the standard error from the inferential task, not from whichever formula gives a more convenient result.

Worked Calculations

Worked Example: Interval and Equality Test With Equal Sample Sizes

A hypothetical study compares whether two independent groups complete a short training module. In group 1, 60 of 100 participants complete it; in group 2, 45 of 100 complete it. Assume independent Bernoulli outcomes within and between groups. Calculate the unpooled standard error and 95% Wald interval for \(p_1-p_2\), then calculate the pooled standard error and test statistic for \(H_0:p_1=p_2\).

Method and estimates. The sample proportions are \(\hat p_1=60/100=0.60\) and \(\hat p_2=45/100=0.45\), so the estimated difference is \(0.60-0.45=0.15\).

Interval calculation. Use the separate sample estimates:

\begin{aligned} SE_{\text{unpooled}} &= \sqrt{\frac{0.60(0.40)}{100}+\frac{0.45(0.55)}{100}}\\ &= \sqrt{0.0024+0.002475}\\ &= \sqrt{0.004875} =0.06982. \end{aligned}

For 95% confidence, \(z_{0.975}=1.96\). The margin of error is \(1.96(0.06982)=0.13685\). Thus the interval is \(0.15\pm0.13685\), or \([0.0132,\,0.2868]\), rounded to four decimal places.

Equality-test calculation. The pooled estimate is \((60+45)/(100+100)=105/200=0.525\). Therefore,

\begin{aligned} SE_{\text{pooled}} &= \sqrt{0.525(0.475)\left(\frac{1}{100}+\frac{1}{100}\right)}\\ &= \sqrt{0.249375(0.02)} = \sqrt{0.0049875} =0.07062. \end{aligned}

The test statistic is \(z=0.15/0.07062=2.124\). For a two-sided 5% large-sample test, the usual standard normal cutoff is 1.96, so this statistic exceeds the cutoff.

Check and common mistake avoided. The pooled proportion \(0.525\) uses all 200 observations and is appropriate for the equality test. The interval instead uses \(0.60\) and \(0.45\) separately. Substituting \(0.525\) into the interval formula would wrongly impose equality while estimating a difference that is allowed to be nonzero.

Worked Example: Unequal Sample Sizes

In a hypothetical independent survey, 32 of 80 members of one group prefer a proposed service feature, while 54 of 120 members of another group prefer it. Treat responses as Bernoulli outcomes with fixed sample sizes. Find both standard errors for the difference \(p_1-p_2\), and calculate the pooled equality-test statistic.

Method and sample proportions. We have \(\hat p_1=32/80=0.40\) and \(\hat p_2=54/120=0.45\). The estimated difference is \(0.40-0.45=-0.05\).

Unpooled calculation.

\begin{aligned} SE_{\text{unpooled}} &= \sqrt{\frac{0.40(0.60)}{80}+\frac{0.45(0.55)}{120}}\\ &= \sqrt{0.003+0.0020625}\\ &= \sqrt{0.0050625} =0.07115. \end{aligned}

Pooled calculation. There are \(32+54=86\) successes among \(80+120=200\) observations, so \(\hat p_{\text{pool}}=86/200=0.43\). Then

\begin{aligned} SE_{\text{pooled}} &= \sqrt{0.43(0.57)\left(\frac{1}{80}+\frac{1}{120}\right)}\\ &= \sqrt{0.2451(0.0208333\ldots)}\\ &= \sqrt{0.00510625} =0.07146. \end{aligned}

The equality-test statistic is \(-0.05/0.07146=-0.6997\). The two standard errors are close, but not identical: the unpooled calculation retains the distinct sample proportions, whereas the pooled one uses the common-proportion estimate under the null.

Check and common mistake avoided. The reciprocal sample-size terms are \(1/80\) and \(1/120\), not two copies of \(1/100\). Replacing unequal sample sizes by their average would give the wrong variance.

Worked Example: A Small Difference Between the Two Standard Errors

A hypothetical pilot study records 28 successes among 40 observations in group 1 and 30 successes among 60 observations in group 2. Assume independent Bernoulli samples. Calculate the interval standard error and the pooled standard error for testing equality, then evaluate the two-sided 5% equality test.

Method and observed difference. The sample proportions are \(28/40=0.70\) and \(30/60=0.50\), giving a difference of \(0.20\).

Unpooled standard error.

\begin{aligned} SE_{\text{unpooled}} &= \sqrt{\frac{0.70(0.30)}{40}+\frac{0.50(0.50)}{60}}\\ &= \sqrt{0.00525+0.0041667\ldots}\\ &= \sqrt{0.0094167\ldots} =0.09704. \end{aligned}

Pooled standard error and test. The combined estimate is \((28+30)/(40+60)=58/100=0.58\). Hence

\begin{aligned} SE_{\text{pooled}} &= \sqrt{0.58(0.42)\left(\frac{1}{40}+\frac{1}{60}\right)}\\ &= \sqrt{0.2436(0.0416667\ldots)}\\ &= \sqrt{0.01015} =0.10075. \end{aligned}

The test statistic is \(z=0.20/0.10075=1.985\). Since \(1.985>1.96\), it just exceeds the two-sided 5% cutoff. The conclusion follows from the pooled test calculation, not from substituting the unpooled standard error into that test.

Check and common mistake avoided. The pooled estimate is \(0.58\), not the unweighted average \((0.70+0.50)/2=0.60\). The samples have different sizes, so combining observations requires weighting by the sample sizes.

Common Exam Traps

The most frequent error is choosing a standard error by habit instead of identifying the question. For example, with \(\hat p_1=0.60,\hat p_2=0.45,n_1=n_2=100\), a wrong interval calculation might pool first and report

\sqrt{0.525(0.475)\left(\frac{1}{100}+\frac{1}{100}\right)} =0.07062

as the interval standard error. That is the pooled equality-test standard error, not the Wald interval standard error. The correction is to use

\sqrt{\frac{0.60(0.40)}{100}+\frac{0.45(0.55)}{100}} =0.06982.

A second trap is to use the unpooled standard error automatically in an equality test. The standard pooled large-sample test estimates variability under the null condition \(p_1=p_2\), so its formula uses \(\hat p_{\text{pool}}\). A third trap is to pool proportions without weighting by sample size. Always compute the pooled estimate as total successes divided by total observations.

Key takeaway: For a Wald confidence interval for \(p_1-p_2\), use the unpooled standard error, with each group’s own sample proportion. For the usual large-sample test of \(H_0:p_1=p_2\), use the pooled standard error, with the combined success proportion. The difference is the model assumption used to estimate variability.

Check Your Understanding

For each question, use the independent Bernoulli model and identify whether the requested calculation is for an interval or an equality test.

In two independent samples of size 50, there are 30 and 20 successes. Find the unpooled interval standard error and the pooled equality-test standard error. (Answer: \(SE_{\text{unpooled}}=0.0980\); \(SE_{\text{pooled}}=0.1000\).)
Sample 1 has 24 successes among 80 observations, and sample 2 has 48 among 120. Calculate both standard errors for comparing \(p_1-p_2\). (Answer: \(SE_{\text{unpooled}}=0.0680\); \(SE_{\text{pooled}}=0.0693\).)
Two samples have 8 successes among 40 and 24 among 60. Find the unpooled standard error for an interval and the pooled standard error for testing equality. (Answer: \(SE_{\text{unpooled}}=0.0894\); \(SE_{\text{pooled}}=0.0952\).)
In two samples of size 100, there are 55 and 45 successes. Calculate the pooled test statistic and state whether it exceeds the two-sided 5% cutoff of 1.96 in absolute value. (Answer: \(\hat p_{\text{pool}}=0.50\), \(SE_{\text{pooled}}=0.0707\), \(z=1.414\); it does not exceed 1.96.)
Sample 1 has 90 successes among 150 observations, and sample 2 has 20 among 50. Compute both standard errors and explain which one belongs in a Wald interval for \(p_1-p_2\). (Answer: \(SE_{\text{unpooled}}=0.0800\); \(SE_{\text{pooled}}=0.0812\); use the unpooled standard error for the interval.)

What You'll Learn