Why Check the Counts in Both Groups?
A normal-based confidence interval for a difference in proportions uses a standard error that treats the two sample proportions as approximately normal. That approximation can be poor when a group has very few observed successes or failures. In that situation, the interval may be calculable, but its usual normal-based coverage is not assured.
For a difference interval, the check must be made within each group. A large total sample, or many successes overall after combining the groups, does not repair a sparse count in one group. This tutorial focuses on that diagnostic: what to count, how to apply a practical rule, and what to conclude when a group does not meet it.
The target is often the difference \(p_1-p_2\), estimated by \(\hat p_1-\hat p_2\). The order matters: group 1 minus group 2 is the opposite signed quantity from group 2 minus group 1. The normal-approximation check, however, is the same whichever order is chosen.
Model and Practical Check
Assume each group consists of independent observations with a binary outcome, such as whether a hypothetical policyholder completed a specified action. Within group \(j\), the outcomes have the same success probability \(p_j\), and the two groups are independent of one another. Under this model, \(x_j\) has a binomial distribution with parameters \(n_j\) and \(p_j\). The usual Wald interval for the difference uses the unpooled estimated standard error
A widely used practical rule for a normal-based interval is to require at least 10 observed successes and at least 10 observed failures in each group:
Some courses and references use 5 rather than 10 as the cutoff. These are rules of thumb, not universal mathematical guarantees. State the rule being used, then apply it separately to both groups. A group that meets the 10-count rule passes this particular diagnostic; it does not prove that the sampling design is sound or that all other interval assumptions hold.
Why Both Counts Matter
For one observation in group \(j\), define an indicator \(I\) that equals 1 for success and 0 for failure. Its mean is \(p_j\), and its variance is \(p_j(1-p_j)\). The sample proportion is the average of \(n_j\) such indicators. Independence therefore gives
When both \(n_jp_j\) and \(n_j(1-p_j)\) are not small, the binomial count has meaningful probability on both sides of its center and the normal approximation is more plausible. The observed counts \(x_j\) and \(n_j-x_j\) are used as practical diagnostics because the true \(p_j\) is unknown. For independent groups, the variance of the difference is the sum of the two variances, which explains why both groups need an adequate approximation.
In a Wald interval, the unknown proportions in the variance are replaced by the sample proportions. If a group has zero observed successes or zero observed failures, its estimated variance contribution is zero. That numerical result is not evidence of certainty: it is a warning that the Wald approximation can behave badly near a boundary.
Worked Checks
Worked Example: A Difference Interval With Adequate Counts
A hypothetical insurer compares the proportion of customers who complete an online account review in two independent randomly selected groups. In group 1, 42 of 120 customers complete it; in group 2, 27 of 100 complete it. Using the 10-count rule, assess the normal approximation and calculate the corresponding 95% Wald interval for \(p_1-p_2\). Assume binary outcomes, independent observations within each group, and independent groups.
Method and count check. Count successes and failures separately. Group 1 has \(x_1=42\) successes and \(120-42=78\) failures. Group 2 has \(x_2=27\) successes and \(100-27=73\) failures. Each of these four counts is at least 10, so both groups pass the stated diagnostic.
Estimate. The sample proportions are \(\hat p_1=42/120=0.35\) and \(\hat p_2=27/100=0.27\). Thus the estimated difference is \(0.35-0.27=0.08\).
Unpooled standard error. Substitute each group’s own sample proportion:
For a 95% interval, use \(z_{0.975}=1.96\). The margin of error is \(1.96(0.062184)\approx0.12188\), using the unrounded standard error. The endpoints are \(0.08-0.12188=-0.04188\) and \(0.08+0.12188=0.20188\). To four decimal places, the interval is \((-0.0419,\ 0.2019)\).
Check and trap avoided. The four counts satisfy the rule, and the interval is centered at \(0.08\): the average of its unrounded endpoints is \(0.08\). The common mistake is to check only the combined total of 69 successes and 151 failures; the correct diagnostic checks both counts in each group.
Worked Example: One Group Has Too Few Successes
A hypothetical workplace study records whether employees complete a voluntary safety module. In group 1, 8 of 80 employees complete it; in independent group 2, 32 of 100 complete it. Under the same binary-outcome and independence assumptions, decide whether the 10-count rule supports using a normal-based difference interval.
Method and count check. In group 1, there are \(8\) successes and \(80-8=72\) failures. In group 2, there are \(32\) successes and \(100-32=68\) failures. Group 1 fails the rule because \(8<10\); group 2 passes because both of its counts are at least 10. Therefore, the check does not support relying on the normal approximation for the difference interval.
The sample proportions are \(8/80=0.10\) and \(32/100=0.32\), so the estimated group-1-minus-group-2 difference is \(-0.22\). A Wald standard error can still be calculated:
That calculation does not make the approximation reliable: it is a computable plug-in standard error, while the count diagnostic has failed in group 1. The common mistake is to see that the standard error is finite, or that group 2 has plenty of observations, and conclude that the interval is automatically appropriate. Neither observation addresses the shortage of group-1 successes.
Worked Example: Large Pooled Counts Conceal a Sparse Group
A hypothetical program compares enrollment in two independently sampled populations. Group 1 has 4 enrollments among 40 people; group 2 has 48 among 400. Determine whether the 10-count diagnostic is met. Also show why combining the groups before checking can mislead.
Method and group-by-group check. Group 1 has \(4\) successes and \(40-4=36\) failures. Group 2 has \(48\) successes and \(400-48=352\) failures. Group 1 fails because it has only 4 successes, even though its failure count is adequate. Group 2 passes. Since every group must pass, the normal-based difference interval is not supported by this check.
The observed proportions are \(4/40=0.10\) and \(48/400=0.12\), giving an estimated difference of \(0.10-0.12=-0.02\). If the outcomes were incorrectly combined, the total would be \(4+48=52\) successes and \(36+352=388\) failures. Both totals exceed 10, but those pooled totals hide the fact that group 1 has only 4 successes.
Trap avoided. A difference in proportions concerns two group-specific rates, so its normal-approximation check is not a check of the combined count. The group sizes are also unequal, but unequal sizes do not change the rule: assess each group’s own successes and failures.
Common Mistakes and Limits of the Check
- Checking only successes. A group with many successes but very few failures can also be near a boundary. Check \(x_j\) and \(n_j-x_j\).
- Pooling the groups. Adequate combined counts cannot compensate for a sparse count in either group. Apply the rule four times: two counts for each of two groups.
- Using the wrong group size. Failures in group \(j\) are \(n_j-x_j\), not the total sample size across both groups minus \(x_j\).
- Treating the cutoff as a theorem. Meeting a count rule is a practical indication, not a guarantee of exact coverage. Failing it signals caution; it does not itself calculate the true coverage error.
- Ignoring the design. These formulas assume independent groups and independent binary observations within each group. Paired observations, repeated measurements on the same individuals, or clustered sampling do not satisfy this model as stated.
- Thinking a calculable interval must be suitable. A formula can return endpoints even when its approximation is questionable. Diagnose the counts before relying on the Wald interval.
Check Your Understanding
For each problem, use the 10-count rule unless the question asks for a calculation. Assume independent groups and binary outcomes where a difference interval is considered.
- A study has 14 successes among 90 observations in group 1 and 28 among 110 in group 2. List each group’s successes and failures, then decide whether both groups pass. (Answer: Counts are 14 and 76 in group 1, and 28 and 82 in group 2; both pass.)
- Group 1 has 3 successes among 50 observations; group 2 has 30 successes among 200. Does the rule support a normal-based difference interval? (Answer: No; group 1 has only 3 successes, although its 47 failures and both group-2 counts, 30 and 170, meet the rule.)
- Group 1 records 24 successes among 160 observations and group 2 records 18 among 120. Check all four counts and calculate the estimated group-1-minus-group-2 difference. (Answer: Counts are 24, 136, 18, and 102, so both pass; the estimated difference is \(24/160-18/120=0.15-0.15=0\).)
- Two groups have 25 successes among 250 observations and 70 among 700. A colleague combines the outcomes and reports 95 successes and 855 failures. Is the pooled count sufficient to establish that both groups pass? Check each group. (Answer: The pooled count alone is not sufficient, but both groups do pass individually: group 1 has 25 successes and 225 failures; group 2 has 70 successes and 630 failures.)
- Group 1 has 6 successes among 500 observations, and group 2 has 24 among 800. Explain whether a large total sample makes the normal-based check pass. (Answer: No; group 1 has only 6 successes, so it fails the 10-count rule despite its 494 failures and the large sample sizes.)