Confidence intervals

Confidence interval for the difference of proportions

Estimate the difference between two independent population proportions, using the Wald or Newcombe-Wilson method.

Calculator

Enter the successes and sample sizes of each group to get the confidence interval for p₁ − p₂.

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Explanation

When comparing two independent groups (for example, conversion in A/B testing, prevalence in two populations, or satisfaction percentage by segment), the confidence interval for the difference of proportions quantifies the size of the difference and its uncertainty.

The calculator estimates \(p_1 - p_2\) from \(\hat{p}_1 = x_1/n_1\) and \(\hat{p}_2 = x_2/n_2\). If the interval contains 0, the observed difference may be compatible with no difference at the chosen confidence level.

Let \(C\) denote the confidence level and \(\alpha=1-C\) the total area outside the interval. For 95% confidence, \(C=0.95\), \(\alpha=0.05\) and \(\alpha/2=0.025\) in each tail.

Wald method (normal approximation)

\( (\hat{p}_1 - \hat{p}_2) \pm z_{\alpha/2} \cdot \sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \)

This is the classic method and is adequate when both groups have enough successes and failures: \(n_i\hat{p}_i \geq 5\) and \(n_i(1-\hat{p}_i) \geq 5\).

Newcombe-Wilson method (recommended)

The Newcombe-Wilson method does not apply the normal approximation directly to the difference \(\hat{p}_1-\hat{p}_2\). It first computes, separately, a Wilson score interval for each proportion: \([L_1, U_1]\) for \(p_1\) and \([L_2, U_2]\) for \(p_2\). The idea comes from Newcombe's (1998) work, which compared several methods for the difference of proportions and highlighted the good performance of Wilson-based methods.

It then combines the distances from each observed proportion to its Wilson limits, keeping the sign of the estimated difference \(d = \hat{p}_1 - \hat{p}_2\). The combination is not simply \([L_1-U_2, U_1-L_2]\): the uncertainties of both groups are added quadratically to build each end of the interval:

\( L = d - \sqrt{(\hat{p}_1-L_1)^2 + (U_2-\hat{p}_2)^2} \)

\( U = d + \sqrt{(U_1-\hat{p}_1)^2 + (\hat{p}_2-L_2)^2} \)

This combination tends to keep the actual coverage of the interval closer to nominal than Wald, especially with small samples or proportions near 0 or 1. That's why it's a good default choice for comparing two independent percentages. This page uses the version without continuity correction, which is the standard form for a fast, reproducible calculator.

Worked example

Group 1 has 84 successes out of 120 observations and group 2 has 57 successes out of 110 observations. \(\hat{p}_1 = 0.70\), \(\hat{p}_2 \approx 0.518\) and the estimated difference is \(0.182\). With 95% confidence (\(C=0.95\), \(\alpha=0.05\), Wald method):

\( 0.182 \pm 1.960 \cdot \sqrt{\frac{0.70\cdot0.30}{120} + \frac{0.518\cdot0.482}{110}} \approx [0.058,\; 0.306] \)

Assumptions for comparing two proportions

  • The two groups must be independent of each other; there must be no unit-to-unit pairing.
  • Within each group, every observation must be classified as a success or failure and have the same probability of success within that group.
  • Observations within each group must be independent.
  • Wald requires enough successes and failures in both groups: \(n_i\hat p_i\ge 5\) and \(n_i(1-\hat p_i)\ge 5\). If this doesn't hold, Newcombe-Wilson is usually preferable.

Wald or Newcombe-Wilson?

Wald is easy to interpret and generally works well with large samples and non-extreme proportions. Newcombe-Wilson tends to be preferable by default if either group has few successes or few failures, because it uses individual Wilson intervals and combines their uncertainties quadratically, not through a direct subtraction of limits. This reduces the coverage problems of the classic normal interval.

How to interpret the result

The interval \([L, U]\) is the plausible range of the difference of population proportions \(p_1 - p_2\) given the chosen confidence level. If the experiment were repeated many times with samples of the same size, a proportion \(C\) of the intervals built with the same method would contain the true value of \(p_1 - p_2\). The width of the interval reflects precision: larger samples or proportions farther from 0.5 produce narrower intervals; lower variability and larger \(n\) lead to more precise estimates.

The key decision hinges on whether 0 belongs to the interval. If \(0 \in [L, U]\), the data do not rule out \(p_1 = p_2\) at the chosen confidence level; equivalently, the two-sided test \(H_0\!: p_1 = p_2\) would not be rejected at level \(\alpha = 1 - C\). If \(0 \notin [L, U]\), there is evidence of a difference: if \(L > 0\) group 1's proportion is significantly higher; if \(U < 0\) group 2's proportion is significantly higher. The calculator automatically states which of the two cases applies, for both methods when shown together.

  • Effect magnitude: beyond significance, the center of the interval (\(\hat{p}_1 - \hat{p}_2\)) indicates the estimated difference in percentage points. A significant CI centered close to 0 may have little practical relevance.
  • Wald vs. Newcombe-Wilson: when Wald's normality condition holds in both groups, both methods usually agree on the conclusion. If either group has few successes or failures, Newcombe-Wilson tends to be more reliable because it uses individual Wilson intervals and combines their uncertainties quadratically.
  • Reading the chart: the green region under the standard normal curve is the confidence zone and the red tails (area \(\alpha/2\) each) mark the critical values \(\pm z_{\alpha/2}\). The chart visualizes the z values compatible with the data; the CI on the original scale is obtained by mapping those values back to the difference of proportions.

Frequently asked questions

  • When does the CI indicate a significant difference? When the value 0 falls outside the interval [L, U].
  • Which difference is calculated? The tool computes p₁ − p₂; if you swap the groups, the sign of the difference changes.
  • What does it mean if the CI includes 0? That at the chosen confidence level you cannot rule out that both population proportions are equal.

References used