Readmission rates and a confidence interval that straddles zero
A hospital quality team compares a transitional-care program with usual care for 30-day readmissions. In the program, 64 of 400 patients were readmitted; in usual care, 100 of 500 were readmitted. The 95% confidence interval for the difference in proportions (program minus usual) is (-0.090, 0.010), and the two-sided test at alpha = 0.05 has p = 0.11. One board memo says, "The interval contains zero, so the program has no effect." Another says, "Readmissions fell from 20% to 16%, a 4-point reduction, so the program works." The quality director wants a confirmatory study sized so that the 95% margin of error for each arm's readmission proportion is about 2 percentage points, assuming a true proportion near 0.18. The board meets in three weeks and wants a defensible interpretation and sample-size estimate.
1. Interpret the confidence interval and p-value. Evaluate both board memos and write a defensible summary statement.
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The 95% CI (-0.090, 0.010) estimates that the program's readmission proportion could be 9.0 percentage points lower or 1.0 point higher than usual care. Because 0 is inside the interval, the data are compatible with no difference at the 0.05 two-sided level; the p-value 0.11 is consistent with that. But "compatible with no effect" is not "proved no effect": the interval also includes a clinically meaningful 9-point reduction. The first memo overstates certainty.
The 4-point estimate (16% vs 20%) is a point estimate with uncertainty. Calling it a 4-point reduction ignores sampling error. A defensible statement is: the observed reduction was 4 points, but the study cannot rule out no difference or a reduction as large as 9 points; a larger trial is needed to resolve the question. Report the absolute difference and its interval, not just the two arm-specific percentages.
2. Calculate the sample size per arm for a 95% margin of error of about 2 percentage points when p is near 0.18. State the assumptions and two practical adjustments.
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For a 95% margin of error E = 0.02 with p near 0.18, use n = z^2 p(1-p)/E^2 = (1.96^2)(0.18)(0.82)/(0.02^2). 1.96^2 = 3.8416; 0.18 x 0.82 = 0.1476; numerator = 0.5670; denominator = 0.0004; n = 1417.6, so 1,418 patients per arm.
This assumes independent patients, random sampling or equivalent, and a known p. If the proportion could be nearer 0.5, use 0.25, which gives n = 2,401 per arm. If patients are clustered by ward or physician, inflate by the design effect 1 + (m - 1)ICC. Adjust for attrition by dividing by the expected follow-up rate. The result is per arm, not total. Also specify whether the margin of error is for each arm separately or for the difference, because that changes the calculation.
3. A colleague says that because the two separate 95% confidence intervals for the arm proportions overlap, the difference cannot be significant. Evaluate this reasoning.
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Overlap of two separate 95% confidence intervals is not a valid test of the difference. Each interval reflects uncertainty for its own group proportion; the difference has its own standard error, sqrt(SE1^2 + SE2^2), which is smaller than the sum and can produce a significant difference even when the individual intervals overlap slightly. Conversely, non-overlap does imply a significant difference, but overlap does not imply non-significance.
Here, the correct interval for the difference is already given: (-0.090, 0.010), which includes zero. So the conclusion of "not significant at 0.05" happens to be correct, but the overlapping-CI reasoning is not. Always test the contrast directly, and prespecify whether the comparison is one- or two-sided. If multiple comparisons are made, adjust the intervals or state the familywise error rate.