3. What does the 95% confidence interval given as “8% shorter stay to 56% longer stay” mean? What is surprising about it and why might it be wrong?
This comes from a 95% confidence interval for the ratio of the geometric means of the length of stay. The geometric mean length of stay in one group is estimated to be between 8% less than the geometric mean length of stay in the other group and 56% greater.
What is surprising is that the full sentence reads:
"Early computed tomography reduced the length of hospital stay by 1.1 days (geometric mean 5.3 days (range 1 to 31) v 6.4 days (1 to 60)), but the difference was non-significant (95% confidence interval, 8% shorter stay to 56% longer stay, P=0.17)."
If early computed tomography shortened stay, the confidence interval should show the estimated shortening at the lower end to be much greater than the estimated lengthening at the upper end.
In the Results section of the paper we read:
"Hospital stay was 1.1 days shorter in the computed tomography arm than in the standard practice arm (geometric mean 5.3 (range 1 to 31) days v 6.4 (1 to 60) days, respectively), but the difference was non-significant (95% confidence interval 0.034 to 0.194; P=0.17). This corresponded to patients in the standard practice arm staying 20% longer than those in the early computed tomography arm (95% confidence interval 8% shorter stay to 56% longer stay)."
When editing this sentence to put it into the abstract, they have cut out "This corresponded to patients in the standard practice arm staying 20% longer than those in the early computed tomography arm . . .". It is the confidence interval for the amount by which length of stay in the standard arm exceeds that in the early arm. In the abstract it appears to be the confidence interval for the amount by which length of stay in the early arm exceeds that in the standard arm.
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Last updated: 8 July, 2008.