# Confidence interval for Mann-Whitney test (Hodges–Lehmann estimator)

Does anybody knows how to calculate the confidence interval using normal approximation for the Hodges-Lehmann estimator (median of Y-X)?

This is how Minitab performs calculation, but I cannot find formula behind it!

Example

Group 1: 0.8, 1.2, 1.4, 0.9, 1.0

Group 2: 1.1, 1.3, 1.5, 0.7, 1.6

And below the result from Minitab which informs a CI of (-0,6998;0,3000) (How?)

Point estimate for $\eta_1$ - $\eta_2$ is -0,2000

96,3 Percent CI for $\eta_1$ - $\eta_2$ is (-0,6998;0,3000)

W = 23,0

Test of $\eta_1 = \eta_2$ vs $\eta_1 \neq \eta_2$ is significant at 0,2017

• Which group corresponds to X and which to Y? – Michael R. Chernick Apr 5 '17 at 20:32
• @MichaelChernick X (group 1) and Y (group 2). – zeferino Apr 5 '17 at 21:42
• What makes you say that there's a normal approximation involved in this confidence interval? – Glen_b Apr 5 '17 at 23:05
• @Glen_b Since IC is calculated over all differences between Y and X, it's not possible to have values with more than one decimal place. As you can see on Minitab output, it shows 4 decimal places for the lower value. Also, since it has ties when we consider all differences and it uses rank, normal aproximation is usually the approach. – zeferino Apr 6 '17 at 0:11
• I don't follow your reasoning at all. I don't see how the fact that Minitab gave a value of -.6998 implies that anything anywhere was normal. When you say " normal aproximation is usually the approach" ... I don't think I've seen a normal approximation used for a Hodges Lehman interval, (certainly not with small n). What's the basis for your statement about Hodges Lehmann intervals? Is it a guess or do you know that's what Minitab did? – Glen_b Apr 6 '17 at 0:15