Biostatistics Study NotesFrom concepts and derivations to interpretation

CHAPTER 08 · TOPIC 02

Mood’s Median Test

Compare two independent groups by classifying observations above or below the pooled median and using exact or large-sample inference.

On this page
  1. Applicable Study Designs
  2. Hypotheses and the 2×2 Table
  3. Hypergeometric Probability
  4. Counts on Either Side of the Pooled Median
  5. Two-proportion approximation for large samples

Mood's median test dichotomizes observations from two independent groups as above or below the pooled median and compares the proportions falling on each side.

Applicable Study Designs#

DesignMethod
Two independent groups; proportions above and below the pooled medianMedian test
Two independent groups; all rank informationRank-sum / Mann–Whitney
One sample or paired dataSign / Signed-rank

Hypotheses and the 2×2 Table#

H0:η1=η2H1:η1η2H_0:\eta_1=\eta_2\qquad H_1:\eta_1\ne\eta_2
Above the pooled medianBelowTotal
Group 1m₁n₁−m₁n₁
Group 2m₂n₂−m₂n₂
TotalMN−MN

Use a prespecified rule for observations equal to the pooled median. They are often excluded, but when ties are numerous the result may depend on this rule, so report it explicitly.

Hypergeometric Probability#

assignments matching the table(n1m1)(n2m2)\binom{n_1}{m_1}\binom{n_2}{m_2}
all assignments(n1+n2m1+m2)\binom{n_1+n_2}{m_1+m_2}
table probabilityP(M1=m1M)=(n1m1)(n2m2)(n1+n2m1+m2)P(M_1=m_1\mid M)=\frac{\binom{n_1}{m_1}\binom{n_2}{m_2}}{\binom{n_1+n_2}{m_1+m_2}}

This is the probability of one observed table, and the two-sided p-value sums the probability of all tables being equally or more inconsistent with H₀. This is the same conditional calculation as Fisher’s exact test.

Counts on Either Side of the Pooled Median#

N is evenm1+m2=N2m_1+m_2=\frac N2
If N is odd, exclude the medianm1+m2=N12m_1+m_2=\frac{N-1}{2}

Two-proportion approximation for large samples#

difference in proportionsp^1p^2=m1n1m2n2\hat p_1-\hat p_2=\frac{m_1}{n_1}-\frac{m_2}{n_2}
pooled proportionp^=m1+m2n1+n2,q^=1p^\hat p=\frac{m_1+m_2}{n_1+n_2},\quad\hat q=1-\hat p
Z statisticZ=m1/n1m2/n2p^q^(1/n1+1/n2)Z=\frac{m_1/n_1-m_2/n_2}{\sqrt{\hat p\hat q(1/n_1+1/n_2)}}