Median Test

Many statistical procedures test for differences in location parameter, where the measure of location pertains to the median (θ), or the 50th percentile. Hypotheses can be tested about a single or multiple medians. [Page 593]

Testing hypotheses about a single sample median presents two problems. First, and perhaps more debilitating, the sample median is not the best unbiased estimate of the population median. (Therefore, the question arises as to the efficacy of testing this hypothesis.)

The Harrell-Davis (θHD) is one statistic, among many, that is a better estimate of the population median than the sample median. The θHD is computed by drawing N ordered deviates from the beta distribution. Sort the original scores (xi), and multiply the first score by the first beta deviate (b1), the second score by the second beta deviate, and so forth. The Harrell-Davis estimator of the population median is

The second problem is that the sampling distribution is intractable. Thus, it is only possible to estimate the standard error of θ. One suitable estimate of the standard error is the Maritz-Jarrett, s, which in turn may be based on the Harrell-Davis estimator of the median.

Thus, for k = 1, the null hypothesis is H0: θ= θ0, where θ is the sample median and θ0 is some hypothesized population value. This is tested against the alternative H1: θ≠ θ0. The test is

where s is the Maritz-Jarrett estimate. Wilcox provided algorithms and computer code to automate computing the Maritz-Jarrett estimate. The significance of Z for a specified α level is determined by a table of the standard normal distribution.

A classical test for a difference in two or more (k ≥ 2) independent medians is due to Mood. The two-sample case is analogous to Fisher's exact test and the k > 2 is based on the chi-square, and they share the assumptions of the analogous tests.

Consider the case of k = 3. The null hypothesis H0: θA = θB = θC, which is tested against the alternative hypothesis Ha: at least one θ is different from the rest. The grand median is computed. The number of scores above and below the median is tabulated. The expected value for each group is half the sample size for that group. The chi-square test is computed, and significance is determined by comparing the obtained chi-square statistic with the critical value based on df = k – 1.

Example

Three groups of test scores are presented in Table 1. The grand median is 65.5 (Fay has provided advice concerning values tied with the grand median).

Count the number of scores greater than 65.5 and the number of scores less than or equal to 65.5 for each group. The frequencies are displayed in Table 1.

Table 1 Test Scores for Groups 1, 2, and 3
Group 1	Group 2	Group 3
63	69	35
99	73	38
89	47	61
44	33	84
88	63	74
70	68	26
66	70	49
84	53	89
93	83	68
66	73	66
96	40	40
37	58	36
46	82	32
78	48	73
95	25	50
49	92	52
61	64	30
97	42	70
95	71	39
75	75	69
65	72	37
70	34	62
44	28	87
67	64	44
48	76	89
54	31	84

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Table 2 Frequencies Above and Below the Grand Median
	Group 1	Group 2	Group 3
>65.5	16	12	11
<65.5	10	14	15

The expected value for each cell is 13. This was found by dividing the sample size for each group (n1 = n2 = n3 = 26). The values for each of the six cells are calculated based on

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