Median Test

Median Test:

Median test is used for testing whether two groups differ in their median value. In simple terms, median test will focus on whether the two groups come from populations with the same median. This test stipulates the measurement scale is at least ordinal and the samples are independent (not necessary of the same sample size). The null hypothesis structured is that the two populations have the same median. Let us take an example to appreciate how this test is useful in a typical practical situation.

Example:  A private bank is interested in finding out whether the customers belonging to two groups differ in their satisfaction level. The two groups are customers belonging to current account holders and savings account holders. A random sample of 20 customers of each category was interviewed regarding their perceptions of the bank's service quality using a Likert-type (ordinal scale) statements. A score of "1" represents very dissatisfied and a score of "5" represents very satisfied. The compiled aggregate scores for each respondent in each group are tabulated be given below:  

 

Current Account Savings Account

79

85

86 80
40 50
50 55
75 65
38 50
70 63
73 75
50 55
40 45
20 30
80 85
55 65
61 80
50 55
80 75
60 65
30 50
70 75
50 62

What are your conclusions regarding the satisfaction level of these two groups?

Next-Analysis and Interpretations                            previous

Analysis and Interpretations:

The first task in the median test is to obtain the grand median. Arrange the combined data of both the groups in the descending  order of magnitude. That is rank them from the highest to the lowest. Select the middle most observation in the ranked data. In this case, median is the average of 20th and 21st observation in the array that has been arranged in the descending order of magnitude. 

Table showing descending order of aggregate score and rank in the combined sample

Descending Order Rank Descending Order Rank
86
85
85
80
80
80
80
79
75
75
75
75
73
70
70
65
65
65
63
62

1
2.5
2.5
5.5
5.5
5.5
5.5
8
10.5
10.5
10.5
10.5
13
14.5
14.5
17
17
17
19
20

61
60
55
55
55
55
50
50
50
50
50
50
50
45
40
40
38
30
30
20

21
22
24.5
24.5
24.5
24.5
30
30
30
30
30
30
30
34
35.5
35.5
37
38.5
38.5
40

Grand median is the average of 20th and 21st observation = (62+61)/2 =61.5. Please note that in the above table, average rank is taken whenever the scores are tied. The next step is to prepare a contingency table of two rows and two columns. The cells represent the number of observations that are above and below the grand median in each group. Whenever some observations in each group coincide with the median value, the accepted practice is to first count the observations that are strictly above grand median and put the rest under below grand median. In other words, below grand median in such cases would include less than or equal to grand median.

Scores of Current Account Holders and Savings Account Holders as compared with Grand Median

  Current Account Holders Savings Account Holders Marginal Total
Above Grand Median 8(a) 12(b) 20(a+b)
Below Grand Median 12(c) 8(d) 20(c+d)
Marginal Total 20(a+c) 20(b+d) 40(a+b+c+d)

= n

Null Hypothesis: There is no difference between the current account holders and savings account holders in the perceived satisfaction level.

alternative Hypothesis: There is difference between the current account holders and savings account holders in the perceived satisfaction level.

The test statistic to be used is given by

The chi-square statistic shown on the left side of the table is the one we would have obtained in  a contingency table with nominal data except for the factor (n / 2) used in the numerator as a correction for continuity . This is because a continuous distribution is used to approximate a discrete distribution. 

on substituting the values of  a, b, c, d, and n we have

 

Critical chi-square for 1 d.f at 5% level of significance = 3.84 (click here for the table). Since the computed chi-square(0.90) is less than critical chi-square(3.84), we have no convincing evidence to reject the null hypothesis. Thus the the data are consistent with the null hypothesis that there is no difference between the current account holders and savings account holders in the perceived satisfaction level.

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