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Median Test

Median Test. Median Test! Yeah its Median Test, but what is it?. The median test is a procedure for testing whether two independent groups differ in central tendencies. More precisely, the median test will give information as to have been drawn from population with the same median.

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Median Test

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  1. Median Test

  2. Median Test! Yeah its Median Test, but what is it? The median test is a procedure for testing whether two independent groups differ in central tendencies. More precisely, the median test will give information as to have been drawn from population with the same median. Null Hypothesis: The two groups are from populations with the same median. Alternative Hypothesis: the median of one population is different from that of the other (two-tailed test) or that the median of one population is higher than that of the other(one-tailed test). The test may be used whenever the scores for the two groups are measured on at least an ordinal scale.

  3. How to perform Median Test? To perform the median test, • We first determine the median score for the combined group (i.e., the median for all scores in both samples). • We then dichotomize both sets of scores at that combined median and cast these data in a 2 x 2 table, like the one below. Median Test: form for data Now if group I and group II are samples from populations whose median are the same, we would expect about half of each group’s scores to be above the combined median and about half to be below. That is, we would expect frequencies A and C to be about equal and B and D also.

  4. It can be shown (Mood, 1950) that, if A is the number of the m cases in group I that fall above the combined median, and if B is the number of the n cases in group II that fall above the combined median, then the sampling distribution of A and B under the null hypothesis (H0 is that the medians are the same) is the hypergeometric distribution P[A, B] = Therefore, if the total number of cases in both groups (m + n) is small, one may use the Fisher exact test to test H0. If the total number of cases is sufficiently large, the chi-square test with df = 1 may be used to test H0. When analyzing data split at the median, the researcher should be guided by these considerations in choosing between the Fisher exact test and the chi-square test for 2 x 2 tables: • When N = m + n is larger than 20, use corrected for continuity.

  5. In the book of Daniel, Wayne W., Applied Nonparametric Statistics, 2nd Edition. The test statistic used is a Normal Approximation. T = Where p =(A + B)/N How it is approximated? It starts from the fisher exact test statistic, which is the PDF of a Hyper- geometric Distribution, then we use Binomial Approximation to it, and which goes to Normal Approximation to the Binomial Random Variables.

  6. 2. When N = m + n = 20 or less, use the Fisher exact test. Decision Rule: For chi-square test statistic, we reject Null Hypothesis if with degrees of freedom v = 1. For normal approximation, we reject Null Hypothesis if Example: In a cross-cultural test of some behavior theory hypotheses adapted from psychoanalytic theory, Whiting and Child studied the relation between child-rearing practices and customs related to illness in various nonliterate cultures. One hypothesis of their study, derived from the notion of negative fixation, was their oral explanations of illness; Illness results from eating poison, from drinking certain liquids , and from verbal spells and incantations performed by others.

  7. Judgments of the typical oral socialization anxiety in any society were based on the rapidity of oral socialization, the severity of oral socialization, the frequency of punishment typical in oral socialization, and the severity of emotional conflict evidenced by the children during the period of oral socialization. Excerpts from ethnological reports of non-literate cultures were used in the collection of the data. By using only excerpts concerning customs relating to illness, judges classified the societies into two groups – those with oral explanations of illness present and those with oral explanations absent.Other judges, using the excerpts concerning child rearing practices, rated each society on the degree of oral socialization anxiety typical in its children. For the 39 societies for which judgments of the presence or absence of oral explanations were possible, these ratings ranged from 6 to 17. Solution: Null Hypothesis: There is no difference between the median oral socialization anxiety in societies which give oral explanations of illness and the median oral socialization anxiety in societies which do not give oral explanations of illness. 5 + = 9

  8. Alternative Hypothesis: The median oral socialization anxiety in societies with oral explanations present is higher than the median in societies with oral explanations absent. • Level of Significance: Let alpha = 0.01 and N is the number of societies for which ethnological information on both variables was available = 39. m the number of societies with oral explanations absent = 16; n is the number of societies with oral explanations present = 23. • Test Statistic • Rejection Region: Since H1 predicts the direction of the difference, the region of rejection is one-tailed. And reject H0 if 6.635 with alpha = 0.01, otherwise do not reject it.

  9. Computation: Oral socialization anxiety and oral explanations of illness Oral socialization anxiety and oral explanations of illness 9.39 • Decision: Since the computed > 6.635, then we reject H0 for alpha = 0.01 • Conclusion: Therefore, we conclude that the median oral socialization anxiety is higher in societies with oral explanations of illness present than is the median in societies with oral explanations absent.

  10. Example 2. Russel et al. reported the stroke-index values shown in Table 3 for patients admitted to the myocardial-infraction research unit of a university hospital. We wish to know whether these data provide sufficient evidence to indicate that the medians of the two populations represented by the sample data are different. Let alpha be 0.05. Table 3. Stroke-index values, milliliters, for patients admitted to the myocardial-infraction research unit of a university hospital. Source: Richard O. Russell, Jr., David Hunt, and Charles E. Rackley, “Left Venticular Hemodynamics in Anterior and Inferior Myocardial infarction.” Amer J. Cardiol., 32 (1973), 8 – 16.

  11. 5 + = 9 Solution: • Hypotheses: H0: H1: • Level of Significance: alpha = 0.05 • Test Statistic: T = ; Where: p = (A + B)/N 4. Rejection Region: Reject H0 if Z < -1.96 or Z > 1.96, otherwise, do not reject H0.

  12. Computation: The median computed from the 48 sample values is (25 + 26)/2 = 25.5. Table 4 shows the number of observations in each sample falling above and below 25.5. Contingency Table of Example 2

  13. Above the median of X = Above the median of Y = 12 12

  14. p = (A + B)/N = (12 + 12)/48 T = == -2.45 • Decision: Since -2.45 < - 1.96, is true. Or -2.45 >1.96 is not true. Then, H0 is rejected since -2.45 is included in the left tail of the normal distribution. • Conclusion: Therefore, the median of the two populations (X and Y) is not equal.

  15. When Ties is involve Although the assumption of continuity underlies the median test, ties do occur in practice, that is, one or more sample observations may be exactly equal to the computed sample median. In fact, if the total number of sample observations ( + = N) is odd, at least one sample observation will be equal to the combined median. We can handle ties in two ways: • If + = N, and if only a few observations are equal to the median, we may discard them before computation. • We may categorize the observations in each sample as either falling above the sample median or not falling above it, so that tied observations fall in the latter category. Power-Efficiency Mood(1954) has shown that, when the median test is applied to data measured in at least an interval scale from normal distribution with common variance (i.e., data that might properly be analyzed by the t test), it has the same power-efficiency as the sign test. That is, its power-efficiency is about 95 percent for m + n as low as 6. This power efficiency decreases as the sample size increases, reaching an asymptotic efficiency of 2/π = 63 percent.

  16. SUMMARY: Median Test is use for testing the null hypothesis that two independent samples have been drawn from populations with equal medians. Assumptions: • The data consist of two independent random samples: , , , . . .,, and , , , . . . , . The first sample is from a population with unknown median, ,and the second is from a population with unknown median, . • The measurement scale employed is at least ordinal. • The two populations have the same shape. • If the two population have the same median, then for each population the probability p is the same that an observed value will exceed the grand median.

  17. Reference: - Siegel, S., et al. Nonparametric Statistics for the Behavioral Science, 2nd edition. McGraw-Hill Publication. - Daniel, Wayne W., Applied Nonparametric Statistics,2nd Edition. Thomson publishing group.

  18. Thanks for listening

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