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THE SIGN TEST FOR THE MEDIAN there are times when a nonparametric test is desirable, particularly if the available measurements are ordinal. The sign test for the median provides a simple, distribution-free test of hypotheses about symmetry of sample data around their median. The sign test gets its name from the fact that pluses and minuses, rather than numerical values, provide the data used in the computations.
an economic geographer consulting for a federal government agency is presented with what is explained to be a random sample of n 20 observations of median family income in Canada The statistical question is, do the sample data support the null hypothesis that the true median family income for this metropolis is $18,000 at the .05 level of significance?As a first step in testing the hypothesis, the geographer determines which tract incomes lie above and which ones lie below the hypothesized median of $18,000. A plus sign is attached to those tract incomes that are above the hypothesized median and a minus sign to those that fall below the median. Observed values that equal the hypothesized median are counted neither as plus or minus, and the sample size is reduced accordingly. From the listing of the sample of household incomes, it may be easily seen that 5 of the tract values are below the hypothesized median of $18,000, and 14 of the values are above the median with one at the median. So the proportion of the observation below the median is p=5/19=.263 The second step in this test is to calculate the probability of obtaining as few or fewer of the less frequently occurring sign, or the probability of obtaining as many or more of the more frequently occurring sign. The critical upper and lower bounds are found with It does not matter which proportion—the most frequently occurring sign or the less frequently occurring sign—is computed, since the test assumption is that half the observations will fall above and half below the median. For a two-tailed test, the area under a symmetrical distribution between the critical upper and lower bounds constitutes an acceptance region Since p falls outside the bounds dictated by the sample size, we reject the null hypothesis and conclude the proportion of negative scores is significantly different from 0.5, or that the median = $18,000. |