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(1) (2) (3) = (1) * 39 [(2) – (3)] 2 / (3)--------------------------------------------------------------------------------------------------------------------------------------------A 0.4 4 15.6 8.63B 0.4 8 15.6 3.70C 0.2 27 7.8 47.26--------------------------------------------------------------------------------------------------------------------------------------------Totals 1.0 39 39.0 2 = 59.59--------------------------------------------------------------------------------------------------------------------------------------------We will accept the null hypothesis that the sites are randomly distributed in space if the computed 2 is 5.991 we will not accept the null hypothesis that the sites arerandomly distributed. This indicates that there may be a relationship between soils and site location. What would be yourinterpretation of the above results? Note, you can test the null hypothesis using other probability levels. A largerprobability level will increases the likelihood of a null hypothesis being accepted. This is because the test is trying toprevent Type I error. The Cramer V for the above example is V = 0.62.ReferencesGibbons, J.D. 1985. Nonparametric Methods for Quantitative Analysis, 2ed. American Sciences Press, Inc.Columbus, Ohio.Liebetrau, Albert M. (1983). Measures of association. Newbury Park, CA: Sage Publications. QuantitativeApplications in the Social Sciences Series No. 32. Rosenberg, M. (1968). The logic of survey analysis. NY:Basic Books.4