View - Kowalewski, M. - Virginia Tech

PALEONTOLOGICAL SOCIETY PAPERS, V. 8, 2002observations in the numerator, increasing the powerof the test, Equation 5 halves the number ofobservations in the denominator, decreasing thepower of the test. A simple computer simulation canresolve this issue. If we draw random samples froma known underlying distribution and use a=5% toreject the null hypothesis, which we know to becorrect in this case, we should reject incorrectly 5%of the tests. Results show that, regardless of samplesize, when Equation 5 is used ca. 5% of tests arerejected and if Equation 6 is used over 11% of testsare rejected (Fig. 4). This simulation indicates thatEquation 5 performs correctly and should beemployed in future studies whereas Equation 6clearly is too powerful and should not be used tocorrect for disarticulated elements.Note here that the example considered aboveassumes the following: (1) the two opposite valvesare equally likely to be preserved; (2) the predatoralways produces a trace in one valve only; (3) thetrace does not weaken the skeleton; and (4) thepredator does not show valve selectivity. All theseassumptions are questionable, and more complexcorrective strategies (most likely, based on theBayesian approach) should be developed in the future.Escalation parameters.—Escalation parametersare estimates that provide some measure of thepredator’s failure. A relative frequency of failedattacks (often referred to as “prey effectiveness”;e.g., Vermeij, 1987; Alexander and Dietl, in press;Kelley et al., 2001), as recorded by tracesdocumenting unsuccessful attacks (e.g., repair scarFIGURE 4—A series of computer simulations testing the statistical power of Equations 5 and 6. In thesimulation, samples of specimens are drawn randomly from an infinite population of disarticulated valvesof bivalve mollusks with a predefined drilling frequency of 50%. The correct null hypothesis (drillingfrequency = 50%) is then tested for each random sample using Fisher’s Exact Test and alpha=0.05.When Equation 5 is used the Type Error I (the erroneous rejection of the correct null hypothesis) variesaround 5% (mean=5.32) indicating that this test performs correctly. When Equation 6 is used over 11% oftests are significant indicating that Equation 6 offers over two times more statistical power than it should.20