View - Kowalewski, M. - Virginia Tech

KOWALEWSKI—ANALYTICAL METHODSFinally, traces of predation may both decreaseand increase the preservational potential of the preyskeleton, biasing quantitative estimates based onfrequency of specimens with traces (e.g., Roy et al.,1994; Hagstrom, 1999; Zuschin and Stanton, 2001;see also below for a discussion of taphonomic biases).How to Correct Predation Frequency Estimatesfor Disarticulated Elements.—The equationsdiscussed in the above section can be applied directlyto taxa with single-element skeletons (gastropodshells, foraminiferan tests, etc.). However, manyprey animals possess skeletons that consist of twoor more elements that tend to disarticulate afterdeath. If a predation event is recorded on one ofthose elements only, then the probability of findingevidence of predation (preserved by only onedisarticulated element) is smaller than the probabilityof finding prey (represented by any of its elements).Consider, for example, a bivalve mollusk killed bya predator that drilled a hole in one of its valves.Assuming that both valves have the samepreservational potential, the probability of findingone of the two valves of the prey is two times morelikely than finding specifically the valve that wasdrilled. Thus, a correction by a factor of 2 is required.It is worth stressing here that, regardless of whetherthe sampling domain is infinite and all sampledvalves are unique or the sampling domain is finiteand some valves come from the same individuals(see Gilinsky and Bennington, 1994), this correctionis required (see also Bambach and Kowalewski,2000; Hoffmeister and Kowalewski, 2001).The issue of correction may appear trivial but itturns out that there are two ways of making thiscorrection and both of them are used in the literature.Equation 5: f d= d/0.5nEquation 6: f d= 2d/n,where f drepresents the estimate of drilling frequency,d is the number of valves in the sample that containat least one successful drill hole, and n is the totalnumber of valves in the sample. These two equationsmay appear synonymous but, from a statisticalperspective, they are not. Equation 6 produces anestimate with a sample size that is two times higherthan an estimate produced by Equation 5.Consequently, Equation 6 offers much more powerthan Equation 5. Table 4 shows a hypotheticalexample of two samples of bivalves. If Equation 6is employed all statistical tests used indicate thatthe two samples differ significantly in drillingfrequency. If Equation 5 is used none of the testsrejects the null hypothesis that the two samplescame from a single underlying population. Whichequation is correct?The answer to this question is not intuitivelyobvious. Whereas Equation 6 doubles theTABLE 4—A hypothetical example illustrating differences in statistical power of the two equations usedto correct frequency estimates for drill holes in bivalved fossils. Symbols: N – total number of valves,D – number of drilled valves, R – drilling frequency, P chi, P G, and P Fisher– The probability estimates (Chisquare,Log-likelihood G, and Fisher’s Exact tests, respectively) for the null hypothesis that the twosamples came from a population with the same drilling frequency. All tests are significant at alpha=0.05level for Equation 6, but none is significant for the more conservative Equation 5. Computer simulations(Fig. 4) show that Equation 5 yields correct estimates of Type I Error.Sample 1 Sample 2Drilling IntensityEquation 5 Equation 6N 1 =30 N 2 =40 N 1 =15, N 2 =20; D 1 =11, D 2 =5 N 1 =30, N 2 =40; D 1 =22, D 2 =10D 1 =11 D 2 =5 R 1 =73.3%, R 2 =25% R 1 =73.3%, R 2 =25%R 1 =36.7% R 2 =12.5% P Chi =0.09, P G =0.08, P Fish =0.13 P Chi =0.02, P G =0.01, P Fish =0.0219