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¦MATH2740 Environmental StatisticsBackground 2 (quadrat analysis)University of LeedsSchool of MathematicsSemester 2Poisson index of dispersion testConsider the following frequency data from a regular 10 10 grid of quadrats:No. of indivs. per quadrat (¡£¢) 0 1 2 3 4 5 6 7 8 9+ Total (¤)Observed no. of quadrats (¥¢) 21 27 22 14 8 2 3 1 2 0 100First find the sample mean and variance:and© ¥¢¡¢ ¤ §§ ¡¨§Then, the test statistic:§ ¤¨ ! ¥¢¡ ¢#"© ¥¢$¡¢&%¤ ' § !()*#"% +' §-,., which follows 57698distribution.a4§0"¤¨% ¦ /§,123The critical values can be obtained by noting¡that§H.2). =A:CB%D§AEF with:G;4?@?Since/:CB P"!:=andlies outside this range we can reject the null hypothesis, that is there is strongevidence that the pattern is non-random.(,) §Note that since the variance is bigger than the mean the data is overdispersed, and so we might considera contagious distribution to be a better model.4goodness-of-fit testConsider the above data again:(KNo. of indivs. (¡I¢) 0 1 2 3 4 5 6 7 8 9+ (¤)Obs. frequencies (J) 21 27 22 14 8 2 3 1 2 0 100Exp. frequencies 14.1 27.6 27.05 17.68 8.66 3.39 1.11 0.31 0.08 0.03 100To calculate the expected frequencies recall that for the Poisson distribution parameterL the is equal tothe mean and hence we can estimate it by the mean,MLN§ sample ¡¨§(from above).¦)The expected frequencies are then simply the probabilities, evaluated for each of¡ value multiplied bythe total frequency, is:¤ that §P¡9%for¡¨§>E33.P"O1

,MATH2740 Environmental StatisticsBackground 2 (quadrat analysis)University of LeedsSchool of MathematicsSemester 2It is a requirement for a reliable test that none of the expected frequencies are too small, and this isusually taken to mean that none should be less than about 5. So we should combine some of the smallfrequencies into a single group, say¡¡ PFwith observed frequency2and expected frequency)..The4test statistics is then:O § "J> K %§where ’s are the observed values and(1’s the expected frequencies. Comparing this with the criticalvalue )7from tablesKtheK theJof4¢6"EF%C§number of frequency classes after pooling)we conclude that we cannot reject the null hypothesis that the data are from a Poisson distribution.with £ § (theBut the index of dispersion test rejected the same null hypothesis quite strongly!Note that the degrees of freedom in the4test is .because there are two restrictions, that the sum ofthe observed and expected frequencies are equal and that the parameterL is equal to the sample mean.£Index of dispersion test vs. goodness-of-fit testConsider the following artificial example:(KNo. of indivs. (¡I¢) 0 1 2 3 4 5 6+ (¤)Obs. frequencies (J) 20 76 0 0 0 5 0 101Exp. frequencies 37.16 27.16 18.58 6.19 1.55 0.31 0.05 101For the index test,¦of §#anddispersion that is the mean and variance are exactly equal, andthe Poisson hypothesis is accepted; but the observed and expected frequencies are clearly different.§ ¡)The4statistic is,.)with4"EF%G§-F. Thepooling the four smallest frequency classes) which should be comparedtest statistic is far larger and so we clearly reject the Poisson assumption.(afterSo either test may detect non-randomness even when the other fails to do so. Note, however, that datalike this, which is generally regular with occasional clusters, is rather unlikely to arise in practice.Note that the index of dispersion test can be used for quite small sample sizes, while the4test requireslarger samples to get sufficient classes with moderate expected frequencies.Fitting a negative binomial distributionConsider the distribution of the number of Plantago Major plants in 100 sq.cm. quadrats 1 :1 J.G. Skellam. (1952). Studies in statistical ecology: I spatial pattern. Biometrika, 39, 346-362.2

andM¡ §§ ¤¦ ¡ MMATH2740 Environmental StatisticsBackground 2 (quadrat analysis)University of LeedsSchool of MathematicsSemester 2give:¤ §P),¦ ¡ §PE2F.F and"¤¨% §No. of plants (¡I¢) 0 1 2 3 4 5 6 7 8 9 (¤)Obs. Frequencies (¥¢) 235 81 43 18 9 6 4 3 0 1 400(E.(F, These leading to:"¤¨% §>E)7,E)72,¡ M% §>E)(Fitting a negative binomial distribution, the expected frequencies are givenM¦E33with"C§¡ and replaced byM¡ andM. If calculated by hand, it is easiest to use the recurrence relationP"Oby¤(see Background 1).No. of plants (¡I¢) 0 1 2 3 4 5 6 7 8 9 (¤)Obs. frequencies (J) 235 81 43 18 9 6 4 3 0 1 400Exp. frequencies 232.2 85.4 40.0 20.1 10.4 5.5 2.9 3.5 400§ ¡9%for¡)To determine whether or not this is a good fit, the4test is performed pooling for¡ values(so as toavoid very small expected frequencies (we might pooling¡¡ consider P).The isOtest statistics based £ categories and degrees of freedom (becausewe have three restrictions, estimation of ¡ and and equality of the sum of the observed and expectedon2frequencies). do not reject the null hypothesis, and so conclude that thenegative binomial distribution is a good§0Ffit.>, ,E) §Since4¤"EF%§ (weTwo-term quadrat variance methodSuppose that we have¤ §-2contiguous quadrats with counts:The average variance is calculated for a range of block sizes, using quadrats in pairs:"3E3,3F3.33@)37%Block size 1 (1,0) (0,3) (3,5) (5,2) (2,1) (1,4) (4,6) Average varianceVariance of pairs 0.5 4.5 2 4.5 0.5 2.5 2 2.64Block size 2 (1+0,3+5) (0+3,5+2) (3+5,2+1) (5+2,1+4) (2+1,4+6) Average varianceVariance of pairs 24.5 8 12.5 2 24.5 14.33

"¡8¡ § ¡% "¡8¢ . ' ¡ ¡% . ' ¦ ¡9% ¡8 ¡ ¡8 ¡ ¡8 §" 3.3E3,3@)33@)3E33E333,E3E3(3F3.3E3333,3(3)3.333 E3.3@)33F3.33E3E33.3@)3(3,E3E3E33E333,E33F3.3E333X§E33,323@)33E3E333E3FE3E3,E33E3E3E3333@)3(3)3.333MATH2740 Environmental StatisticsBackground 2 (quadrat analysis)University of LeedsSchool of MathematicsSemester 2Block size 3 (1+0+3,5+2+1) (0+3+5,2+1+4) (3+5+2,1+4+6) Average varianceVariance of pairs 8 0.5 0.5 3Block size 4 (1+0+3+5,2+1+4+6) Average varianceVariance of pairs 8 8The average variances per sampling unit (i.e. per quadrat) are then: . (1F3,,The peak of the average variance per quadrat is at block size 2, this is suggestive of clustering with meancluster radius of 2 quadrats (or, equivalently, a gap between cluster centers of 4 quadrats).§-. ) §32 § §-.)3), .)Notes:1. It is useful to recall that, for any pair of numbers"¡83@¡%the sample variance of the pair can becalculated as:"¡¢£2. Whilst the method can provide variances for any block size to8uprestrict maximum block to8sizelarger block sizes.¤¤ (¤ = F) or8' § ."¡8 ¡% .8¤£¤ (¤¦¥ ),. 'as(as above), it is usual toestimates lack precision at¤A less trivial example! An insect ecologist studying the spatial pattern of fire ant hills in a pasturein Florida samples the pasture with a transect 10m wide and 1280m long, in 10m sections giving 128sampling units. The vectors of observations of the number of ant hills observed in each quadrat was:4% E33.3@)3(3F3,33E3E3E3.E33,E323) E3E33E3F3(3.33E3E3E333@)33,3E3E3333E3E3.3F32E3,E3.

8¤£ .2¡ 8MATH2740 Environmental StatisticsBackground 2 (quadrat analysis)University of LeedsSchool of MathematicsSemester 2The average sample variance per sampling unit is calculated for blockusing quadrats in pairs. i.e.:sizes3.33Block size 1:Block size 2:Block size 3:etc." ,,In each case, the sample variance of the pair is calculated, hence giving the average variance for eachblock size. This average is divided by the block size, and then plotted against the block size. A peak inthe graph indicates the block size at which clustering occurs..3 ,%3". E3, )%3" ,3@) %3 "3.%3".37%3"E3,7%3",3)%3E3, ) %3". ,3@) )%3" , )3 ) 7%3 .Variance per sampling unit2 6 10 142 4 6 8 10 12Block sizeFig. 1: Plot of variance per sampling unit against block size for fire ant hill dataFig. 1 is suggestive of clustering of fire ant hills, with mean cluster radius of around 4 quadrats (or gapbetween clusters of 80m).5