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1、(一）空间分布 型1.意义 种群生态特性：空间是聚集 分布还是 随机分布，解决抽样方法，提供理论依据。2.分类 随机分布：泊松(Poisson)分布 聚集分布：负二项分布(negative binomial distribution)奈曼分布(Neyman)泊松二项分布第1页/共94页 The simplest view of spatial patterning can be obtained by adopting an individual orientation,and asking the question,Given the location of one individual,w
2、hat is the probability that another individual is nearby?There are three possibilities:1.This probability is increasedaggregated pattern 2.This probability is reduceduniform pattern 3.This probability is unaffectedrandom pattern第2页/共94页RandomAggregatedUniformFigure4.3 Three possible types of spatial
3、 patterning of individual animals or plant in a population.第3页/共94页3.频次分布理论公式 （1）泊松（普阿松）分布例：蝗蝻的田间分布02050101200112第4页/共94页(1)普阿松分布（Poisson 分布）第5页/共94页例：对公共汽车客流进行调查，统计某天上午10301147左右每隔20秒钟来到的乘客批数，共得到230个记录。来到批数i0123 4总共频数ni100813496230频率0.430.350.150.040.030.420.360.160.050.01第6页/共94页普阿松分布的意义已经发现许多随机现象
4、服从普阿松分布 （1）社会生活，服务行业 如：电话交换台中来到的呼叫数 公共汽车站来到的 乘客数 （2）物理学 放射性分裂落到某区域的质点数 （3）昆虫个体的空间分布第7页/共94页普阿松分布的特点以交换台电话呼叫数为例 （1）平衡性 在t0,t0+t中来到的呼叫数只与时间间隔长度t有关，而与时间起点T0无关 （2）独立增量性（无后效性）在t0,t0+t内来到k个呼叫这一事件与时刻T0发生的事件独立 （3）普通性 在充分小的时间间隔中，最多只来到一个呼叫第8页/共94页例：蝗蝻分布型调查，共取样408个虫数 x频率 f f*x02250113013024080310304312408252计算
5、方法计算方法第9页/共94页另样的理论数 n*p0=408*0.5391=219.09有一头虫的样本的理论数 n*p1=135.9 第10页/共94页观察值与理论值比较虫数 x观察值(o)理论值(c)0225219.90.111130135.90.2624042.20.093108.70.21 431.32.222.89自由度自由度=n-2=3=n-2=3，失去两个自由度，失去两个自由度（1 1）用来限制实际样本数）用来限制实际样本数N N (2)(2)用来估计用来估计第11页/共94页 意味不是一个小概率事件（p0.05),没有理由否定假设第12页/共94页要求各组内的预计数都不少于5，当某
6、组的Y少于5时，须把它和相邻的一组或几组合并直到Y大于5，然后再用上式计算 x2值。第13页/共94页 检验的理论与方法1 公式 O为实际观测值，E为理论推算值。其基本原理是应用理论推算值与实际观测值之间的偏离程度来决定其 值的大小。是理论分布总体的频数 是观察分布总体的频数 两个样本来自不同的总体第14页/共94页2分布的特点 df=1 df=3 df=5（1）分布于区间1，），偏斜度随自由度降低而增大，当自由度df=1时，曲线以纵轴为渐近线。（2）随自由度df增大，分布趋左右对称，当df30时，分布接近正态。第15页/共94页3 检验的基本步骤（1）建立检验假设，确定检验水平。（2）计算检
7、验统计量第16页/共94页（3）确定概率P值，计算自由度dfk-1 由 和自由度查统计表 的临界值（4）判断结果 临界值检验假设的关系 值 P 假设 判断 0.05 不拒绝 差异无显著性 0.05 拒绝 差异有显著性第17页/共94页例：假定某地婴儿出生的男女比例为1：1。研究者抽取了一个含10，000名婴儿的样品，男孩5100，女孩4900，问他是否证实了假设或否定了假设。某地婴儿出生性比为1：1 拒绝 婴儿性比不为1：1第18页/共94页注：在自由度df1时，需进行连续性矫正，其矫正的 为：适合性检验 比较观测数与理论数是否符合的假设检验叫适合性检验。例如在遗传学上，常用 检验来测定所得的
8、结果是否符合孟德尔分离规律，自由组合定律等。第19页/共94页例 有一鲤鱼遗传试验，以荷包红鲤（红色）与湘江野鲤（青灰色）杂交，其 代获得如表5-2所列得体色分离尾数，问这一资料的实际观察值是否符合孟德尔的青：红=3：1一对等为基因的遗传规律？表 鲤鱼遗传试验 F2观察结果 体 色 青 灰 色 红 色 总数 F2观测尾数 1503 99 1602第20页/共94页（1）鲤鱼体色 分离符合3：1比率。（2）取显著水平（3）计算 青灰色理论数 红色理论数第21页/共94页（4）差 值表。df=1时，故否定 ，接受 即鲤鱼体色 分离不符合3：1比率。第22页/共94页（2）负二项分布正二项分布是(p
9、+q)n 的展开式的各项,其中n为个体总数，p,q为分成对比两类期望的比例。Student(1907).展开上述式子，于是一个样本单位有r个个体的概率为可以估算出p,k。矩法第23页/共94页由此可以推出第24页/共94页（二）分布型指数第25页/共94页上述蝗蝻例子中说明上述蝗蝻属Poisson分布。第26页/共94页2.David&Moore(1954)方法第27页/共94页 Index of Dispersion Test.We define an index of dispersion I to be For the theoretical Poisson distribution,t
10、he variance equals the mean,so the expected value of I is always 1.0 in a Poisson world.The simplest test statistic for the index of dispersion is a chi-squared one:where I=Index of dispersion(as defined in equation 4.3)n=Number of quadrats counted =value of chi-squared with(n-1)degrees of freedom.第
11、28页/共94页0 41 82 23 54 25 36 1 虫数 频率25例：取了25个样，调查蚯蚓的田间分布。第29页/共94页由于 observed chi-squared 所以，我们接受原假设：蚯蚓田间分布符合Poisson分布。第30页/共94页3.Waters(1959)提出 负二项分布中的Kk的特性：当种群密度因为随机死亡而减小时，k保持不变，表示种群空间分布的内在特点，而与密度无关第31页/共94页4.Tayloz(1961,1965,1978)方法密度越高，种群分布越均匀，（聚集度越低）第32页/共94页5.平均拥挤度指标Lloyd,M.(1967)例：a 1b 0c 2d 3
12、X1=1;x2=0X3=2;x4=3n=4第33页/共94页A：一头“独居”1*(1-1)B:没有邻居C:有两头，各以对方为邻居；2*(2-1)=2D:每个有两个邻居，3*（3-1）=6，总共“邻居”数为：0+0+2+6=8 平均每个个体有1.33个邻居第34页/共94页Lloyd定义聚集度指标：第35页/共94页Iwao 发现第36页/共94页 The idealized index should have three properties.(Elliott 1977)1.It should change in a smooth manner as moves from maximum un
13、iformity to randomness to maximum aggregation.2.It should not be affected by sample size(n),population density(),or by variation in the size and shape of the sampling quadrat.3.It should be statistically tractable,so that a confidence belt can be specified and comparixons between samples can be test
14、ed for significance.第37页/共94页Morisitas Index of Dispersion (1)Morisitas index of dispersion 样本大小 sum of the quadnat counts=Morisita(1962)证明 随机分布的假设下：第38页/共94页Standardized Morisita IndexUniform index=(2)Clumped index=(3)=Value of chi-squared from table with(n-1)degrees of freedom that has 97.5%of the
15、 area to the right.第39页/共94页WhenWhenWhen第40页/共94页When 取值以-1.0到+1.0带着95%置信区间随机分布聚集分布均匀分布第41页/共94页In a simulation study Myers(1978)found the standardized Morisita index to be one of the best measures of dispersion because it was independent of population density and sample size.第42页/共94页例：E.Sinclair 在
16、26个10公顷的样点调查大象的数量，其中一个样点有20头，令一样点30头，还有一样点10头，其他23点为零。（1）计算 Morisitas index第43页/共94页（2）以公式(2)(3)中计算临界点。当自由度=n-1=25,Uniform index Clumped index第44页/共94页（3）计算 Standardized Morisita index:由于（4）因为 于是我们得到结论：在置信水平95%下，在我们取样区大象是聚集分布的。第45页/共94页（三）Sample and Experimental Design Sampling and experimental desi
17、gn are statistical jargon for the three most obvious questions that can occur to a field ecologist:Where should I take my samples,how should I collect the data in space and time,and how many samples should I try to take?第46页/共94页抽样理论及在生态学中的应用W.Gosset 1908年以“Student”笔名将“t-检验”发表于biometrika上,文章中说:“任何实验
18、可以作为是许多可能在相同条件下作出的实验的总体中的一个个体.一系列的实验则是以从这个总体中所抽得的一个样品”1.总体与抽样 设一块棉田有N株棉株,每株上某种害虫数分别为X1,X2.XN,第47页/共94页从总体N中,随机抽取n株(nN)样本,每株虫数分别为X1,X2,Xn.目的:通过样本对总体做出推断第48页/共94页 抽样误差估计及t分布1908年,“Student”发表了t分布第49页/共94页例:棉田中随机调查50株棉株,以估计该棉田中害虫的数量.第50页/共94页Sample Size for Continuous Variable理论抽样数模型第51页/共94页例:洪泽湖蝗区虫数样本
19、数(f)fx0170153532183631030428100127第52页/共94页 如果，我们引入变异系数（coefficient of variation）这儿，=标准差 =观察平均数 那么，绝对误差 可写成相对误差 ，（以百分比形式）（方程1）第53页/共94页 两个平均数的比较例如，我们要比较两个池塘中同一种鱼的重量是否有差异，典型的方法是个抽取一定数量的样本用t检验来检验两样本平均数是否有差异。但是，如何在抽样前回答应该取多少样？Snedecor and Cochran(1967,113)提出了如下的近似公式：一般 这儿 =从两个种群中的每一个抽取的样本大小；=水平为 的标准正态离
20、差值 （）第54页/共94页 =水平为 的型错误概率下的标准正态离差值（见下表）=测量的方差。（已知，或推测）。=你希望以 概率能检测出的两平均值的最小差异。Type error Power Two-tailed 0.40 0.60 0.25 0.20 0.80 0.84 0.10 0.90 1.28 0.05 0.95 1.64 0.01 0.99 2.33 0.001 0.999 2.58第55页/共94页例.如果上例中我们希望检测出的平均数差异是：（从以前的研究中知道）如果，则 条。第56页/共94页2.SAMPLE SIZE FOR DISCRETE VARIABLES Counts
21、of the numbers of plants in a quadrat or the numbers of eggs in a nest differ from continuous variables in their statistical properties.The frequency distribution of counts will often be described by either the binomial distribution,the Poisson distribution or the negative binomial distribution(Elli
22、ott 1977).The sampling properties of these distributions differ,so we require a different approach to estimating sample sizes needed for counts.(1)Proportions and Percentages Proportions like the sex ratio or fraction of juveniles in a population are described statistically by the binomial distribut
23、ion.All the organisms are classified into two classes,and the distribution has only two parameters:Proportion of types in the population Proportion of types in the population第57页/共94页If sample size is above 20,we can use the normal approximation to the confidence interval:Where Observed proportion V
24、alue of Students t-distribution for n-1 degrees of freedom Standard error of Thus the desired margin of error is Solving for n,the sample size required is 第58页/共94页 where n=Sample size needed for estimating the proportion p d=Desired margin of error in our estimate As a first approximation for we ca
25、n use We need to have an approximate value of p to use in this equation.Prior information,or a guess,should be used;the only rule-of-thumb is that when in doubt,pick a value of p closer to 0.5 than you guess.This will make your answer conservative.As an example,suppose you wish to estimate the sex r
26、atio of a deer population.You expect p to be about 0.40,and you would like to estimate p within an error limit of with .From equation第59页/共94页(2)Counts from a Poisson DistributionSample size estimation is very simple for any variable that can be described by the Poisson distribution,in which the var
27、iance equals the mean.From this it follows thatorThus from equation,(1)assuming :where Sample size required for a Poisson variable Desired relative error(as percentage)Coefficient of variation=第60页/共94页For example,if you are counting eggs in starling nests and know that these counts fit a Poisson di
28、stribution and that the mean is about 6.0,then if you wish to estimate this mean with precision of (width of confidence interval),you have:nestsEquation(2)can be simplified for the normal range of relative errors as follows:For precision 第61页/共94页3.STATISTICAL POWER ANALYSIS DecisionState of real wo
29、rld Do not reject null hypothesis Reject the null hypothesisNull hypothesis is Correct decision Type error actually true (probability=1-)(probability=)Null hypothesis is Type error Correct decision actually false (probability=)(probability=(1-)=power)Most ecologists worry about ,the probability of a
30、 Type error,but there is abundant evidence now that we should worry just as much or more about ,the probability of a Type error(Peterman 1990;Fairweather 1991).第62页/共94页 Power analysis can carried out before you begin your study(a priori,or prospective power analysis)or after you have finished(retro
31、spective power analysis).Here we discuss a priori power analysis as it is used for the planning of experiments.Thomas(1997)discussed retrospective power analysis.The key point you should remember is that there are four variables affecting any statistical inference:sample sizeProbability of a Probabi
32、lity of a Type error Type error Magnitude of the effect=effect sizeThese four variables are interconnected,and once any three of them are fixed,the fourth is automatically determined.Looked at from another perspective,given any three of these,you can determine the fourth.第63页/共94页第64页/共94页第65页/共94页第
33、66页/共94页SUMMARYThe most common question in ecological research is,how large a sample should I take?This chapter attempts to give a general answer to this question by providing a series of equations from which sample size may be calculated.It is always necessary to know something about the population
34、 you wish to analyze unless you use guesswork or prior observations.You must also make some explicit decision about how much error you will allow in your estimates(or how small a confidence interval you wish to have).For continuous variables like weight or length,we can assume a normal distribution
35、and calculate the required sample sizes for means and for variances quite precisely.For counts,we need to know the underlying statistical distributionbinomial,Poisson,or negative binomialbefore we can specify sample sizes needed.第67页/共94页 Power analysis explores the relationships between the four in
36、terconnected variables (probability of Type error),(probability of Type error),effect size,and sample size.Fixing three of these automatically fixes the fourth,and ecologists should explore these relationships before they begin their experiments.Significant effect sizes should be specified on ecolog
37、ical grounds before a study is begun.第68页/共94页Sampling Designs:Random,Adaptive and Systematic Sampling(1)Simple Random Sampling(2)Stratilied Random Sampling(3)Adaptive Sampling(4)Systematic Sampling第69页/共94页 Simple random sampling is the easiest and most common sampling design.Each possible sample u
38、nit must have an equal chance of being selected to obtain a random sample.All the formulas of statistics are based on random sampling,and probability theory is the foundation of statistics.Thus you should always sample randomly when you have a choice.In some cases the statistical population is finit
39、e in size,and the idea of a finite population correction must be added into formulas for variances and standard errors.These formulas are reviewed for measurements,ratios,and proportion.第70页/共94页 Often a statistical population can be subdivided into homogeneous subpopulations,and random sampling can
40、 be applied to each subpopulation separately.This is stratified random sampling,and represents the single most powerful sampling design that ecologists can adopt in the field with relative ease.Stratified sampling is almost always more precise than simple random sampling,and every ecologist should u
41、se it whenever possible.Sample size allocation in stratified sampling can be determined using proportional or optimal allocation.To use optimal allocation,you need rough estimates of the variances in each of the strata and the cost of sampling each strata.Optimal allocation is more precise than prop
42、ortional allocation,and is to be preferred.Some simple rules are presented to allow you to estimate the optimal number of strata you should define in setting up a program of stratified random sampling.第71页/共94页 If organisms are rare and patchily distributed,you should consider using adaptive cluster
43、 sampling to estimate abundance.When a randomly placed quadrat contains a rare species,adaptive sampling adds quadrats in the vicinity of the original quadrat to sample the potential cluster.This additional nonrandom sampling requires special formulas to estimate abundance without bias.第72页/共94页 Sys
44、tematic sampling is easier to apply in the field than random sampling,but may produce biased estimates of means and confidence limits if there are periodicities in the data.In field ecology this is usually not the case,and systematic samples seem to be the equivalent of random samples in many field
45、situations.If a gradient exists in the ecological community,systematic sampling will be better than random sampling for describing it.第73页/共94页第74页/共94页第75页/共94页第76页/共94页第77页/共94页 Step 1.Calculate the average abundance of each of the networks:(8.35)where =Average abundance of the i-th network =Abund
46、ance of the organism in each of the k quadrats in the i-th network =Number of quadrats in the i-th netwrok Step 2.From these values we obtain an estimator of the mean abundance as follows:(8.36)where Unbiased estimate of mean abundance from adaptive cluster sampling Number of initial sampling units
47、selected via random sampling第78页/共94页If the initial sample is selected with replacement,the variance of this mean is given by:(8.37)where Estimated variance of mean abundance for sampling with replacement and all other terms are as defined above.If the initial sample is selected without replacement,
48、the variance of the mean is given by:(8.38)where N=Total number of possible sample quadrats in the sampling universe第79页/共94页The example shown in Figure 8.3.in the initial random sample of n=10 quadrats,from equation(8.36).plants per quadratSince we were sampling without replacement,we use equation(
49、8.38)to estimate the variance of this mean:第80页/共94页We can obtain confidence limits from these estimates in the usual way:For this example with n=10,for 95%confidence limits ,and the confidence limits become:or from 0.0 to 0.171 plants per quadrat.When should one consider using adaptive sampling?Muc
50、h depends on the abundance and the spatial pattern of the animals or the plants being studied.In general the more clustered the population and the rarer the organism,the more efficient it will be to use adaptive cluster sampling.Thompson(1992)shows,in Figure 8.2,that adaptive sampling is about 12%mo