Sampling Methods Part IV: Census Techniques PDF

SAMPLING METHODS PART IV: CENSUS TECHNIQUES Biodiversity Management 1B  3. SAMPLING Most widely used of the three census techniques Two basic approaches are considered AREA BASED NON-AREA BASED SAMPLING SAMPLING Usually involves Usua...

SAMPLING METHODS PART IV: CENSUS TECHNIQUES Biodiversity Management 1B  3. SAMPLING Most widely used of the three census techniques Two basic approaches are considered AREA BASED NON-AREA BASED SAMPLING SAMPLING Usually involves Usually involves sampling during a sampling at more single time period than one time period  3. SAMPLING – DECISION PATH SAMPLING APPROACHES AREA BASED NON-AREA BASED SAMPLING SAMPLING PLOTS PLOTLESS CHANGE IN RATIO REGRESSION QUADRATS NON-STANDARD STRIP SELECTIVE REMOVAL STANDARD STRIP SAMPLING SAMPLING MARK-RECAPTURE  AREA BASED SAMPLING Whether plot or plotless sampling is used, some form of the basic equation P = D x A applies, where P = Population number D = Density (number of animals per unit area) A = Surface area (i.e. total area occupied by animals) The method (plot/plotless) determines how “D” is calculated  AREA BASED: PLOT SAMPLING For plot sampling “D” is calculated using the equation: 𝑦 𝐷= 𝑎 Where: D = number of animals per unit area y = number of animals in a sample plot a = area of the sampling plot Shape of plot indicates whether Quadrats or Standard Strip Sampling is used Remember we already covered Quadrats (the same applies here) How do we calculate density using quadrats? What two aspects could affect the accuracy of estimates?  AREA BASED: PLOT SAMPLING STANDARD STRIP SAMPLING Narrow, long plots with a fixed width (i.e. fixed transect lines) → same as Belt Transects Equation to calculate density is as follows: 𝑦 𝐷= 𝐿𝑊 Where: D = number of animals per unit area y = number of animals in a sample plot L = length of strip W = width of strip NB! Remember “y” is not a total number but a mean calculated from each transect or plot  AREA BASED: PLOT SAMPLING STANDARD STRIP SAMPLING Example: Consider a ranger having to estimate the population of Zebra within a game reserve that has an area of 3000 ha. He uses 10 randomly placed strip samples, each measuring 1000 m x 1000 m (100 ha). His results are as follows: Strip 1 2 3 4 5 6 7 8 9 10 Number 5 8 3 11 16 6 7 13 18 21 of Zebra What is the estimated density or population of zebra in the reserve?  AREA BASED: PLOT SAMPLING STANDARD STRIP SAMPLING Example: Estimating Zebra density in a game reserve 𝑦 𝐷= 𝐿𝑊 y = mean number of animals = 108/10 = 10.8  AREA BASED: PLOT SAMPLING STANDARD STRIP SAMPLING Example: Estimating Zebra density in a game reserve 𝑦 𝐷= 𝐿𝑊 y = mean number of animals = 108/10 = 10.8 Sampled Area = 100 ha  AREA BASED: PLOT SAMPLING STANDARD STRIP SAMPLING Example: Estimating Zebra density in a game reserve 𝑦 𝐷= 𝐿𝑊 y = mean number of animals = 108/10 = 10.8 Sampled Area = 100 ha Density = 10.8/100 = 0.108 zebra per 100 ha  AREA BASED: PLOT SAMPLING STANDARD STRIP SAMPLING Example: Estimating Zebra density in a game reserve 𝑦 𝐷= 𝐿𝑊 y = mean number of animals = 108/10 = 10.8 Sampled Area = 100 ha Density = 10.8/100 = 0.108 zebra per 100 ha Population in reserve = 0.108 x 3000 = 324 Therefore it’s estimated that there are 324 Zebra in the reserve.  3. SAMPLING – DECISION PATH SAMPLING APPROACHES AREA BASED NON-AREA BASED SAMPLING SAMPLING PLOTS PLOTLESS CHANGE IN RATIO REGRESSION QUADRATS NON-STANDARD STRIP SELECTIVE REMOVAL STANDARD STRIP SAMPLING SAMPLING MARK-RECAPTURE AREA BASED: PLOTLESS SAMPLING Plotless methods do not involve counting animals within a sample plot of given dimensions The area per animal rather than the animals per area is sampled NON-STANDARDISED STRIP SAMPLING also referred to as the Line Intercept method NB: Many different ways to measure and calculate densities from the Line Intercept method AREA BASED: PLOTLESS SAMPLING Non-standardised Strip Sampling or Line Intercept method Example 1: Using one line in a small open reserve AREA BASED: PLOTLESS SAMPLING Non-standardised Strip Sampling or Line Intercept method Example 1: - Four animals were observed from the line AB → 1 animal on 1st perpendicular line (10 m) → 1 animal on 2nd perpendicular line (15 m) → 2 animals on 3rd perpendicular line (15 m) - 15 m was the widest distance that animals were observed so take twice 15 m multiplied by the length of the line AB (100 m) - Area of measurement = 100 m x 15 m x 2 = 3000 m2 AREA BASED: PLOTLESS SAMPLING Non-standardised Strip Sampling or Line Intercept method Example 1: - Area of measurement = 100 m x 15 m x 2 = 3000 m2 - Area per animal is therefore 3000/4 = 1 per 750 m2 - If the area of the reserve is 4500 m2 then we could estimate that there are 6 animals in the reserve How was this value concluded? 4500/750 AREA BASED: PLOTLESS SAMPLING Non-standardised Strip Sampling or Line Intercept method Example 2: In a large reserve, more than one line transect can be used AREA BASED: PLOTLESS SAMPLING Non-standardised Strip Sampling or Line Intercept method Example 2: - For each of the four transect lines, the same method as in the previous example is used - All perpendicular distances are added and the mean distance is calculated as Y - All animals recorded are denoted as N - Total length of the four transect lines are added up and denoted as X The Density = N/2XY and this can be extrapolated to total area as was done previously  3. SAMPLING – DECISION PATH SAMPLING APPROACHES AREA BASED NON-AREA BASED SAMPLING SAMPLING PLOTS PLOTLESS CHANGE IN RATIO REGRESSION QUADRATS NON-STANDARD STRIP SELECTIVE REMOVAL STANDARD STRIP SAMPLING SAMPLING MARK-RECAPTURE  NON-AREA BASED SAMPLING Does not take area into account when estimating population numbers Involves two separate periods of measurement (i.e. sampling is undertaken once and after a certain amount of time has passed, another sampling session occurs) X X X NON-AREA BASED: MARK-RECAPTURE Based on trapping/catching, marking and releasing a known number of marked animals into the population At another time period, individuals are recaptured from the population The ratio of marked to unmarked individuals in the sample is then used to estimate the total population NON-AREA BASED: MARK-RECAPTURE Assumptions: 1. All individuals in the population have an = chance of capture 2. Ratio of marked to unmarked individuals remains the same from time of capture to time of recapture 3. Marked individuals redistribute themselves homogeneously throughout the population wrt unmarked ones 4. Marked individuals do not lose their marks 5. The population is closed (i.e. no emigration or immigration) NON-AREA BASED: MARK-RECAPTURE The Lincoln index is used to calculate the relative population size: 𝑛𝑋𝑀 𝑁= 𝑅 Where N = Estimated population, M = animals marked in first survey, n = total animals counted in second survey and R = number of marked animals in the second survey NON-AREA BASED: MARK-RECAPTURE EXAMPLE: Sample of 200 fish were tagged in a lake. After allowing time for the fish to mix with the population, a second sample of 450 fish were caught and only 15 of them were found to be tagged. Using the Lincoln Index, estimate the population of fish in the lake. 𝑛𝑋𝑀 𝑁= 𝑅 450 × 200 N= 15 90 000 N= 15 N = 6000 NON-AREA BASED: MARK-RECAPTURE Estimates are subject to sampling error. Therefore, to get a better estimate we can calculate the standard error as: 𝑀2 ×(𝑛 −𝑅) 2002 × (450 − 15) 𝑆𝐸 = 𝑆𝐸 = 𝑅3 153 40 000 × (435) 𝑆𝐸 = 3 375 17 400 000 𝑆𝐸 = 3 375 𝑆𝐸 = 5 155.55556 𝑆𝐸 = 71.8021975 = 71.80 NON-AREA BASED: MARK-RECAPTURE Next step is to calculate a 95% confidence limit (CL): 95% CL = Estimated population (N) ± SE x 1.96 CL at the upper limit = 6000 + 71.80 X 1.96 = 6 140.728 = 6 141 CL at the lower limit = 6000 – 71.80 x 1.96 = 5 859.272 = 5 859 Therefore, we are 95% confident that the lake has between 5 859 and 6 141 fish.  NON-AREA BASED: EXERCISE QUESTION 1: Mark-recapture A total of 150 tortoises were caught on Dassen Island and were marked. They were released back into the habitat. A week later, a total of 180 tortoises were caught and only 25 of them were marked. Estimate a) the population of tortoises on the Island using 𝑛𝑋𝑀 the formula 𝑁 = 𝑅 b) Indicate the upper and lower population estimates by M2 ×(n −R) calculating the standard error (SE = R3 ) and 95% confidence limits (N ± SE x 1.96) PRACTICE THIS EXAMPLE NON-AREA BASED: SELECTIVE REMOVAL Involves removal of individuals from the population which often influences ratios of age, sex or two types of animals This method therefore considers changes in ratios, involving two periods of sampling 1st survey 2nd survey Male Female Male Male Female Male Male Male Female Male Male Male Certain number of individuals removed NON-AREA BASED: SELECTIVE REMOVAL Assumptions: 1. The population is essentially stationary 2. Probability of capture during trapping period is equal for each animal exposed to capture 3. Probability of capture remains constant from trapping to trapping NB: Last assumption requires that the animals are not trap- shy or trap-prone and requires that bait acceptance, weather conditions and differences in sex and age will not affect the probability of capture Challenging to meet these assumptions NON-AREA BASED: SELECTIVE REMOVAL The first (initial) survey determines the ratio of the variable and the second survey involves determining the changed ratio between recording periods The ratios are used in a formula to estimate the population at the time of the first survey (N1 ) (𝑇𝑥 − 𝑉2 × 𝑇 ) N1 = (𝑉2 −𝑉1 ) Where: N1 = Population estimate at first survey time V1 = Ratio of one variable during first survey V2 = Ratio of same variable at second survey Tx = Number of animals of variable x that were added or removed between surveys T = Tx + Ty , where Ty = number of animals of variable y (second gender or age etc.) that were added or removed between surveys NON-AREA BASED: SELECTIVE REMOVAL EXAMPLE: A researcher was interested in estimating how many impala there are in a game reserve. He decides to use the Selective Removal method using gender as the variable. A first survey of 100 impala found that there were 64 ewes and 36 rams. The proportion of ewes is thus 0.64 (V1) He then removed 50 ewes (Tx = – 50) and 10 rams (Ty = –10) Both the values of 50 and 10 are negative because the animals are being removed NON-AREA BASED: SELECTIVE REMOVAL EXAMPLE cont.: A second survey was conducted a week later and the ewes now only formed 51% of the population. Using the information provided, estimate the initial population, N1. (𝑇𝑥 − 𝑉2 × 𝑇 ) N1 = (𝑉2 −𝑉1 ) (−50− 0.51×−60 ) N1 = (0.51−0.64) −50 −(−30.6) N1 = −0.13 −19.4 N1 = = 149.23 = 149 Impala −0.13 NON-AREA BASED: SELECTIVE REMOVAL EXAMPLE cont.: What about N2? Because its known that 50 ewes and 10 rams (total of 60) were removed after the first survey, 149 – 60 = N2 = 89 Of the 89 animals, 51% were ewes, therefore a total of 45 ewes and 44 rams remained.  NON-AREA BASED: EXERCISE QUESTION 2: Selective removal The same researcher decided to use selective removal to estimate how many rabbits were inhabiting Dassen Island. During the first survey, 100 rabbits were caught. The proportion of males was 0.6 and the proportion of females was 0.4. Then, a total of 40 males and 15 females were removed. A week later, another survey was undertaken and it was found that there was a proportion of 0.48 males. Using this data, calculate: (𝑇𝑥 − 𝑉2 × 𝑇 ) a) N1, the initial population, N 1 = (𝑉2 −𝑉1 ) b) N2, the population estimate at the time of the second survey and c)the total of males and females remaining. PRACTICE THIS EXAMPLE Finally, the end of sampling methods... We will start with an Introduction to Statistics in our next lecture. Any questions?