# Estimating Population Size Using Mark and Recapture Method

**Topics:**Arithmetic mean, Standard deviation, Sample size

**Pages:**5 (1699 words)

**Published:**July 26, 2013

Table 1: Uncertainties of apparatus used in the experiment.

Apparatus| Uncertainties|

Stopwatch| ±0.01s|

Table 2: Formulae and sample calculations involved in processing data in the experiment. Calculations| Formula| Sample Calculation|

Mean

( x )| x = 1n i=1naiWhere, 1. n refers to the total number of values. 2. ∑ refers to the addition of all values starting with the first value, denoted by i = 1, and ending off with the last value, denoted by n. 3. ai refers to the values in sequence from i = 1 to the nth term or last term.| Mean Lincoln Index: x= 15114+127+194+163+172≈154 | Standard Deviation ( σ )| σ=1ni=1nfi(xi-x)2Where, 1. n refers to the total number of values. 2. ∑ refers to the addition of all values starting with the first value, denoted by i = 1, and ending off with the last value –denoted by n. 3. fi refers to the frequency of that exact term being calculated. 4. (xi-x)2 refers to the square of the term, denoted by xi, subtracted by the mean value of the terms, denoted by x.| Standard Deviation for Lincoln Index: σ= 15(114-1542+127-1542+194-1542+163-1542+(172-154)2)≈33.0| Lincoln Index (Total Population)| Total Population =x1-x2x3 Where, 1. x1 refers to the number Of white beads in the first capture. 2. x2 refers to the total number of white and red beads in the second capture. 3. x3 refers to the number of red beads in the second capture| Total Population of white beads (1st replicate):19×122=114| Chi Squared Test| x2 = i=1n(Oi-Ei)2EiWhere, 1. x2 refers to the Pearson’s cumulative test statistic. 2. ∑ refers to the addition of all values generated. 3. i refers to the degrees of freedom. 4. n refers to the number of different types of appearance. 5. Oi refers to the observed frequencies. 6. Eirefers to the expected frequencies.| Chi Squared Value for the use of Lincoln index to calculate total population of white beads:x2=159-1542154=0.162Taking the degrees of freedom in this case to be 4, compare the chi squared value with a probability table to obtain the percentage reliability of the null hypothesis; There is no significant different between the observed number of white beads and the expected number of white beads.Probability Value: 3.36|

Table 3: Number of white beads in 1st capture, total number of white and red beads in 2nd capture and number of red beads recaptured used to obtain a mean expected total population calculated using the Lincoln index. Replicate| Number of white beads in 1st capture| Total number of white and red beads in 2nd capture| Number of red beads in the 2nd capture (recaptured)| Lincoln Index*| 1| 19| 12| 2| 114|

2| 33| 23| 6| 127|

3| 36| 43| 8| 194|

4| 35| 42| 9| 163|

5| 31| 39| 7| 172|

Mean| -| -| -| 154|

Standard Deviation| -| -| -| 33.0|

Chi-Squared Value| | | | 33.2|

*Values in 3 significant figures to provide a proper comparison with the actual value of 159. Graph 1: Number of White Beads Observed and Expected, calculated using the Lincoln Index.

Discussion of Results and Conclusion

From Graph 1, we can see that the expected and observed numbers of white beads are very close and the error bar of the expected population size overlaps with the observed value, hence we say that this mode of calculation is rather accurate.

However, it can be seen that the size of the error bar is very big and covers a large range of values, which may show that the Lincoln Index is not a very precise mode of calculation as it results in a large range of values.

The Chi-Squared value is much lower than the critical value for 4 degrees of freedom at a 95% confidence level, hence we accept the null hypothesis; there is no significant difference between the observed and expected values.

The Lincoln Index provides a way to measure population sizes of individual animal...

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