# hw5_sol - MS&E 226 Small Data Solutions for Problem Set 5...

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MS&E 226 Solutions for Problem Set 5 “Small” Data Problem 1. Often, biologists are interested in estimating the size of an animal’s population size. Frequently it is impossible to count all animals, so we have to resort to estimation. A popular method is called capture-recapture. First, a portion of the population is captured, marked and released. At a later time, another portion is captured and the number of marked individuals in this second sample is counted. We assume that all animals have the same probability of being captured is identical. Let N be the total number of animals in the population, and let n denote the number of animals marked on the first visit. Furthermore, let K be the number of animals captured on the second visit and k be the number of marked recaptured animals. For this problem, we assume that in both samples the animals are sampled without replacement. A estimator of the total population is the Lincoln-Peterson estimator . We expect that the pro- portion of marked individuals ( k/K ) should be approximately equal to the total population that is marked ( n/N ). Hence, the estimator then becomes: ˆ N LP = Kn k . However, it turns out to be biased in small samples. Instead, we consider the Chapman estima- tor : ˆ N C = ( K + 1)( n + 1) k + 1 - 1 . For more information, see the Wikipedia page Mark_and_recapture . a) Suppose you are given n , K and k . Explain how you can use the bootstrap to create confi- dence intervals for the estimator ˆ N C . b) Suppose initially 200 animals are captured and of 200 recaptured animals, 21 were marked. What is the estimated population size? Also, use the bootstrap to estimate a 95% confidence interval around this estimate, both by using the quantile approach, and bootstrapped standard error estimates. Comment on which of the two approaches seems more appropriate in this specific context.

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SOLUTION: a) We can create a bootstrap sample by sampling with replacement from the original data set. Since we have K total animals and k of them are marked in the original data set, every data point in the bootstrap sample has probability k/K of being marked and ( K - k ) /K of being unmarked. Then, taking bootstrap samples from the original data is equivalent to sampling from a bino- mial distribution where the probability of getting a marked animal data point is p = k K . Then, for each bootstrap sample, we can apply the Chapman estimator. We will denote the Chapman estimator for the ith bootstrap sample as ˆ N ( i ) C . We can create a 95% confidence interval for the estimator ˆ N C with either 1) a bootstrap standard error interval by taking the mean, μ B , and standard deviation, σ B , of the ˆ N ( i ) C values and giving the interval ( μ B - 1 . 96 * σ B , μ B + 1 . 96 * σ B ) or 2) a quantile interval by taking the 0 . 025 quantile of the ˆ N ( i ) C values, denoted q B, 0 . 025 and the 0 . 975 quantile of the ˆ N ( i ) C values denoted q B, 0 . 975 and giving the interval ( q B, 0 . 025 , q B, 0 . 975 ) More generally, if we want a (1 - α )100% confidence interval, we can give ( μ B - z α/ 2 * σ B , μ B + z α/ 2 * σ B ) Where z α/ 2 is the (1 - α 2 )
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