One method for estimating abundance of animals is known as line-intercept sampli

Question

One method for estimating abundance of animals is known as line-intercept sampling. the theory of this method, when applied to Alaskan wolverines, predicts that p = 0.453 of attempts to locate wolverine tracks should be successful. Suppose that biologists will make 100 attempts to locate wolverine tracks in random locations in Alaska. Show that this sample size is large enough to justify using the normal approximation to the sampling distribution of p. What is the mean of the sampling distribution of p if the theory of line-intercept sampling is correct? What is the standard deviation of the sampling distribution of p if the theory of line-intercept sampling is correct? If the theory of line-intercept sampling is correct, what is the probability that a sample proportion, p, would differ from p = 0.453 by as much as

Explanation / Answer

a. As n>30, also np=45.3>10 also npq=24.78>10

Hence distribution can be approximated to normal

b. Mean=np=45.3

c. Standard deviation =sqrt(npq)=4.98

d. P(-0.05<x<0.05)=P(-0.05/sqrt(pq/n)<z<0.05/sqrt(pq/n))=P(-1<z<1)=0.6827

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