Borrowing owls sometimes build their nests in holes that were dug by prairie dog

Question

Borrowing owls sometimes build their nests in holes that were dug by prairie dogs, coyotes, or badgers and have been since abandoned. They sometimes line the nests with cattle or horse dung. Why? Perhaps it insulates the nest from temperature extremes. A possibility suggested by biologist Dennis Martin is that the owls use the dung to keep predators away. To test this theory, biologist Gregory Green observed lined and unlined owl nests in the Columbia River basin. He recorded whether the nest was raided or not in both samples. For now consider only the lined nests. Green identified n=25 nests lined with dung and observed that only 2 were raided.

(a) If the probability that a next is raided is 0.6, is Green's finding something that would have much chance of happening? What would be that chance? Explain and show work. Also, provide the R code that you would use to find that chance using R.

(b) What should Green do to get a more accurate picture of how often lining the nest with dung prevents raiding by predators?

Explanation / Answer

Solution      

Let X = number of nests raided. Then, X ~ B(n, p), where n = sample size, which is 25 in the present case and p = probability of a raid, which is given to be 0.6

Under (a), we want to find, P(X = 2). This can be either calculated applying the pmf of

B(25, 0.6) or using Excel Function.

Using Excel Function, P(X = 2) = 0.000000076 which is insignificantly small.

For Part (b), we need to find the sample size required to estimate the occurrence more closely. The formula for this is: n = p(1 - p)Z2/2/E2, where 100(1 – ) is the confidence level required, p is the initial estimate of p, (0.6 in the present case), Z/2 is the upper /2 percent point of Standard Normal Distribution and E is the error margin.

Since these values are not specified in the question, an exact answer cannot be given. However, just to illustrate the use of the formula, if we want to estimate p with 95% confidence within a margin of ± 0.1, the required n = (0.6 x 0.4 x 1.962)/0.01 = 92.19 ~ 93

DONE

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