In Problems 3.6.16 and 6.2.33, you studied Edwards and Eberhardt's data on 135 cottontail rabbits observed in a protected enclosure. They were studying various capture-recapture techniques for estimating the size of wildlife populations. Here's how one of the techniques works:In Problem 6.2.33 you showed that the geometric model which had been proposed by Edwards and Eberhardt on the basis of several different analyses is not unreasonable. Assuming a geometric model with Y as the number of rabbits which will be recaptured X times, the proportion of rabbits which will be captured X = x times is pqx” , So if the model is valid, the expected value of Y is Npqx“, We want N. Using logarithms;' we can isolate N this way: In(Y) = In(Np) + x In(q) . This is a linear equation in x and you can use your observed data to “fit” this equation and then estimate N. First, transform the observed data from (x, y) to (x, lny):When you look at the original data (Problem 3.6.16), you'll see we've omitted the “observation” for X = 5. Because no rabbit was caught five times, there was no such observation. Note that the 59 from the original data is also not part of your data because you do NOT know how many rabbits were never captured.(a) We will use a regression model for In(Y) with X as explanatory variable. What are α; and β? (b) Fit a least squares line to the transformed data.(c) Use the line in part (a) to estimate In(Np). Then estimate exp[ln(Np)] =Np. (d) Use the regression line in part (a) to estimate In(q), then q, and, finally, p. (e) Use parts (c) and (d) to estimate N . Compare your estimate with what in this study we know to be the true value: N = 135. (f) Note that when X = 0, the model says Y = N p. In other words, Np is the number of rabbits never captured. So then  But you would know that number from the data; here, N q = 76. So there are two more ways to estimate N. What are they? (g) Suppose in part (e) you had estimated p by its maximum likelihood estimator (MLE) as you did in Problem 6.2.33(b). What happens and why? (h) Critique the analysis of this problem.Problem 3.6.16:Various methods have been proposed for estimating the size of wildlife populations. One technique is to trap animals periodically in their home range. When an animal is caught, it's marked and then released. As the trapping proceeds, a record is kept of the number of times each animal is caught. W.R. Edwards and L.L. Eberhardt (1967) did an experiment with 135 cottontail rabbits in a protected enclosed 40-acre area where they repeated the trapping seven times. Here's their data:For example, 16 of the 135 rabbits were caught on two occasions. The number 59 was not observed, of course; it was inferred from the rest of the data by assuming none of the 135 rabbits died or escaped the area or in some other way were lost from the population.In a study of an extensive set of live-trapping data, Eberhardt together with T.]. Peterle and R. Schofield (1963) had given several arguments suggesting a geometric distribution as a good model for the capturerecapture experiment with X being “the number of times a particular animal istrapped.” This proposes a purely abstract model because there's no Bernoulli trial here to be repeated. Because X starts at zero, the geometric random variable would be, let's say, W = X + 1. So P(W = w) = pqw-1 which means P(X = x) = pqx. Edwards and Eberhardt (1967) estimated p to be 0.4424. We'll see later  how they obtained this value and how they estimate an unknown population size.Compare Edwards and Eberhardt's data with what would be expected if indeed the geometric model is valid. One decimal place of accuracy is adequate for comparison purposes.Problem 6.2.33:(a) There are eight cells and one parameter which was estimated. In fact, there can be technical difficulties depending on how the parameter was estimated. Let's ignore that technicality. So we have six degrees of freedom for x2 and obtain a p-value greater than ten percent. Based on that analysis, there is no reason to doubt the model.(b) MLE = 1/1.8684Here X2 = 12.8 with six degrees of freedom, giving a p-value a bit lessthan 5%. This does not seem to challenge the model.Still the model from part (a) with a much smaller p-value is not necessarily better: In a X2 test, the parameters should be estimated by the so-called “minimum chi-squared estimator” which is often well approximated by the MLE. But the estimator we used in part (a) is neither of these!