Question

Conservation of whooping cranes Whooping Cranes are large, heron-like birds that live in the west/mid-west of America (Alberta and Wisconsin) in the summer, and migrate south (Texas and Florida, respectively) in the winter. A combination of hunting and habitat loss nearly to drove them to extinction in the 1930s. In 1967, the Whooping Crane was declared endangered, and it has been the subject of intense conservation e↵orts ever since. The conservation e↵ort has had limited success due, in part, to the longevity of the birds (they can live to be over 25 years), and their long, di cult migration. Below is a table showing the number of wild Whooping Cranes each decade from 1940 to 2010. It comes from a paper by Butler, Harris and Strobel, published in the journal Biological Conservation in 2013. year time since start(years) number of Cranes 1940 0 26 1950 10 32 1960 20 37 1970 30 58 1980 40 79 1990 50 146 2000 60 180 2010 70 283 (a) Plot the raw data with the following commands: t=seq(from = 0, to = 70, by = 10) N=c(26,32,37,58,79,146,180,283) plot(t,N) (b) We begin by investigating whether the population exhibits exponential growth. We will plot the data on a semilog graph and find the best fit line to log10(N) vs. t. Assuming you have already entered t and N into R (if not use the code above), the commands for making the semilog plot with the data and best fit line are as follows: logN = log10(N) F = lm( logN~t ) plot(t,logN) lines(t,fitted(F)) (c) The equation of the best fit line defines a model; i.e. an equation which gives an approximation to the number of cranes as a function of time. To get the equation for the best-fit line, type in “F” (what we named the output from the fitting command) at the prompt in R. If the linear fit is of the form y = mx + b, then the coe cient called “intercept” is b and the other coe cient is the slope. Let N(t) represent the model equation for the number of whooping cranes as a function of time (where time is measured in years since 1940). - Use the equation of the best fit line to find the equation for N(t). - Plot the original data and a graph of N (t) on the same plot. The commands for plotting the data are given in part (a). To add the model, evaluate your expression for N (t) at the same time points, and use the lines command to add the curve to the plot. (d) Discuss and answer the following (in complete sentences): - Could you predict the number of cranes 10 years in the future? 100 years? Would you have confidence in these predictions? - Even though we don’t have data for the years between our 10 year mea- surements, the model predicts the number of cranes at any year. In fact, the model predicts the number of cranes down to the minute. Do you think the model is reliable in its predictions of the number of cranes over very short intervals? (e) The number of cranes at 10 year intervals forms a sequence, N0 = 26, N1 = 32, N2 = 37, etc. Use your model from part (c) (i.e. your equation for N(t)) to generate a sequence of the predicted number of cranes, P0, P1, . . . P7, at 10 year intervals. (f) Find a recursion, of the form Pi+1 = rPi (where r is a constant that you must determine) that describes the predicted number of cranes. Don’t forget that your model in part (c) assumes that time is measured in years while the pre- dicted data P0,P1,...,P7 is given in 10 year increments. Give a biological interpretation for the constant r. Check your answer by performing the recursion with a “for” loop in R. Here’s an example (with made up numbers for P0 and r) to illustrate how it works. Pretend my recursion from is Pi+1 = 2Pi, and suppose that P0 = 10. Then, the following code will generate the sequence {Pi}, stored in the variable P. Note that the first entry of P will be P0, the second P1 etc. P=10 for(i in 1:7){P[i+1]=2*P[i]} P

Answer 1

(a)

(b)

a((c) y = mx+b

F = logN(t)= N(t)= 0.1539t+1.33881

(d) From the above equation

predicted values of F are 1.33881, 1.49271, 1.64661, 1.80051, 1.95441, 2.10831, 2.26221, 2.41611

predicted values of N(t) are 21.81775, 31.09639, 44.32105, 63.16987 , 90.03472, 128.32462, 182.89844,  260.68137

Actual values of N(t) are 26,32,37,58,79,146,180,283

The errors are 4.1822480, 0.9036082, -7.3210459, -5.1698725, -11.0347162 17.6753761, -2.8984395, 22.3186268

As most of the errors are significantly away from 0, then the fitted model may not be good.

(e) the recursion formula for predicted values of N(t) is

N(1)=21.81775

N(t+1) =1.425279184xN(t) for t = 1,2,...,7.

q=21.81775
> for(i in 1:7){q[i+1]=1.425279184*q[i]}
> q
[1] 21.81775 31.09638 44.32103 63.16984 90.03466 128.32453 182.89828
[8] 260.68111