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CSU FW 662 - FW 662 Midterm Exam

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February 12, 1999FW 662 Midterm ExamThis exam is a take-home, open-book exercise. There are 3 questions; you must answer all ofthem, including multiple parts. You may use any reference material (class notes, assignedreading, library material, etc.). Under NO circumstances are you to discuss this exam withclassmates or any other individual. You are to work independently and you should not conferwith others. If you need clarification on a question, please see the instructor, or send email withyour question to [email protected] . This exam is to be turned in by 8:00 am Monday,15 February, at the start of class. Turn in this sheet with your written answers and a disk thatholds the spreadsheet models on which your answers are based. All questions require a writtenanswer. In addition, some questions also require you to provide a spreadsheet demonstratinghow you obtained your answer. Typed, short, concise answers will be graded more generouslythan hand-written, long, rambling responses. Your spreadsheets on a disk will be used to verifythat your answers were obtained in a logical fashion, and provide you with partial credit in caseswhere you got the wrong solution, but just made a simple mistake in the spreadsheet. Identifyyour answer sheets and disks with your SSN only. Only put your name (via your signature)on this sheet.By my signature below, I certify that I have not collaborated with anyone concerning anymaterial related to this examination. SSN Signature DateFW662 Midterm Exam 1999 21. For an extensively-studied mule deer population, adult female over-winter (1 December until15 June) survival averages 0.853, and over-winter fawn survival averages 0.444. Assume that noadditional mortality occurs in adult females from the period 15 June until 1 December. Nohunting mortality was inflicted on the female and fawn portions of the population during the timeperiod of interest. Fawns per 100 adult females during the last 20 years follows the function:fawns/100 adult females = 2952.8 - 1.4543 year ,×where year is from 1978 to 1998. This relationship was developed from helicopter surveysduring December of each year.Biological background. Mule deer are born approximately 15 June each year. Mule deer do nothave fawns until their second birthday, and animals that are 1 year and 6 months old during theDecember counts cannot be distinguished from older animals. Thus, the December ratios offawns per 100 does provide the newly produced animals in terms of all animals in the population,not just the animals that actually had young.A. (25 pts) In what year did the population’s rate of increase decline to 0, i.e., in what year did8 = 1? Describe how you answered this question, and provide me with your spreadsheet on adisk that you used to perform the calculations.B. (15 pts) What assumptions did you have to make to obtain this estimate?C. (15 pts) Nobody believes that adult and fawn survival is constant as described above. Estimates of temporal variation of this population are: adult female survival mean = 0.853,SD = 0.034; fawn over-winter survival mean = 0.444, SD = 0.217; with fawns/100 does assumedto have no temporal variation (other than the linear function given above) for the purposes of thisquestion.. How would your conclusions about the year that 8 = 1 change from part A above ifyou put this level of temporal variation in your model? Even if the mean value of 8 is close to 1,does the population maintain itself? Again provide me with the spreadsheet you used to generateyour answer. If you are unable to incorporate this temporal variation into your model, at leastspeculate on what you think the effect of temporal variation will be.D. (5 pts) No demographic variation has been incorporated into either of your models. Do youthink that demographic variation is needed to answer this question? Why or why not.2. One group of researchers published an article on the decline of northern spotted owls. Theybased their conclusion that the population was declining at the rate of 8 = 0.809 per year on thefollowing Leslie matrix for the female segment of the population:FW662 Midterm Exam 1999 3L '0 0 0.25 0.25 0.250.83 0 0 0 00 0.83 0 0 00 0 0.83 0 00 0 0 0.83 0The parameter values they used were adult survival = 0.83, juvenile survival (from fledgling until1-year old) = 0.25, and 1 female chick per nest. A second group published their model of owlpopulation dynamics based on the same parameter estimates as the first group. However, theyconcluded that the population was only declining at the rate of 8 = 0.958 per year. They alsoused a Leslie matrix:L '0 0 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.250.83 0 0 0 0 0 0 0 0 00 0.83 0 0 0 0 0 0 0 00 0 0.83 0 0 0 0 0 0 00 0 0 0.83 0 0 0 0 0 00 0 0 0 0.83 0 0 0 0 00 0 0 0 0 0.83 0 0 0 00 0 0 0 0 0 0.83 0 0 00 0 0 0 0 0 0 0.83 0 00 0 0 0 0 0 0 0 0.83 0A.(15 pts) Were the 8 values published by each group correct? Use a spreadsheet to substantiateyour answer, and provide me with the spreadsheet. If both groups were correct, how do youreconcile the differences in their estimates of 8?B.(10 pts) What would you estimate as the correct value of 8? In other words, construct yourown Leslie matrix model for this population. Provide your solution on a disk.C.(5 pts) Are these matrices based on a pre-birth census or a post-birth census? Why?3. (10 pts) Do the following Leslie matrices produce the same value of 8 if S1 = S2 = S3? Showyour reasoning.S0b1S1b2S1b3S00 00 S1S1S0b1S1b2S2b3S3b3S00 0 00 S10 00 0 S2S3FW662 Midterm Exam 1999 4Helpful hints:For the beta distribution, mean = , variance = s2 = , mode = for ."" % $"$(" % $)2(" % $ % 1)"& 1" % $ & 2"$ 1Given a mean (µ) and standard deviation (s),, and ." '&µ(µ2& µ % s2)s2$ '(µ & 1)(µ2& µ % s2)s2February 17, 1999FW 662 Midterm Exam – Answers1. A. Year was 1984.861, obtained by optimizing the spreadsheet to get a value of the year thatmade lambda exactly 1. I got this value by removing male fawns after their first year, i.e.,only putting female yearlings into the population. Most of you didn’t do the optimizationthat I show in my answer, but just modeled fawn ratios as th linear function I gave you,and picked out the year when 8 became <1. This approach is okay as long as your modelhad reached stable age distributions before 8 < 1.B.1. System is assumed to not change from 1978-1998, e.g., no habitat changes thatwould change survival.2. Deterministic, no process


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