For this exam, you will render this Rmarkdown document into a PDF as you have done with your labs. You can write your answers below each question and show your work using the code chunk. Note that small coding errors will cause your render to fail and Rmarkdown is not as easy to debug as a typical R script that can be run one line at a time. We recommend (a) rendering your exam before you start, and (b) rendering after each answer, and/or (c) using a separate script to debug your code and then transferring the working code into the Rmarkdown code chunk.
Read each question carefully and make sure that you answer completely. SHOW ALL YOUR WORK to get partial credit for incorrect answers. Use graphs and illustrations any time, but it will be clear if you need to use them. USE APPROPRIATE AXIS LABELS on your graphs. If you aren’t sure about a question, give it your best shot and explain what you did and why.
##A Lincoln-Peterson mark recapture analysis is a mark-recapture method that is used to estimate population size. This is done by capturing and marking organisms during a sampling collection. Then there is a second sampling collection where more animals are collected and marked. Estimates of population size are determined by the the proportion of marked animals in the surveys compared to the proportion of marked animals in the population. The equation N/M = C/R -> N = M*C/R is used to determing this where N=population estimate, M=marked on 1st survey, R=recaptured individuals, and C=total sample for the second survey. The three critical suumptions for this analysis are there are no births or deaths in the population, no immigrating or emigrating individuals to the population, and that marking has no affect on mortality.
##The p-value is the probability that there is not a relationship between the x and y axis values or is the slope statistically different from 0. R2 is how well the regression line fits the data. The closer the data is to the regression line then the closer the R2 value is to 1. R2 is not dependent on slope but is a measure of how far the data points are from the regression line.
300*(.88)^1 ##after 1 year
## [1] 264
300*(.88)^7 ## after 7 years
## [1] 122.6027
elk=c(140, 170, 190, 228, 215, 251)
lambda=elk[2:6]/elk[1:5]
lambda
## [1] 1.2142857 1.1176471 1.2000000 0.9429825 1.1674419
mean(lambda)
## [1] 1.128471
100*exp(.07)*1
## [1] 107.2508
107.2508*exp(.07)
## [1] 115.0274
115.0274*exp(.07)
## [1] 123.3678
100*exp(.07)^20
## [1] 405.52
##16 years
#100*exp(.07)^t
#100*exp(.07)^16
#knitr::include_graphics("Exam1_Q9a.png")
##The population is declining and also the population does not fit the linear regression line that is plotted well. While the population looks to be declining there is a lot of variability from year to year and the data doesn’t fit the line well.
#plot(xlab=population,y=time)
reg_log=lm(log(SuckerCatch\(catch)~SuckerCatch\)Year) summary(reg_log)
#Call: # lm(formula = log(SuckerCatch\(catch)
~ SuckerCatch\)Year) # # Residuals: # Min 1Q Median 3Q Max #
-1.3275 -0.8013 -0.0525 0.7962 1.3376 # # Coefficients: # Estimate Std.
Error t value Pr(>|t|)
# (Intercept) 538.66373 199.56816 2.699 0.0244 # SuckerCatch$Year
-0.26813 0.09978 -2.687 0.0249 # — # Signif. codes: 0
‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1 # #
Residual standard error: 1.047 on 9 degrees of freedom # Multiple
R-squared: 0.4452, Adjusted R-squared: 0.3835 # F-statistic: 7.221 on 1
and 9 DF, p-value: 0.02491
coef(reg_log)[2] # SuckerCatch$Year # -0.2681318
summary(reg_log)$coefficients[2,4] # 0.02491362
confint(reg_log) # 2.5 % 97.5 % # (Intercept) 87.2091862 990.11826873 # SuckerCatch$Year -0.4938588 -0.04240484
#### d) What are lambda and the percent change per year? Show your work. (2 points)
## the percent change is -26.81% a year.
```r
##coef(reg_log)[2]
# SuckerCatch$Year
# -0.2681318
##We should give the population special conservation status because we are statisically confident the population is decreasing. We need to know what QET is for this species to help determine if protection is needed and support a conservation listing.
##The population abundances with the highest growth rates are ones that are at half the carrying capacity. The population abundance associated with the lowest overall growth rate are populations that have exceeded carrying capacity.
##No because we don’t know what the carrying capacity is.
#knitr::include_graphics("Exam1_Fig1.png")
##I think the breeding success would start to level off. As more individuals are competeing to breed then breeding success will decrease.
##There is more competition for mates. More competition means that less individuals will have success in being able to breed per capita. The term for this is density dependent. The density of the population affects the breeding success.
The dataset includes the number of individuals caught each day (Catch), the number of those that are recaptures (Recaptures), and number of individuals newly marked (Newmarks). Hint: Does the Schnabel estimate use the NEW marked each day or some other metric? Make sure your results make sense given the data as a sanity check.
As a reminder:
n = sum(Mt*Ct) / sum(Rt)
Where: n = an estimate of the true number (N) of individuals in the population at the time of initial marking. Mt = the number of individuals in the population that have been marked at the time the t sample was taken. Ct = the number of individuals captured at the sampling that occurred at time t. Rt = the number of recaptured, marked individuals taken at time t.
ct<-eulachon\(Catch Rt<-eulachon\)Recaptures nmk<-eulachon$Newmarks mt<-c(0,cumsum(nmk)[1:4]) mt [1] 0 29 49 67 79 N_hat<-sum(mt+ct)/sum(Rt) N_hat #[1] 8.780488
#eulachon <- data.frame(Month=1:5,Catch=c(29,34,28,20,25),Recaptures=c(0,14,10,8,9),Newmarks=c(29,20,18,12,16))
#ct<-eulachon$Catch
#Rt<-eulachon$Recaptures
#nmk<-eulachon$Newmarks
#mt<-c(0,cumsum(nmk)[1:4])
#mt
#N_hat<-sum(mt+ct)/sum(Rt)
#N_hat
# N_hat
#[1] 8.780488
Risk and IUCN status criteria. The International Union for the Conservation of Nature (IUCN) uses the following general criteria for “listing” species: http://www.iucnredlist.org/technical-documents/categories-and-criteria/2001-categories-criteria
“>=50% chance of extinction within 10 years or three generations, whichever is the longer (up to a maximum of 100 years) = list as ‘Critically Endangered’”
“>=20% chance of extinction within 20 years or five generations, whichever is the longer (up to a maximum of 100 years)= list as ‘Endangered’”
“>= 10% chance of extinction in 100 years = list as ‘Vulnerable’”
“<10% chance of extinction in 100 years = no listing”
Species or populations are always listed at the highest level for which they meet the criteria.
##I would list this population at endangered because it has a greater that 20% chance of falling below QET within 20 years. By less than 50 years almost 100% of the population would be below QET.
#knitr::include_graphics("Exam1_Q12a_TS.png")
#knitr::include_graphics("Exam1_Q12a_QET.png")
##I would give this population a vulnerable listing because there is between 10 and 20% of the population below QET at 100 years.
#knitr::include_graphics("Exam1_Q12b_TS.png")
#knitr::include_graphics("Exam1_Q12b_QET.png")