I will explicitly deduct points if these instructions are not followed and for typos.
In 2009, the government of India started a sanctuary for tigers (Panthera tigris). There are plenty of prey available, poaching is strictly prohibited, and there is no other factor to disturb the intrinsic growth of the population. The data are as follows:
| Year | Abundance |
|---|---|
| 2011 | 59 |
| 2012 | 78 |
| 2013 | 91 |
| 2014 | 93 |
| 2015 | 157 |
What was the geometric growth rate per time step?
What was the average geometric growth rate between 2011 and 2015?
Using the average geometric growth rate, what do you predict the population size would be in 2016?
What is the exponential growth rate per time step?
What is the average exponential growth rate between 2011 and 2015?
What is the arithmetic mean of tiger abundance between 2011 and 2015?
What is the geometric mean of tiger abundance between 2011 and 2015?
Compare the arithmetic mean and the geometric mean. Which is the more appropriate one to use in this case? Why?
Canada lynx (Lynx canadensis) are highly dependent on snowshoe hare (Lepus americanus). Although hare breed four times per year and lynx only breed once per year, you are interested in comparing growth rates of the two species in Ontario, Canada. You sample abundance of each annually for 6 years and come up with the following numbers:
| N of hare | N of lynx |
|---|---|
| 599 | 81 |
| 732 | 82 |
| 691 | 79 |
| 642 | 75 |
| 728 | 88 |
| 756 | 92 |
Calculate λ and r for both species.
Provide
To create these vectors, use the methods used for creating vectors for survivorship in the last homework. For example, define the the vector for the hare’s lambda values:
hare.lambda = c()then population it with values for each time step:
hare.lambda[1] = hare[2]/hare[1]
hare.lambda[2] = hare[3]/hare[2]Finally, ask R to recall the vector
hare.lambdaUsing both lambda and r, what are the average growth rates for the two species?
Which population is growing faster?
Hints: Note that the breeding scales are different. Thus, do you want to compare average or total growth?
You took over a program that manages a population of lion-tailed macaque (Macaca silenus) in the Western Ghats of South India. This primate primarily eats indigenous fruits, leaves, buds, insects and small vertebrates in virgin forest. These animals are disappearing due to loss of habitat. Let us suppose human settlement was withdrawn from a small area in the Western Ghats and, slowly, the number of trees, insects and small vertebrates began to increase. Your predecessor passed on 26 years of data and you have taken an additional three years of data.
| Year | N |
|---|---|
| 1968 | 266 |
| 1969 | 602 |
| 1971 | 756 |
| 1972 | 789 |
| 1974 | 1389 |
| 1976 | 1548 |
| 1979 | 1789 |
| 1980 | 1892 |
| 1982 | 1911 |
| 1983 | 2045 |
| 1984 | 2089 |
| 1985 | 2197 |
| 1986 | 2245 |
| 1988 | 2268 |
| 1990 | 2547 |
| 1993 | 3289 |
| 1995 | 3491 |
| 1997 | 3507 |
| 1999 | 3624 |
| 2000 | 4001 |
| 2004 | 4466 |
| 2005 | 4555 |
| 2007 | 4798 |
| 2009 | 4912 |
| 2010 | 4901 |
| 2011 | 5199 |
| 2013 | 5157 |
| 2014 | 5119 |
| 2015 | 5881 |
You will use the DA method to estimate average population growth over this 25-year period. Why use the DA method?
Showing your calculations, transform by hand your abundance and year values through the third year.
Now you will transform the entire data set. You can do that at once with a function for x. First, create a vector for year and a vector for abundance. Call them mac.year and mac.N, respectively. Now use the code below exactly as I’ve written it and you should get new vectors for abundance and year.
Here is the function for x:
x = c()
for (i in 1:length(mac.year)-1) {x = c(x, sqrt(mac.year[i+1]-mac.year[i]))}
xand the function for y:
lnL = c()
for (i in 1:length(mac.N)-1) {lnL = c(lnL, log(mac.N[i+1]/mac.N[i]))}
y = lnL/x
yUsing the following code to plot y against x.
plot(y~0+x,ylab="transformed population change", xlab = "transformed time
intervals")
abline(lm(y~0+x))Regress y onto x with:
summary(lm(y~0+x)) What is your estimated average growth rate?
What is your SE of the slope?
Is the population changing significantly? How do you know?
Using the t-table in the homework folder, calculate the 90% confidence interval around your estimated slope.
If you want to try to calculate your t value in R (vs pulling it from a table), use this code to calculate the 90% CI:
est = [you’ll fill this in from the output]
sterr = [you’ll fill this in from the output]
upper = (est + ((qt(.95, (length(x)-1))*sterr)))
upper
lower = (est - ((qt(.95, (length(x)-1))*sterr)))
lowerWhat does the confidence interval tell you about your estimate?
You are concerned that the establishment of an invasive species of snail has drastically altered carrying capacity and thus population growth of a local snail species. You estimate that the carrying capacity of this habitat for the invasive is 493 and the intrinsic rate of population growth is 4.910^{-4} (measured in months).
What is the equation for the logistic growth model for the invasive snail?
What is the abundance of the invasive snails when the population is growing the fastest?
What is the rate of change of the population when it is growing the fastest? Be sure to state your answer in terms of months.
When do you think would be the most effective time to intervene to halt the growth of the invasive snail population?
to calculate the natural log of x:
log(x)