NRES 488 HW 1
SOLUTIONS
1. Remember that the estimate of the proportion successes will just be the average
proportion success across the 500 trials. Calculate the average for each column
(50 trials), using the average function. Calculate the variance of the proportion
successes for each column using the variance function. Remember that E(x/n) =
, var(x/n)= (1-)/n. Thus, an estimate of n is given by:
n
 
θ̂ 1 - θ̂
.
x
var  
n
Estimate θ̂ from the mean value for each column. I estimated the overall mean θ̂ =
0.62. Mean variance (across columns) = 0.0094. Mean n = 26.1. I have plotted the
distribution for θ for all 500 trials below.
Frequency
120
100
80
60
Frequency
40
20
0
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9 More
For sheet 2 the calculations are:
Mean θ̂ = 0.75, mean variance = 0.004, mean n = 51.6. The histogram below shows
the distribution for θ across all 450 trials.
Frequency
180
160
140
120
100
Frequency
80
60
40
20
0
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9 More
The estimates of the mean variance indicate that trials with 52 tosses had lower
variance than those with 26 tosses. Lower variance is also apparent in the distribution
of the θ s from trials on sheet 2 versus those on page 1. You can access my version
of the spreadsheet by following the link spreadhseetHW1.
2. Using notation from lecture,
M = 100
C = 80
m = 30.
Our estimate of N̂ 
MC 100  80

 267
m
30
95% CL for the estimate of m/C is given by:
 
 
m
m
m   1  f  C 1  C
  z
C  
C 1
 

1 
 2C 

where f = m/M, z = 1.96. I calculate the result to be 0.375  0.0956.
3. In the formulation of the Lincoln-Petersen estimator shown relies on the
assumption that the population is closed (no individuals enter or leave the
population), so the proportion marked at the time of the original capture equals
N̂ C
 .
that at the time of recapture. Thus:
M m
Alternatively, we can use the data to estimate the capture (encounter) probability.
Our estimate of capture probability from the data is:
p̂ 
m
.
M
The number captured in the second sample is C, while the probability of capturing an
individual is p̂ . Thus our estimate of population size:
N̂ 

C
C

p̂ m/M
N̂ C

M m