In this practical, we plot a histogram of line transect data and estimate a detection function. The data were collected during a line transect survey of duck nests in Monte Vista National Wildlife Refuge, Colorado, USA: twenty lines of 128.75 km were specified and a distance out to 2.4m was searched and the perpendicular distances of detected nests were recorded and summarised (Table 1).
Distance.band | Frequency |
---|---|
0.0-0.3 | 74 |
0.3-0.6 | 73 |
0.6-0.9 | 79 |
0.9-1.2 | 66 |
1.2-1.5 | 78 |
1.5-1.8 | 58 |
1.8-2.1 | 52 |
2.1-2.4 | 54 |
The aim of this exercise is to plot a histogram of the perpendicular distances to the detected duck nests and estimate (by eye) a detection function and hence estimate density of duck nests, i.e. the number of nests per square metre or per square kilometre (be careful of units). Work through the following steps:
\[ Area_{rectangle} = \] \[ Area_{curve} = \]
\[ \hat{P}_a = \frac{Area_{curve}}{Area_{rectangle}} = \]
\[ \hat{N}_a = \frac{n}{\hat{P}_a} = \]
\[\hat{D} = \frac{\hat{N}_a}{a} = \frac{\hat{N}_a}{2wL} = \]
\[ \hat \mu = \]
\[\hat{D} = \frac{n}{2\hat \mu L} = \]
How does it compare to your estimate from part 4?
\[\hat{D} = \frac{n \hat f(0)}{2L} = \]
How does it compare with your previous estimates?
Solution 1. Line transect detection functions by hand
\[ Area_{curve} = (75+74+72+70+66+62+58+53) \times 0.3 = 530 \times 0.3 = 159 \] There are lots of other ways to work out the area under a curve, e.g. counting the number of grid squares under the curve on your graph paper or using the trapezoidal rule.
\[Area_{rectangle} = height \times width = 75 \times 2.4 = 180\]
Hence, my estimate of the proportion of nests detected in the covered region is:
\[\hat P_a = \frac{159}{180} = 0.883\]
\[ \hat N_a = \frac{n}{\hat P_a} =\frac{534}{0.883} = 604.7 \textrm{ nests in the covered area}\] This estimate is for a covered area of \(a = 2wL = 2 \times (\frac{2.4}{1000}) \times 2575 = 12.36\) km\(^2\).
\[\hat D = \frac{\hat N_a}{2wL} = \frac{604.7}{12.36} = 48.9 \textrm{ nests per km}^2\]
The green vertical dashed line shows my estimated effective strip half-width of 2.13 metres; I estimate that the area below my curve to the right of 2.13 is the same as the area above the curve to the left of 2.13.
Using this estimate of the effective strip half-width, the effective area covered is estimated as \(2\mu L = 2 \times \frac{2.13}{1000} \times 2575 = 10.97\) km\(^2\). Therefore,
\[\hat{D} = \frac{n}{2\hat \mu L} = \frac{534}{10.97} = 48.7 \textrm{ nests per km}^2\]
Substituting this into the formula gives a density estimate of:
\[\hat{D} = \frac{n \hat f(0)}{2L} = \frac{534 \times (0.472 \times 1000)}{2 \times 2575} = \frac{252048}{5750} = 48.9 \textrm{ nests per km}^2\]
(Note, I had to multiply \(f(0)\) by 1,000 to convert from m\(^{-1}\) to km\(^{-1}\).)
Another way to estimate \(f(0)\) is \(f(0)=1/\mu\): using this method I’d get the same density estimate as in Part 5.