Lesson 13: Line and Point Transects

In Section 13.1, we introduce line and point transects and the objectives of density estimation methods. Then in sections 13.2-13.4, we will discuss three types of density estimation methods for line transects problems. In section 13.2, we will talk about the narrow strip method. In section 13.3, we will discuss the smooth-by-eye method and in section 13.4, we will discuss parametric methods.

Lesson 13: Ch. 17.1-17.4 of Sampling by Steven Thompson, 3rd edition.

Objectives

Upon completion of this lesson you should be able to:

Apply the various density estimation methods for line transects to estimate population density,
Estimate the density of line transects using the narrow strip method,
Estimate the density of line transects using the smooth by eye method, and
Estimate the density of line transects using parametric methods.

13.1 - Density Estimation Methods for Line and Point Transects

Line transect sampling is usually used to sample objects for which detectability depends on location relative to the observer.

The objective is to estimate the density of objects in the study region.

Examples include birds, mammals, and plant species.

Here is a picture for line transect method.

In the following sections, we will introduce the Narrow-strip method, the smooth-by-eye method and parametric methods.

13.2 - Narrow-Strip Method

General Idea

The detectability of the objects usually becomes smaller as the distance from the transect line increases. However, there may be some narrow strip along the line in which detectability is virtually perfect. Therefore, we can use the observations within the narrow strip to do the estimation.

Notations

\(y_i\) - number of objects observed from the ith transect
\(n\) - number of transects selected
\(\tau\) - total number of objects in the study region
\(A\) - area of the study region
\(D=\tau /A\) - density of the objects -namely, number of objects per unit area
\(L\) - length of the transect
\(w_0\) - maximum distance from the line to which detectability is assumed perfect
\(2w_0\) - width of the strip
\(2w_{0}L\) - area of the strip
\(y_0\) - number of objects detected within the narrow strip

Formulas for Estimation

Density Estimation
\(\hat{D}=\frac{y_0}{2w_{0}L}\)
Estimation of Total Number of Objects in the Study Region
\(\hat{\tau}=A\hat{D}=\frac{Ay_0}{2w_{0}L} \)

Note!

There are several ways to choose \(w_0\), the perfect detection distance. We will illustrate one method in the example later.

Burnham et al. (1980, p. 33) suggest that the data should include at least 40 detections to provide a reliable estimation. We will have a smaller number just for illustrative purposes.

Example 13-1:

( Reference: text p. 231)

On a line transect of length L=100 meters, a total of y = 18 birds were detected at the following distances (in meters) from the transect line:

0, 0, 1, 3, 7, 11, 11, 12, 15,15, 18, 19, 21, 23, 28, 33, 34, 44

Please estimate the density of birds in the study region.

How to choose \(w_0\)?

From the data, we know that

5 birds were seen within 10 meters
7 birds were seen between 10 and 20 meters
3 birds were seen between 20 and 30 meters
2 birds were seen between 30 and 40 meters
1 bird was seen between 40 and 50 meters

We plot the above information in a histogram.

We can see from the above histogram that the relative frequency of observing the birds drops off sharply (from 7 to 3) after 20 meters from the transect line. Thus we choose \(w_0\) = 20.

Given \(w_0\) = 20, we know that \(y_0\) = 12

\(\hat{D} =\dfrac{y_0}{2w_{0}L} =\dfrac{12}{2(20)(100)}=0.003\)

So, the density estimate is 0.003 birds per square meter or 30 birds per hectare.

Limitation of Narrow-Strip Method

Not all observations obtained are used.
The determination of the width of the narrow strip seems somewhat subjective.
The detectability may, in fact, decrease smoothly with distance so that the narrow strip with perfect detectability really has a width of zero.

(We will introduce other line transect sampling methods in the following sections.)

13.3 - Smooth-by-Eye Method

Motivation of the Smooth-by-Eye Method

Recall that in 17.2, one main limitation of the Narrow-Strip Method is that it assumes the detectability of objects is perfect within the strip. But, in reality, the detectability decreases as the distance to the transect line increases.

In this section, the main idea of the Smooth-by-Eye Method is to use the histogram for distance x from the transect line to approximate the density of the detectability function f (x). We will approximate f (x) as a decreasing function, which agrees with the situation in real life.

How to Approximate the Detectability Function f (x)

The first step is to construct a histogram for the distance x from the transect line.

The formula for the height of the histogram for a given distance x is

\( \hat{f}(x)=\dfrac{y_x}{y \times w_x}\)

\(y_x\) - number of observations in the interval containing x
\(y\) - total number of observations
\(w_x\) - interval width

Example 13-2:

( Reference: Section 17.2 page 231 of the text)

y = 18 (total number of observations)

5 birds were seen within 10 meters
7 birds were seen between 10 and 20 meters

Choose the interval width of the histogram to be 10 meters.

Thus,

the height for the 1st interval is \(5/[18(10)] = 0.0028\)
the height for the 2nd interval is \(7/[18(10)] = 0.039\)

Similarly,

the height for the 3rd interval is 0.017
the height for the 4th interval is 0.011
the height for the 5th interval is 0.006

Knowing the heights of each interval, we got the following histogram:

Notice that the smooth-by-eye curve intersects the vertical axis at 0.048

Notice from the histogram that the detectability actually increases a little bit in the second interval compared with that in the first interval. However, the true density of detectability decreases smoothly with distance. We regard the increase in the irregularities in the histogram due to random chance and the small number of observations.

We still fit a smooth, decreasing curve irrespective of the irregularity. The curve is fitted by eyeballing and each person may have a slightly different fit.

According to the “height of histogram” formula given before, we know:

\(f_0=\dfrac{y_0}{y \times w_0} \)

Combining with the formula in 17.2:

\(\hat{D}=\dfrac{y_0}{2w_{0}L} \)

We got a new formula to estimate the density:

\( \hat{D}=\dfrac{\hat{f}(0) \times y}{2L}\)

\(\hat{f}(0)=0.048\) (the smooth-by-eye curve intersects the vertical axis at 0.048)

Thus,

\( \hat{D}=\dfrac{\hat{f}(0) \times y}{2L}=\dfrac{0.048 \times 18}{2 \times 100}=0.00432 \)

So, the estimated bird population density is 0.00432 birds per square meter or 43.2 birds per hectare.

13.4 - Parametric Methods

We will introduce a new concept here: Effective Half-Width of the Transect: \(\textbf{w}\)

Imagine an equivalent strip plot, with perfect detectability out to some distance \(\textbf{w}\), in which the same number of animals would be seen, on average, as are seen from the transect with decreasing detectability. We have:

\(f(0)=\dfrac{1}{w}\)

In terms of effective half-width, the density estimate based on an estimate of \(\textbf{w}\) is:

\(\hat{D}=\dfrac{y}{2L\hat{w}} \)

Given the above formula, one may proceed either to estimate \(f(0)\) or to estimate \(\textbf{w}\).

In the parametric method, we assumed that the distribution of the detection distance \(x\) will have the same shape as the detectability function.

There are numerous density functions \(g(x)\), we introduce two of them:

The exponential detectability function:
\(g(x)=\text{exp}(-x/w)\)

The corresponding maximum likelihood estimator for \(\textbf{w}\):

\(\hat{w}=\bar{x}\)
The half-normal detectability function:
\(g(x)=\text{exp}\left( \dfrac{-\pi x^2}{4w^2} \right)\)

The corresponding maximum likelihood estimate for w:

\(\hat{w}=\sqrt{\dfrac{\pi}{2y}\sum_{i=1}^{y}x_{i}^{2}}\)

Example of Exponential Detectability Function

(with the data from the bird example)

\(\bar{x}=(0+0+1+ \cdots + 44)/18=16.39\)

\(\hat{D}=\frac{y}{2L\hat{w}}=\frac{18}{2 \times 100 \times 16.39}=0.055\)

Example of Half-Normal Detectability

(with the data from the bird example)

\((1/n)\Sigma x_{i}^{2}=(1/18)(0^2 + \cdots + 44^2)=417.5\)

\(\hat{w}=\sqrt{\frac{\pi}{2y}\sum_{i=1}^{y}x_{i}^{2}}=\sqrt{\left(\frac{3.1417}{2}\right)(417.5)}=25.61\)

\(\hat{D}=\frac{y}{2L\hat{w}}=\frac{18}{2 \times 100 \times 25.61}=0.0035\)

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