The Spotted Handfish 19992001 Recovery Plan
BD Bruce and MA Green
Spotted Handfish Recovery Team, March 1998
ISBN 0 643 061657
Appendix A
Power of tests for changes in density of handfish
For Mark Green and Barry Bruce, CSIRO Division of Marine Research, Hobart
By Kathy Haskard, CSIRO Mathematical and Information Sciences
Background
You have performed a survey in a small specified area near Hobart in order to estimate the density and abundance of handfish (B. hirsutus). You will repeat this survey, possibly annually, or more often if this seems desirable, in order to examine whether the density changes. With the information so far obtained, you wish me to estimate power for detecting changes in density.
Survey Design
Following previous discussions, your survey design identifies an approximately rectangular region, with one long side at approximately the 5 m depth contour, and extending out at right angles to near or beyond the 10 m contour. Twentyfive equally spaced parallel transect lines were described, running across the short dimension of the rectangle, so running from depth 5 m to 10 m. For an individual survey 12 of these were selected at random, and a strip 2 m wide, centred on each selected transect, was examined by two divers swimming together, each carrying a 1 m pole to delimit their search area. Handfish seen greater than 1 m either side of the transect line were not included in the estimates.
Rather than covering a true rectangle, transects were defined to be between the 5 m depth contour, or beyond this (deeper than 5 m) if a rocky reef is present, since this is believed to be unsuitable habitat for handfish, and the 10 m depth contour, so that the area covered by the survey is not a true rectangle, and the transects have unequal lengths.
For the purpose of calculating an index of abundance, it would be best to examine the same transects at each repeated survey. However for simply obtaining the best overall estimate of abundance (for example if the density did not change over time) it is desirable to randomly select the transects separately each survey, so that a bigger part of the region is ultimately examined. The procedure used here is a compromise between these two. There will be partial overlap between surveys.
Estimated density and standard error
From a previous consultation, I provided you with the following formulae. Because the twelve individual transects are of different lengths, I recommended a weighted average, with the transect lengths as weights.
Suppose there are k transects, transect i has length l_{i} metres, and n_{i} handfish are observed on transect i, i=1...k. You wish to estimate D = average density of handfish in the survey area. Let L be the total of the lengths of the k (=12) transects. Each transect i covers area 2l_{i} square metres, and the total area sampled is 2L. The estimate of D is the weighted average
where N = total number of handfish seen, , and .
The estimated variance of is
The standard error (SE) of is simply the square root of the variance.
This will give estimated density in units of numbers per square metre, which will be very small numbers to work with. You might prefer to use, for example, numbers per 100 m^{2}. To do this, simply multiply the estimated density and its SE by 100, and multiply variance by 10,000.
Your question
Having performed the survey once and done the above calculations, you have asked me to estimate the power of tests to determine if handfish abundance has significantly changed from one time to another. As I explained at our meeting, for a particular statistical test, power depends on several interrelated things:
 the significance level at which the test will be declared significant, commonly 0.05
 the variability within the data, e.g. between transects
 sample sizes, in this case number of transects or total area of transects
 the size of the true difference
 the probability with which a test will be significant given a difference of that size  this is the power of the test. Power of around 80% or more is generally regarded as useful.
Less variability, greater sample sizes, and larger true differences all increase the power of a test, as does a larger significance level a, e.g. a=0.10. We will consider only a=0.05. The more variable data are, the greater the sample size will need to be, or the larger the underlying difference would have to be, to achieve a given power. This is because with more variability in the data, we need more evidence, or a bigger effect, to have reasonable confidence that an observed difference is not simply due to chance.
What test will we use?
Before we can estimate power, we must decide what statistical test will be used. If data can be assumed to approximately follow a normal distribution, this is usually straightforward.. However, the data from your first survey using this design reveal very low numbers of handfish, between 0 and 3 per transect, with a total of 15 handfish from the 12 transects of total length 1506 m, covering area 3012 m^{2}, giving estimated density of 0.00498 handfish m^{2} with variance 1.412*10^{6}, and SE 0.001188 handfish m^{2}.
Converting to more convenient, though unconventional, units, this is 49.8 handfish per hectare (10000 m^{2}), with variance 141.2 and SE 11.88.
The study area to which this applies has size 66,270 m^{2}, giving an estimated abundance of 330 handfish, with estimated standard error 70.
With such small numbers, we are counting occurrences of a rare event, and the Poisson distribution may be appropriate here. This assumes that the fish are essentially independently located, with no clustering nor the opposite which would lead to more uniform spreading out of the individuals. This appears to be a reasonable assumption in this case, and it enables us to form a test.
We assume that the number observed on a given transect has a Poisson distribution with mean equal to the overall density times the area of that transect, i.e. 2l_{i}D. Assuming each transect is independent, this would imply the total number of handfish N would have a Poisson distribution with mean 2LD. A property of the Poisson distribution is that the variance is equal to the mean, i.e. var(N) should equal 2LD, and the variance of the density estimate =N/(2L) would be var(N)/(2L)^{2} = D/(2L). Using our estimate = 0.00498 we obtain =1.653*10^{6} as the estimated variance of , compared with 1.412*10^{6} calculated directly from the weighted average, without using the Poisson assumption. These are very consistent, and support the Poisson assumption.
Thus, I proceeded on the assumption that the total number N of handfish seen in traversing transects of lengths totalling L follows a Poisson distribution with mean 2LD. We now must devise a procedure to test whether two surveys suggest a difference in density D between them. Two different approaches lead to the same test.
Derivations of test
One approach is to simply estimate the expected number of handfish seen in each of two surveys, given the total of the two surveys, proportionately according to the total transect lengths in each, and compare these to the actual numbers in each survey, using a wellknown chisquare test,
where the sum is over two cells corresponding to the two surveys. Provided the two expected values are not too small (greater than 5 is sufficient) X^{2} has approximately a c^{2} distribution on 1 df.
Given total numbers N_{1} and N_{2} seen in the two surveys respectively, and total transect lengths L_{1} and L_{2} in the two surveys respectively, and writing N = N_{1}+N_{2} and L=L_{1}+L_{2}, under the null hypothesis that both densities are the same the estimated common density is . Then, given the total N, the expected number seen for survey 1 is , and for survey 2 the expected number is . (Observedexpected)^{2} is the same for both cells,
and
and in the special case when L_{1} = L_{2}, this simplifies to
For example, suppose the next survey found a total of 10 handfish in transects totalling 2000 m in length, giving an estimated density of 10/(2*2000) = 0.0025 handfish m^{2}, half that of survey 1. Would we conclude the density had changed from the previous survey with estimated density 15/(2*1506) = 0.00498? Given a total of 15+10=25 handfish, and total transect lengths 1506 and 2000 m, totalling 3506 m, if the densities were the same we would have expected to observe 25*1506/3506 = 10.74 handfish in the first survey and 25*2000/3506 = 14.26 handfish in the second survey. Compared with the observed values of 15and 10 respectively, we obtain X^{2} = 4.26^{2}/10.74 + 4.26^{2}/14.26 = 2.964 and comparing this to a chisquare distribution with 1 df, we obtain p = 0.085, not significant at the 5% level.
The alternative approach uses the fact that a Poisson distribution with sufficiently large mean can be approximated by a normal distribution with mean and variance equal to the Poisson mean. So writing N_{1} and N_{2} for the total numbers seen in surveys 1 and 2 respectively, and L_{1} and L_{2} for the total length of transects in the two surveys respectively, we have
N_{1} ~ approximately N(2L_{1}D_{1}, 2L_{1}D_{1}),
N_{2} ~ approximately N(2L_{2}D_{2}, 2L_{2}D_{2})
and hence
~ approximately N(D_{1}, D_{1}/ (2L_{1}) ),
~ approximately N(D_{2}, D_{2}/ (2L_{2}) ),
and
~ approximately N(D_{1}  D_{2}, ),
which provides an approximate test based on the normal distribution, namely that under the null hypothesis H_{0}: D_{1} = D_{2} = D,
~ approximately N(0 , ),
~ approximately N(0 , 1 ).
This would be our ideal test statistic. However, we do not know the true D, and we substitute our best estimate , giving our test statistic
~ approximately N(0 , 1 ) (a poorer approximation than for Z).
Now the square of standard unit normal variable has a chisquare distribution, and the square of T is
which can be shown to be algebraically equal to X^{2} in the first approach above (my notes, 28/10/97, workbook p. 55), with the approximate c^{2} distribution with 1 d.f. Both these expressions can be written more directly for calculation, based on numbers of handfish seen and transect lengths, as
Note that if L_{1} = L_{2}, this simplifies even further to
and the corresponding Tstatistic becomes
The test
To test H_{0}: D_{1} = D_{2} against the alternative hypothesis H_{1}:D_{1} ¹ D_{2}, calculate
and compare with a c^{2} distribution with 1 d.f., i.e. declare significant at the 5% level if X^{2} is greater than 3.84. This gives a twosided test, i.e. you are looking for differences and don't care whether they are increases or decreases in density.
For a onesided test, e.g. if you don't believe density would ever increase (could be a dangerous assumption), and are only concerned with detecting a decrease, test H_{0}: D_{1} £ D_{2} against the alternative hypothesis H_{1}:D_{1} > D_{2}, calculate
and compare to a standard normal distribution, rejecting H_{0} only if T is large and positive. That is, reject H_{0} in favour of H_{1} (conclude that the density has decreased from survey 1 to survey 2) at the 5% level if T is greater than 1.645. This will be a stronger test if you are happy to neglect the possibility of increases in density.
If you reject when T is bigger than 1.96 or smaller than 1.96, then this is equivalent to the twosided test above (because X^{2} = T^{2} and 3.84 = 1.96^{2}).
Calculating power
We ask: for a given change in density, and using the information we have from the first survey which gives us some information about variability of these data, what is the probability of rejecting the null hypothesis that the densities are equal at each time?
Some further assumptions must be made. In particular, we don't know in advance what the total length of transects in the next survey will be. We would expect it to be similar to L_{1}, and calculate the power on this basis. We must also decide how we will specify a change in density. It is convenient to express the D_{2} as a fraction or multiplier of D_{1}. I.e. suppose D_{2} = c D_{1}, where c ³ 0 and c = 1 for no change, c 1 for an increase.
The question is: if D_{2} = c D_{1}, for a specified value of c, what is the probability that X^{2} will be greater than 3.84?
It is convenient to work through the theory again, beginning with the assumed Poisson distributions for N_{1} and N_{2} and using L_{1} = L_{2}. We assume that N_{1} ~ Poisson with mean m_{1} and N_{2} ~ Poisson with mean m_{2}. Since L_{1}=L_{2}, m_{1} = 2L_{1}D_{1} and m_{2} = 2L_{1}D_{2} = 2L_{1}cD_{1} = c m_{1}. Then we say
N_{1} ~ approximately N(m_{1}, m_{1}) and N_{2} ~ approximately N(c m_{1}, c m_{1}) and hence
N_{1}  N_{2} ~ approximately N(m_{1}(1c), m_{1}(1+c) ), and
~ approximately normally distributed with mean and variance 1.
Z is not exactly the test statistic used, rather we use T which has N_{1} +N_{2} in the denominator as an approximation to m_{1}(1+c). However, it is difficult or impossible to obtain the distribution of T, so for calculating power we use this 'idealised' test statistic Z. Further, in the expression for the mean of Z, we see m_{1}, which we do not know , so we calculate power using N_{1} as a substitute for m_{1} in the mean, another approximation.
This enables us to calculate the power for any specified value of c.
Probability of rejecting H_{0} is
= probability that T > 1.96 or T
@ probability that Z > 1.96 or Z
= probability that or
= probability that a N(0,1) variable is or
@ probability that a N(0,1) variable is or
Power results
We can calculate these probabilities for a range of values of c. I have done this in a Microsoft Excel spreadsheet which I can provide to you. Some results are presented in Table 1. The formulas are shown in Table 2. All tests are assumed to be at the 5% significance level.
Recall that c = 1 means no change in true density between survey 1 and survey 2; c = 0.5 means density in survey 2 was half that of survey 1; and c = 1.333 means the density increased by one third. When c = 1 the power is 0.05, as we would expect with a test at the 5% significance level  if there is truly no difference, there is a 5% chance of declaring a significant difference.
Table 1. Power of twosample 5%level test of equal densities vs densities not equal, and for onesided 5% tests. We assume that D_{2} = cD_{1}, and m_{1} is the expected number of handfish seen in survey 1. For this table m_{1} is set to the estimate 15, the number observed in the first survey.
For the twosided test, upper tail is power when , , and lower tail is power when , . Total power is the sum of these two.
Two columns show power for the onesided test. The first applies when density decreases between surveys 1 and 2 (0 £ c , and the second is for onesided tests when the density increases between surveys 1 and 2 (c > 1), .
In all these expressions, Z refers to a standard unit normal variable, and all tests are assumed to be at the 5% significance level. Note that when there is no difference in density (c = 1, shaded cells), power equals significance level, 0.05.
m_{1} = 15 
Twosided 
test 
Onesided 
tests 

c 
Upper tail 
Lower tail 
Total power 
Testing vs D_{2}1 
Testing vs D_{2} > D_{1} 
0 
0.972 
0.000 
0.972 
0.987 
0.000 
0.1 
0.887 
0.000 
0.887 
0.936 
0.000 
0.2 
0.733 
0.000 
0.733 
0.826 
0.000 
0.3 
0.550 
0.000 
0.550 
0.670 
0.000 
0.4 
0.382 
0.000 
0.382 
0.506 
0.000 
0.5 
0.252 
0.001 
0.252 
0.362 
0.002 
0.6 
0.161 
0.002 
0.162 
0.249 
0.004 
0.7 
0.101 
0.004 
0.105 
0.168 
0.010 
0.8 
0.063 
0.008 
0.071 
0.112 
0.019 
0.9 
0.040 
0.015 
0.055 
0.075 
0.032 
1 
0.025 
0.025 
0.050 
0.050 
0.050 
1.1 
0.016 
0.038 
0.054 
0.034 
0.072 
1.2 
0.010 
0.054 
0.064 
0.023 
0.098 
1.3 
0.007 
0.073 
0.080 
0.016 
0.127 
1.4 
0.005 
0.094 
0.099 
0.011 
0.159 
1.5 
0.003 
0.118 
0.121 
0.008 
0.192 
1.6 
0.002 
0.143 
0.145 
0.006 
0.226 
1.7 
0.002 
0.170 
0.171 
0.004 
0.261 
1.8 
0.001 
0.197 
0.198 
0.003 
0.295 
1.9 
0.001 
0.224 
0.225 
0.002 
0.329 
2 
0.001 
0.252 
0.252 
0.002 
0.362 
2.1 
0.000 
0.279 
0.279 
0.001 
0.393 
2.2 
0.000 
0.306 
0.306 
0.001 
0.424 
2.3 
0.000 
0.332 
0.332 
0.001 
0.453 
2.4 
0.000 
0.357 
0.358 
0.001 
0.480 
2.5 
0.000 
0.382 
0.382 
0.000 
0.506 
2.6 
0.000 
0.406 
0.406 
0.000 
0.530 
2.7 
0.000 
0.428 
0.428 
0.000 
0.553 
2.8 
0.000 
0.450 
0.450 
0.000 
0.575 
2.9 
0.000 
0.471 
0.471 
0.000 
0.596 
3 
0.000 
0.491 
0.491 
0.000 
0.615 
3.1 
0.000 
0.509 
0.510 
0.000 
0.633 
3.2 
0.000 
0.527 
0.527 
0.000 
0.649 
3.3 
0.000 
0.544 
0.544 
0.000 
0.665 
3.4 
0.000 
0.561 
0.561 
0.000 
0.680 
3.5 
0.000 
0.576 
0.576 
0.000 
0.694 
3.6 
0.000 
0.591 
0.591 
0.000 
0.707 
3.7 
0.000 
0.604 
0.604 
0.000 
0.719 
3.8 
0.000 
0.618 
0.618 
0.000 
0.730 
3.9 
0.000 
0.630 
0.630 
0.000 
0.741 
4 
0.000 
0.642 
0.642 
0.000 
0.751 
5 
0.000 
0.733 
0.733 
0.000 
0.826 
6 
0.000 
0.790 
0.790 
0.000 
0.869 
7 
0.000 
0.828 
0.828 
0.000 
0.896 
Table 2. Microsoft Excel formulas used for Table 1.
Col Row 
A 
B 
C 
D 
E 
F 
G 
1 
m_{1}= 
15 
Twosided 
test 

2 
c 
Upper tail 
Lower tail 
Total Power 

3 
0.5 
=1NORMSDIST(1.96(1B3)*SQRT(B$1)/(1+B3)) 
=NORMSDIST(1.96(1B3)*SQRT(B$1)/(1+B3)) 
=SUM(C3:D3) 
.. 
.. 

4 
2.0 
=1NORMSDIST(1.96(1B4)*SQRT(B$1)/(1+B4)) 
=NORMSDIST(1.96(1B4)*SQRT(B$1)/(1+B4)) 
=SUM(C4:D4) 
Col Row 
A 
B 
C 
D 
E 
F 
G 
1 
m_{1}= 
15 
Onesided 
tests 

2 
c 
Testing D_{2} = D_{1} vs D_{2}1 
Testing D_{2} = D_{1} vs D_{2} > D_{1} 

3 
0.5 
.. 
.. 
.. 
=1NORMSDIST(1.645(1B3)*SQRT(B$1)/(1+B3)) 
=NORMSDIST(1.645(1B3)*SQRT(B$1)/(1+B3)) 

4 
2.0 
=1NORMSDIST(1.645(1B4)*SQRT(B$1)/(1+B4)) 
=NORMSDIST(1.645(1B4)*SQRT(B$1)/(1+B4)) 
Table 1 also shows power for onesided tests. The first column is for a test of H_{0}: D_{2} ³ D_{1} vs H_{1}: D_{2}1 (sensible if you only expect density to decrease, i.e. only expect values of c £ 1) and the second is for tests of H_{0}: D_{2} £ D_{1} vs H_{1}: D_{2} > D_{1} (sensible if you only expect density to increase, i.e. only expect values of c ³ 1).
Discussion
You can see that the power is quite low unless there are massive changes in density. This is partly because so few handfish were observed. With such small numbers, seeing just a few individuals less or more could happen quite easily, but could produce quite different density estimates. In other words, the density estimates are not very precisely estimated with such small numbers of handfish seen. If you don't know densities very precisely, you can't be very confident that two densities are different; thus power is low.
It is commonly accepted that powers of 80% or more are desirable. To reach this power with the current level of sampling (area covered by transects), your density would have to decrease to less than one sixth of the density at survey 1, or increase more than 6fold.
To increase power you would need to increase the area you sample. Table 3 shows some other examples of power if m_{1} was 30 (obtained by approximately doubling the area covered by transects), 45, 60, 75, 90 or 105, rather than 15. Clearly very much more effort is required to substantially improve the power, i.e. so that moderate changes in density have a good chance of being detected. Even to have a good chance of detecting a 50% change (c=0.5 or c=2) in density from its current estimated value, you would require about 5 times the sampling area, to get m_{1}, the expected number of individuals seen, = 75.
Bear in mind that these power calculations are only approximate. Although you observed N_{1} = 15 in the first survey, the true density and hence m_{1} may have been larger or smaller than indicated by this. If larger, the power will be a little greater, if smaller the power will be even less.
Table 3. The same information as Table 1, but with larger values of m_{1}.
m_{1} = 30 
Twosided 
test 
Onesided 
tests 

c 
Upper tail 
Lower tail 
Total power 
Testing vs D_{2}1 
Testing vs D_{2} > D_{1} 
0 
1.000 
0.000 
1.000 
1.000 
0.000 
0.1 
0.994 
0.000 
0.994 
0.998 
0.000 
0.2 
0.955 
0.000 
0.955 
0.978 
0.000 
0.3 
0.839 
0.000 
0.839 
0.904 
0.000 
0.4 
0.651 
0.000 
0.651 
0.759 
0.000 
0.5 
0.447 
0.000 
0.447 
0.572 
0.000 
0.6 
0.277 
0.000 
0.278 
0.391 
0.001 
0.7 
0.160 
0.002 
0.162 
0.249 
0.005 
0.8 
0.088 
0.005 
0.093 
0.150 
0.012 
0.9 
0.047 
0.012 
0.060 
0.087 
0.027 
1 
0.025 
0.025 
0.050 
0.050 
0.050 
1.2 
0.007 
0.072 
0.079 
0.016 
0.126 
1.4 
0.002 
0.148 
0.150 
0.005 
0.232 
1.6 
0.001 
0.243 
0.244 
0.002 
0.352 
1.8 
0.000 
0.346 
0.347 
0.001 
0.468 
2 
0.000 
0.447 
0.447 
0.000 
0.572 
2.2 
0.000 
0.537 
0.537 
0.000 
0.659 
2.4 
0.000 
0.616 
0.616 
0.000 
0.729 
2.6 
0.000 
0.682 
0.682 
0.000 
0.785 
2.8 
0.000 
0.737 
0.737 
0.000 
0.829 
3 
0.000 
0.782 
0.782 
0.000 
0.863 
3.5 
0.000 
0.861 
0.861 
0.000 
0.919 
4 
0.000 
0.908 
0.908 
0.000 
0.950 
m_{1} = 45 
Twosided 
test 
Onesided 
tests 

c 
Upper tail 
Lower tail 
Total power 
Testing vs D_{2}1 
Testing vs D_{2} > D_{1} 
0 
1.000 
0.000 
1.000 
1.000 
0.000 
0.1 
1.000 
0.000 
1.000 
1.000 
0.000 
0.2 
0.994 
0.000 
0.994 
0.998 
0.000 
0.3 
0.951 
0.000 
0.951 
0.975 
0.000 
0.4 
0.820 
0.000 
0.820 
0.891 
0.000 
0.5 
0.609 
0.000 
0.609 
0.723 
0.000 
0.6 
0.389 
0.000 
0.389 
0.513 
0.000 
0.7 
0.219 
0.001 
0.220 
0.322 
0.002 
0.8 
0.112 
0.003 
0.116 
0.184 
0.008 
0.9 
0.054 
0.010 
0.064 
0.098 
0.023 
1 
0.025 
0.025 
0.050 
0.050 
0.050 
1.2 
0.005 
0.088 
0.094 
0.012 
0.150 
1.4 
0.001 
0.200 
0.201 
0.003 
0.299 
1.6 
0.000 
0.340 
0.340 
0.001 
0.461 
1.8 
0.000 
0.483 
0.483 
0.000 
0.607 
2 
0.000 
0.609 
0.609 
0.000 
0.723 
2.2 
0.000 
0.711 
0.711 
0.000 
0.808 
2.4 
0.000 
0.789 
0.789 
0.000 
0.868 
2.6 
0.000 
0.846 
0.846 
0.000 
0.909 
2.8 
0.000 
0.888 
0.888 
0.000 
0.937 
3 
0.000 
0.918 
0.918 
0.000 
0.956 
3.5 
0.000 
0.961 
0.961 
0.000 
0.981 
4 
0.000 
0.981 
0.981 
0.000 
0.991 
Table 3 continued. Power for two and onesided tests at 5% significance level, with different expected number m_{1} of handfish seen in survey 1.
m_{1} = 60 
m_{1} = 75 

c 
Twosided 
One 
sided 
c 
Twosided 
One 
sided 
0 
1.000 
1.000 
0.000 
0 
1.000 
1.000 
0.000 
0.1 
1.000 
1.000 
0.000 
0.1 
1.000 
1.000 
0.000 
0.2 
0.999 
1.000 
0.000 
0.2 
1.000 
1.000 
0.000 
0.3 
0.986 
0.994 
0.000 
0.3 
0.997 
0.999 
0.000 
0.4 
0.913 
0.953 
0.000 
0.4 
0.960 
0.981 
0.000 
0.5 
0.733 
0.826 
0.000 
0.5 
0.823 
0.893 
0.000 
0.6 
0.491 
0.615 
0.000 
0.6 
0.581 
0.698 
0.000 
0.7 
0.277 
0.390 
0.001 
0.7 
0.333 
0.454 
0.001 
0.8 
0.138 
0.216 
0.006 
0.8 
0.161 
0.247 
0.005 
0.9 
0.069 
0.108 
0.020 
0.9 
0.074 
0.117 
0.018 
1 
0.050 
0.050 
0.050 
1 
0.050 
0.050 
0.050 
1.2 
0.108 
0.009 
0.173 
1.2 
0.123 
0.008 
0.196 
1.4 
0.252 
0.002 
0.362 
1.4 
0.303 
0.001 
0.420 
1.6 
0.432 
0.000 
0.557 
1.6 
0.515 
0.000 
0.638 
1.8 
0.600 
0.000 
0.715 
1.8 
0.697 
0.000 
0.797 
2 
0.733 
0.000 
0.826 
2 
0.823 
0.000 
0.893 
m_{1} = 90 
m_{1} = 105 

c 
Twosided 
One 
sided 
c 
Twosided 
One 
sided 
0 
1.000 
1.000 
0.000 
0 
1.000 
1.000 
0.000 
0.1 
1.000 
1.000 
0.000 
0.1 
1.000 
1.000 
0.000 
0.2 
1.000 
1.000 
0.000 
0.2 
1.000 
1.000 
0.000 
0.3 
0.999 
1.000 
0.000 
0.3 
1.000 
1.000 
0.000 
0.4 
0.982 
0.992 
0.000 
0.4 
0.992 
0.997 
0.000 
0.5 
0.885 
0.935 
0.000 
0.5 
0.927 
0.962 
0.000 
0.6 
0.660 
0.766 
0.000 
0.6 
0.726 
0.820 
0.000 
0.7 
0.388 
0.512 
0.000 
0.7 
0.440 
0.565 
0.000 
0.8 
0.184 
0.277 
0.003 
0.8 
0.207 
0.306 
0.003 
0.9 
0.079 
0.126 
0.016 
0.9 
0.084 
0.134 
0.014 
1 
0.050 
0.050 
0.050 
1 
0.050 
0.050 
0.050 
1.2 
0.139 
0.006 
0.217 
1.2 
0.154 
0.005 
0.238 
1.4 
0.353 
0.001 
0.475 
1.4 
0.401 
0.000 
0.525 
1.6 
0.591 
0.000 
0.707 
1.6 
0.657 
0.000 
0.764 
1.8 
0.774 
0.000 
0.857 
1.8 
0.833 
0.000 
0.900 
2 
0.885 
0.000 
0.935 
2 
0.927 
0.000 
0.962 
Confidence intervals for density and abundance
Another indication of the degree of precision with these low numbers of handfish is given by obtaining a confidence interval for the survey 1 density. We can use either (a) a normal approximation based on the Poisson distribution for N_{1}, CI for m_{1} is N_{1} ± 1.96Ö N_{1}, and D = m_{1}/2L_{1}, or (b) a tdistribution with 11 degrees of freedom, using the weighted mean of the twelve transect densities and the corresponding variance, which does not require the Poisson assumption CI for D is . Both approaches give very similar results. Recall estimated mean density was 0.00498; estimated abundance was 330.
Approximate 95% confidence interval for density at the time of survey 1:
by method (a)0.00246 to 0.00750
by method (b) 0.00236 to 0.00760
Approximate 95% confidence interval for abundance in the survey region of 66,270 m^{2}, at the time of survey 1:
by method (a)163 to 497
by method (b) 157 to 503
Exact Confidence intervals for a Poisson mean
Exact confidence intervals for the mean of a Poisson distribution, given a single observation N, can be obtained very easily using a link between the c^{2} distribution and the Poisson distribution, which says that, if N is distributed as a Poisson variable with mean m, then for positive integers x,
Pr(N ) = Pr(a c^{2} variable with 2x degrees of freedom is > 2m).
Reference: Armitage, P. and Berry, G. (1987) Statistical methods in medical research, 2^{nd} edition. Blackwell, Oxford.
Then a 100(1 a)% confidence interval for the mean of a Poisson distribution, given a single observation N, is between
m_{L} = (1/2)*(lower 100a/2 % point of c^{2}(2N) distribution)
m_{U} = (1/2)*(upper 100a/2 % point of c^{2}(2N+2) distribution)
The formulas for use in Microsoft Excel, for 95% confidence intervals, i.e. a = 0.05, are
m_{L} = CHIINV(0.975,2*N)/2
m_{U} = CHIINV(0.025,2*N+2)/2
(Note that the CHIINV function gives upper 'tails' of the c^{2} distribution.)
On the next page are calculated values for a range of possible N values, with corresponding confidence intervals for the abundance, based on area sampled = 2*1506 m^{2} and total area 66270 m^{2}. I have also shown the percentage below and above the estimated mean for this exact method (e) and method (a) of the previous section.
Again you can see that quite large numbers are required to obtain a reasonably narrow confidence interval. You may wish to experiment with using 90% confidence intervals (a =0.10).
Provided the Poisson assumption is reasonable, these exact confidence intervals are best, although there is little difference between all three in this case. If we do not wish to assume the Poisson distribution, we cannot so easily predict the width of confidence intervals for future surveys with different N, because we need to assume something about how the variance of N, the number observed, changes as N (or, strictly, its expected value or mean) changes. That is, for confidence intervals by method (b) we cannot produce a table such as given overleaf for methods (e) and (a), without making some extra assumption.
We certainly cannot assume that the variance of N would be constant if the mean of N changes, and simply use the estimated variance of 1.412 x 10^{6} for estimated density. We would expect the variance of N to increase for larger expected values of N. The Poisson assumption automatically answers this question for us; it says the variance of N is proportional (in fact equal) to the mean of N. This means that the standard error as a proportion of the expected mean of N would decrease as N increased, by the square root, SE(N)/E[N] = , so that with larger N, we get narrower confidence intervals, and hence more precise estimates, relative to the size of the mean.