Modern Wild Faunal Resource System Management
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We'd expect n = f (1/d). The greater the variance (s2) the more the samples needed. The greater the confidence you must have in the answer (related by your choice of the alpha level and a value for t) the more samples needed. The relationships are:
n = s2 t2 / d2.
Where an average has been found (for a test situation) of 12 and a variance of 16, and when the wildlife manager decides that t can be about 1.7 for large samples and alpha of 0.10, and accuracy level can be plus or minus 0.05, then
n = (16) (1.72) / ((12) (0.05))2 =128.
In wildlife work it may not be possible to get 128 samples or to afford analyses on each one at $35 each.
What is the change in n if your relaxed your accuracy requirement to 10% not 5%? The answer:
n = (16) (1.72) / ((12) (0.1))2 = 32.
Decisions about alpha levels and risk and accuracy can be very influential.
There is a big difference in the costs of processing 32 samples and processing 128. Decisions about alpha levels and risk and accuracy can be very influential. The manager's job is to maximize benefits to a desired level and reduce the costs (and risks) of units changed.
The manager cannot count as personal success the animals present, only those added (or reduced) above the natural population (or population present when the manager first arrived to begin work.)
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Last revision January 18, 2004.