Statistics Review - Mathematics and Statistics for the Environment - page 117

it is a measure of the variation among sample means, just as the standard deviation is a measure of
the variation among individuals. The standard error of the mean may be used to compute confidence
limits for a population mean.
The computation of the standard error of the mean (often symbolized by s x ) depends on the
manner in which the sample was selected. For simple random sampling without replacement (i.e., a
given unit cannot appear in the sample more than once) from a population having a total of N units
the formula for the estimated standard error of the mean is

2

s
n

n
N

s

x =

1 −

In a forestry example, if we had n = 10 and found that s = 1.414 and s 2 = 2 in the population that
contains 1000 trees, then the estimated mean diameter ( X = 9.0) would have a standard error of

2
10

10
1000

=

s x = 1 − 0 198 . = 0 445 .

Note: The term (1 - n / N ) is called the finite population correction , or fpc. The fpc is used when the
sampling fraction (the number of elements or respondents sampled relative to the population)
becomes large. The fpc is used in the calculation of the standard error of the estimate. If the
value of the fpc is close to 1, it will have little impact and can be safely ignored.

7.8 COVARIANCE
Very often, each unit of a population will have more than a single characteristic. In forestry practice,
for example, trees may be characterized by their height, diameter, and form class (amount of taper).
The covariance is a measure of the association between the magnitudes of two characteristics. If
there is little or no association, the covariance will be close to zero. If the large values of one char-
acteristic tend to be associated with the small values of another characteristic, the covariance will
be negative. If the large values of one characteristic tend to be associated with the large values of
another characteristic, the covariance will be positive. The population covariance of X and Y is often
symbolized by σ xy ; the sample estimate by s xy .
Let's return to a forestry practice example. Suppose that the diameter (inches) and age (years)
have been obtained for a number of randomly selected trees. If we symbolize diameter by Y and age
by X , then the sample covariance of diameter and age is given by

− (

)( )

∑∑

XY
N

∑

XY

s

xy =

n − 1

This is equivalent to the formula

∑ (

XXYY
n

− )( − )

s

xy =

− 1

If n = 12 and the Y and X values were as follows:

Y = 4 + 9 + 7 + 7 + 5 + 10 + 9 + 6 + 8 + 6 + 4 + 11 = 86

X = 20 + 40 + 30 + 45 + 25 + 45 + 30 + 40 + 20 + 35 + 25 + 40 = 395