Geoscience Reference
In-Depth Information
■
EX AMPLE 7. 5
Problem:
The sample mean volume per forested acre for a 10,000-acre tract is
X
= 5680 board feet
with a standard error of
s
x
= 632 (so
s
2
= 399,424). The estimated total volume is
L
=
10 000
,
(
X
)
=
56 800 000
,
,
board feet
The variance of this estimate would be
=
()
=
2
2
2
s
(,
10 000
)
s
39 942 400 000 000
,
,
,
,
L
x
Since the standard error of an estimate is the square root of its variance, the standard error of the
estimated total is
2
s
==
s
6 320 000
,
,
L
■
EX AMPLE 7.6
Problem
: In 1995, a random sample of 40 1/4-acre circular plots was used to estimate the cubic foot
volume of a stand of pine. P
lo
t centers were monumented for possible relocation at a later time. The
mean volume per plot was
X
1
= 225 ft
3
. The plot variance was
s
x
1
= 8281 so that the variance of
the mean was
s
x
2
= 8281/40 = 207.025. In 2000, a second inventory was made using the same plot
ce
nters. This time, however, the circular plots were only 1/10 acre. The mean volume per plot was
X
2
= 122 ft
3
. The plot variance was
s
x
2
= 6084, so the variance of the mean was
s
x
2
= 152.100. The
covariance of initial and final plot volumes was
s
x
1,2
= 4259, making the covariance of the means
s
x
12
= 4259/40 = 106.475.
,
Solution:
The net periodic growth per acre would be estimated as
3
GXX
=
10
−
4 0 122
=
(
)
−
4 225
(
)
=
320 ft /acre
2
1
By the rule for linear functions the variance of
G
would be
2
2
2
2
2
s
=
()
10
s
+ −
()
4
s
+
2 10
( ()
,
−
4
s
G
x
12
x
x
2
x
1
=
100
(
152 100
.
)
+
16 207 025
(
.
)
−
80 106 475
(
.
)
=
10 004 4
,
.
In this example there was a statistical relationship between the 2000 and 1995 means because the
same plot locations were used in both samples. The covariance of the means (
s
x
1,
x
2
) is a measure of
this relationship. If the 2000 plots had been located at random rather than at the 1995 locations, the
two means would have been considered statistically independent and their covariance would have
been set at zero. In this case the equation for the variance of the net periodic growth per acre (
G
)
would reduce to
2
2
2
2
2
s
=
()
10
s
+ −
()
4
s
=
100 152 100
(
.
)
+
16 207
(
025
)
=
18 522 40
,
.
G
x
2
x
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