Date 1/05
Title
Name Edward Kannel
Pg
Standard error of estimates for measured and computed values
Standard error of measured values
Random error in our measurements will always occur, but we can establish the bounds within which we
expect the true value of a measured value to be. The standard error of a set of measurements is the
parameter we calculate to identify the expected limits of error. The standard error of the estimate is
based on assumptions that most of the errors are small and that the errors are normally distributed. For
distances or angles that are directly measured, the standard error of the mean of those measurements,
based on the observations from a single set of measurements of size n, is calculated as::
Standard error of mean = [ ∑ (Xi – Xmean)/ (n -1) ] / √ n
The term in brackets [ ] is the standard deviation of the sample, and n is the sample size.
In CE 111 the instructors expect you will use your calculator or a spreadsheet to determine the standard
deviation of the observed measurements.
From statistical theory, under the assumption that we completed a large number of observations of
sample size n, using the z- statistical distribution which is based on having a large number of samples ,
we state, with 95% confidence, that the true distance or angle is within +/-2 standard errors of the
measured mean of our sample. An example is:
Line AB measured twice: X1 = 355.77’ X2 = 355. 89’
Xmean =
355.83
Standard deviation =
0.085
Standard error of mean =
0.060
With 95% confidence, the true mean is within the range of
355.83 – 2 * (0.06)
and 355.83 + 2* 0.06
The standard error of the mean is also used to compute the relative precision of the measurements, or the
accuracy ratio as follows:
1 / X = standard error / mean
or 1 / X = 0.06 / 355.83
The resulting accuracy ratio is 1 / 5,900. This accuracy ratio is within accepted limits for measuring
lines by taping for our general our general small area surveys.
Standard error of computed values
Many values we use are not directly measured, but are calculated based on values that are
measured. The standard error of the computed value must incorporate the errors of the measured
components. Areas and volumes are common in our surveying .
Consider Area of a rectangle = length (X ) * width (Y)
From Calculus and the theory of error propagation, the general expression for error of the computed
value is:
Std Error of A = A = SQRT [(A/X)2X2 + (A/Y)2 Y2 ]
The above expression would be expanded for any number of additional variables included in the
formula.
Example: Find the standard error of the Area of a rectangle
A= X*Y
X = 300.10 +/- 0.06m ,
Y = 400.03m +/- 0.08m
Since
[(A/X) = Y
Area
=
Area
=
and (A/Y) = X
SQRT[Y2 * (0.06)2
+ X2 * (0.08)2]
34 m2
so Area = 120,050 m2 +/- 30 m2
The +/- 30 m2 assumes a one standard error limit. If Two standard error range were desired (95%
confidence) we would have said
Area = 120,050 +/- 70 m2
Note: If we had just taken the product of X * Y, the calculator yields 120,049.003. . Recording the
answer, even to the nearest square meter, would seem out of line when the standard error of the Area is
over 30 meters. The surveyor should consider this when reporting a final result. It is also of interest to
note the rules of significant digits indicates only 5 digits should be retained in the final answer (300.10,
400.03 used as X,Y). The 5 significant digits is reasonable.
Alternate strategies for estimating the Standard Error of AREA
CASE 1
Often the Area does not have a nice function that can be readily integrated or differentiated. An
indication of the standard error of the area can be found as follows:
1)
add 1 standard error for the linear measurement to each line,
2)
use the new line lengths to calculate the area just as would be done for the original line length
3)
calculate difference between the Area found using the means and the Area using the means plus
the standard error.
4)
divide the difference by SQRT (2) = 1.41 to obtain the estimate of standard error of the area
EXAMPLE FROM PREVIOUS Problem
X = 300.10 +/- 0/06
Y = 400.03 +/- 0.08
AREA with one standard error added for each line :
Area = (300.10 + 0.06) * ( 400.03 +.08) = 120,097 or 120,100 m 2
Difference = 120,097 – 120,049 = 48
Area (est) = 48/sqrt(2) = 34 m2
This is the same value found from the exact solution that used differential calculus.
Case 2
Relative error in Traverse Closure is known
If estimates of standard error of linear measurements and angles of a traverse are not known, but the
accuracy ratio of a closed traverse has been computed, the standard error of the area can be
estimated by
a) dividing the area by the accuracy ratio.
b) Multiplying the above result by SQRT (2) = 1.41
Example : For a rectangle:
X = 300.10 Y = 400.03
Area = 120,050 m2 and the accuracy ratio for a closed traverse was found to be 1/5,000
a) Estimated standard error of area =
Area = 120,050/5000
= 24 m2
b) Multiply above by SQRT(2) = 24 *1.4 = 34 m2
_________________
Special observation on Standard error of the mean:
If only two measurements have been made, the standard error of the mean is always
Standard error = 1/2 *
Check it out.
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(difference in measurements)
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