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Proof of Eq. (8-21) is left to the reader [Hint: XXXX ii ]. Taking expectations of Eq. (8-21) in accordance with Eq. (8-16), we get

2n

1i

2

i

n

1i

2

i XnEXXXXE

(8-22)

Noting from Eq. (5-29) that 22X , and from Eqs. (8-19), (8-20), and (5-29)

that n

X22 , we get

222n

1i

2

i 1nnnnXXE

(8-23)

It therefore follows that

2n

1i

2

i

2XX

1n

1ESE

(8-24)

Which shows that2

S is an unbiased estimator of 2 . This is the reason why the term

1n is used for the sample variance. If the sum of the squares of the deviations were to be divided by n rather than by 1n , the resulting estimator would be biased.If the population from which the sample is drawn is normally distributed, it can be shown

that2

S is also a consistent estimator of2

and that the distribution of2

S is related to the

chi-square distribution; specifically

22 S1n

Y

(8-25)

has a chi-square distribution with 1n degrees of freedom.

8.7. CONFIDENCE INTERVAL FOR THE MEAN

If a random sample of size n is drawn from a normally distributed population with mean

and variance 2 , the sample mean X has a normal distribution with mean and

variancen

2 . Thus, according to Eq. (5-23), the quantity

n

XZ

(8-26)

has a standard normal distribution with zero mean and unit variance, and it follows that

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1Z2Z

N

XZP

(8-27)

where Z is the value of the standard normal distribution function, obtainable fromTable I of Appendix B.Rearranging the inequality inside the brackets of Eq. (8-27), we get

1Z2n

zX

n

zXP

(8-28)

which reads as follows: The probability that lies betweenn

zX

andn

zX

is 2 1z . When a specific numerical value x is provided for x , the foregoing

probability statement becomes a confidence statement. The valuesn

zX and

n

zX are known as confidence limits, the interval between them is known as a

confidence interval, and 2 1z is known as the degree of confidence, or confidencelevel, often stated as a percentage. The construction of a confidence interval for a

particular distribution parameter, such as u, is known as interval estimation.

EXAMPLE 8-5

In Example 8-3, the sample mean of 20 independent measurements of a distance was

calculated to be 537.615 m. If the standard deviation of each measurement (i.e., thestandard deviation of the population) is known to be 0.033 m, construct a 0.95 (95%)

confidence interval for the population mean, .

Solution

For 95.01z2 , 975.02

195z . From Table I of Appendix B, 96.1z .

Therefore, the confidence limits are

m601.53720

)033.0(96.1615.537

n

zx

and

m629.53720

)033.0(96.1615.537

n

zx

Thus, we can say with 95% confidence that u lies in the interval 537.601 m to 537,629 to.In Example 8-5 the standard deviation v of the population was known. More often,however, o is unknown and must be estimated, usually by the

212 Analysis and Adjustments of Survey Measurements

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sample standard deviation S. Thus, instead of using Eq. (8-26), we must use

nS

XT (8-29)

It is easily shown from Eqs. (8-7), (8-25), (8-26), and (8-29) that T has a t distribution with

n -1 degrees of freedom. Thus, instead of Eq. (8-28), we have

1)t(F2n

tSX

n

tSXP

(8-30)

When specific numerical values x and s are provided for X and S, we obtain a confidence

interval with confidence limits ntSX and ntSX and degree of confidence 1)t(2 .

EXAMPLE 8-6

With reference to Example 8-3, in which the sample mean of 20 independentmeasurements of a distance is calculated to be 537.615 m, and to Example 8-4 in whichthe sample standard deviation of the same 20 measurements is calculated to be 0.035 m,

construct a 0.95 (95%) confidence interval for the population mean, .

Solution

For 975.0295.1)t(F,95.01)t(2 Degrees of freedom 191201n . From Table

III, Appendix B, 209t19,975.0

. Therefore the confidence limits are

m599.537

20

)035.0)(09.2(615.537

n

tSX

and

m631.537

20

)035.0)(09.2(615.537

n

tSX

Thus, we can say with 95% confidence that lies in the interval 537.599m to 537.631m.

In establishing a confidence interval for the mean of a distribution it has been assumed that

the random sample is drawn from a normal distribution. If the population distribution is

not normal, but the sample size is large, X will have a distribution that is approximately

normal, and Eqs. (8-28) and (8-30) are still valid for all practical purposes.Finally, as the sample size increases, we see that the values of t approach thecorresponding values of z . Indeed, for a sample size of 30 or larger, the t


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distribution can be approximated very well by the standard normal distribution.

8.8. CONFIDENCE INTERVAL FOR THE VARIANCE

When a random sample of size n is drawn from a normal population, the relationship given

by Eq. (8-25), that 2S21n has a chi-square distribution with n-1 degrees offreedom, can be used to construct a confidence interval for the population variance,

2 .

Thus,

a bXS)1n(

XP 21n, b2

22

1n,a

(8-31)

where 2

1n,aX and 2

1n, bX are the th and b:h percentiles, respectively, of the

chi-square distribution with n-1 degrees of freedom.

From Eq. (8-31) it follows that:

a bX

S)1n(

X

S)1n(P

2

1n,a

2

2

1n, b

22

(8-32)

and when a specific numerical value2s is provided for

2S , we obtain a confidence

interval with limits 21n, b

X2S)1n(

and 2

1n,aX

2S)1n(

and degree of confidence

.a b

In constructing an appropriate confidence interval for 2 , it is customary to make the two

percentiles complementary, i.e., .1a b

If a confidence interval for the standard deviation is desired, positive square roots of the

confidence limits for 2 are taken.

EXAMPLE 8-7

With reference once more to Example 8-4, in which a sample variance of 0.001212

m iscalculated from 20 independent measurements of a distance, construct a 0.95 confidence

interval for2

and the corresponding confidence interval for .

Solution

For 95.0a b and 1 ba we get 025.0a and 975.0 b

Degrees of freedom 191n . From Table II of Appendix B 91.82

19,025.0X and

9.322

19,975.0X Therefore the confidence limits are

2

2

19,975.0

2

m00071.09.32

)00121.0)(19(

X

s)1n(


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and

.m00258.091.8

)00121.0)(19(

X

s)1n( 22

19,025.0

2

Thus, we can say with 95% confidence that2

lies in the interval 0.000702

m to 0.00258

2m .The corresponding 95% confidence interval for o is 0.026 m to 0.051 m.

8.9. STATISTICAL TESTING

It is often desirable to ascertain from a sample whether or not a population has a particular

probability distribution. The usual course of action that is taken is to make a statement

about the probability distribution of the population, and then test to see if the sampledrawn from the population is consistent with the statement.The statement that is made about the probability distribution of the population is called a

statistical hypothesis. If the hypothesis specifies the probability distribution completely, itis known as a simple hypothesis; otherwise, it is known as a composite hypothesis.

For every hypothesis 0H there is a complementary alternative 1H . 0H and 1H , are often

called the null hypothesis and alternative hypothesis, respectively.An hypothesis is tested by drawing a sample from the population in question, computingthe value of a specific sample statistic, and then making the decision to accept or reject the

hypothesis on the basis of the value of the statistic. The statistic used for making the test iscaned the test statistic.

Testing of a statistical hypothesis 0H is not infallible, since it is based upon a sample

drawn from a population rather than upon the entire population itself. Four possible

outcomes can occur:

1. 0H is accepted, when 0H , is true.

2. 0H , is rejected, when 0H is true.

3. 0H , is accepted, when 0H is false.

4. 0H is rejected, when 0H , is false.

If outcome (1) or outcome (4) occurs, no error is made in that the correct course of actionhas been taken. Outcome (2) is known as a Type I error; outcome (3) is known as a Type II

error.

The size of the Type I error, designated , is defined as the probability of rejecting 0H

when 0H is true, i.e.,


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.trueisHwhenH jetReP 00 (8-33)

When is fixed at some level for 0H , and is expressed as a percentage, it is known as thesignificance level of the test. Although the choice of significance level is arbitrary,

common practice indicates a significance level of 5% as "significant", and 1% as "highlysignificant".

8.10. TEST OF THE MEAN OF A PROBABILITY DISTRIBUTION

Under certain conditions we may expect the mean of a probability distribution to have a

specific value0

. The hypothesis that.0

, can be tested by drawing a sample of size n

and using the sample mean X as the test statistic. Specifically, we have

00 :H

00 :H

X is assumed to be normally distributed, or at least approximately normally distributed.

Under the hypothesis that 0 , the following probability statement can be derived from

Eq. (8-27), assuming is known:

,1)z(2)c(X)c(P 00 (8-34)

where .nzc

If is unknown, the following probability statement can be derived from Ea. (8-30):

,1)t(F2)c(X)c(P 00 (8-35)

where .ntsc

0H , is accepted if x , the specific value of X calculated from the sample, lies

between c0 and c0 ; otherwise, 0H , is rejected. The regions of acceptance and

rejection are shown in Fig. 8-4. If is the probability that 0H , is rejected when it is true,

then 1 must e the probability that 0H is accepted when it is true. It follows, then, that

1)z(21 for know (8-36a)

1)t(F2 for know (8-36b)


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Fig. 8-4.

Solving for )z( , or )t(f , we get:

21)z(

(8-37a)

or

21)t(F

(8-37b)

Thus, the value of z or t is obtained from the significance level of the test, . Specifically,

Table I of Appendix B is used to evaluate z ; Table III of Appendix B is used to evaluate t .

[Note: In Table III, 1n, ptt , where )t(F p .]

EXAMPLE 8-8

An angle is measured 10 times. Each measurement is independent and made with the same

precision, i.e., the 10 measurements constitute a random sample of size 10. The sample

mean and sample standard deviation are calculated from the measurements: x =

42°12'14.6", s = 3.7". Test at the 5%o level of significance the hypothesis that , the

population mean of the measurements, is 42°12'16.0" against the alternative that is not

42°12'16,0",

Solution

0 42°12'16,0",

For 5% level of significance, a = 0.05. Thus,


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,975.02

1)t(F p

Degrees of freedom ,91n From Table III of Appendix B, 26.29,975.0tt . Thus,

´´´´

6.210

)7.3)(26.2(

n

tsc

and so´´´º

0 4.131242c

and´´´º

0 6.181242c

Since 12'14.6"42x lies between c0

and c0

, the hypothesis that µ=42°12'16.01 is

accepted at the 5% level of significance.

8.11. TEST OF THE VARIANCE

OF A PROBABILITY DISTRIBUTION

Under the assumption that a population is normally distributed, we can test the null

hypothesis 0H that the population variance is 2

0 ; against the alternative that it is not 2

0 ,

using the sample variance 2S as the test statistic.

Noting that 20

2S)1n( is distributed as chi-square with n-1 degrees of freedom, we can

make the following probability statement:

,a bxS)1n(xP 2 1n, b2022 1n,a (8-38)

from which we get

,a b)1n(

xS

)1n(

xP

2

0

2

1n, b2

2

0

2

1n,a

(8-39)

0H is accepted if2

s , the specific value of2

S calculated from the sample, lies

between )1n(x 2

0

2

1n,a and )1n(x 2

0

2

1n, b ; otherwise, 0H is rejected. Theregions of acceptance and rejection are shown in Fig. 8-5.

Again, 1 is the probability that 0H is accepted when it is true. Thus,

,a b1 (8-40)


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Fig. 8-5.

and for 1 ba (complementary percentiles). We obtain.

2a

(8-41)

and

,2

1 b (8-42)

This test procedure can, of course, be used to test the standard deviation as well as thevariance.

EXAMPLE 8-9

Referring to the data in Example 8-8, test at the 5% level of significance the hypothesis

that , the population standard deviation of the measurements, is 2.0" against the

alternative that is not 2.0".

Solution

.0.2 ´´20


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From Example 8-8, the sample standard deviation, ´´7.3s .Now 05.0a .

Thus 025.02aa , and b= I – (a/2) = 0.975. Degrees of freedom= n – 1=9. From Table

II, Appendix B, 70.22 9,025.0X , and 0.192

9,975.0X . Thus,

20.19

)0.2)(70.2(

1n

X 2´´2

9,025.0

(seconds of arc) 2

and

44.89

)0.2)(0.19(

1n

X 2´´29,975.0

(seconds of arc)

2

Now 7.132)7.3(

2s (seconds of arc)

2. Since

2s does not lie between 1.20 and 8.44,

the hypothesis that

´´

0.2 is rejected at the level of significance.

8.12. BIVARIATE NORMAL DISTRIBUTION

The probability distribution of two jointly distributed random variables was discussed ingeneral terms in Chapter 5. We shall now look at a particular joint distribution of two

random variables — the bivariate normal distribution. This distribution is very useful whendealing with planimetric (x, y) positions in surveying.

The joint density function of two random variables X and Y which have a bivariate normaldistribution is

,

2

y

yy

y

yy

x

xx2

2

x

xx

)

2

1(2

1exp

2

1yx2

21

)y,x(f }{

(8-43)

in whichx

, andx

are the mean and standard deviation, respectively, of X;y

, and

y are the mean and standard deviation, respectively, of Y; and is the correlation

coefficient of X and Y as defined by Eq. (5-55),This density function has the form of a bell-shaped surface over the x, y coordinate plane,

centered yy,xx , as shown in Fig. 8-6. The marginal density functions for X and Y

are, respectively,

}{

2

xx2

1

exp2

1

)x(f x

(8-44)


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Fig. 8-6.

And

,exp2

1

)y(f }{

2

y

yy

2

1

y

(8-45)

which are the usual density functions for individual normally distributed random variables.

The two marginal density functions are also shown in Fig. 8-6.A plane that is parallel to the x, y coordinate plane will cut the bivariate density surface inan ellipse (see Fig. 8-6). The equation of this ellipse is obtained by setting f (x, y) in Eq.(8-43) equal to the height K of the intersecting plane above the x, y plane, and simplifying.

The result is

,c1yyx

2x 22

2

y

y

y

y

x

x

2

x

x

(8-46)

where

1

)2

1(2

y

2

x

2K 2

4ln2c

, a constant.


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EXAMPLE 8-10

The parameters of a bivariate normal distribution are 4 , 5 , 4x

, 4y

,

and 5.0xy

. A plane intersects the density function at 1.0K above the x, y coordinate

plane. Evaluate and plot the ellipse of intersection.

Solution

60.21

)25.01(2)5.0(

2)1(

2)1.0(

24ln

2c

,95.1)60.2)(25.01(c)1( 2222 Thus, the equation of the ellipse of intersection is

,95.1

5.0

5y

5.0

5y

1

4x)5.0(2

1

4x 22

Simplifying, we get

.95.1)5y(4)5y)(4x(2)4x( 22

Letting 4xu , and 5yv , we have

i2v4uv2

2u

Solving for u in terms of v, we get

.v395.1vu 2

Thus,

2)5y(395.15y4x or

2)5y(395.11yx

Values for x and y are listed in Table 8-4, and the ellipse is plotted in Fig. 8-7

It can be shown through appropriate differentiation of Eq. (8-46) that the extreme points ofthe ellipse (A, B, C, and D in Fig. 8-7), have the following coordinates:

X Y

A xc

x y

cy

B xc

x y

cy

C xc

x ycy

D xc

x y

cy


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Table 8 .4

x

4.2 1.03 and 3-374-4 2-47 4-33

4.6 2-39 4.81

4.8 2-45 5-15

5-0 2.60 5.40

5.2 2.85 5.55

5-4 3.19 5.61

5.6 3.67 5.53

5.8 4.63 4.97

Fig. 8-7.

If the ellipse is enclosed within an imaginary box, indicated by the broken lines

in Fig- 8-7, we see that the half-dimensions of the box are xc , and yc . We can

also see that acts as a proportioning factor in locating A, B, C, and D.

Points E and G (Fig. 8-7) on the ellipse can be located by setting x in Eq- (8-46)equal to zero and solving for y:

.1cy 2

y

6


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Similarly, points F and H (Fig. 8-7) are located by setting y in Eq- (8-46) equal to zero andsolving for x:

.2

1xcz

When yx , and 0 , Eq- (8-46) reduces to

,cyx 222y2

x (8-47)

which is the equation of a circle with radius c .

8.13. ERROR ELLIPSES

In the previous section, the general case of the bivariate normal distribution wasconsidered. This is the usual model which applies to survey measurements. If we wish to

focus on the random error components only, we can set x , and y , equal to zero and get

probability distribution that centers on the origin of the x, y coordinate system.

When 0yx , Eq. (8-43) reduces to

2

yyx

2

x

22

yx

yyx2

x

12

1exp

12

1)y,x(f

(8-48)

and Eq.(8-48) becomes

.c1yyx2x 222

yyx

2

x

(8-49)

Equation (8-49) represents a family of error ellipses centered on the origin of the x, y

coordinate system. When 1c , Eq. (8-49) is the equation of the standard error ellipse.

The size, shape and orientation of the standard error ellipse are governed by the

distribution parameters x , y and . Six examples illustrating the effects of different

combinations of distribution parameters are shown in Fig. 8-8.

A typical standard error ellipse is shown in Fig. 8-9. Since 1c , the imaginary box

(broken line) that encloses the ellipse has half-dimensions x , and y , In general, the

principal axes of the ellipse, x´ and y´, do not coincide


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with the coordinate axes x and y; the major axis of the ellipse, x', makes an angle with

the x-axis.

A positional error is expressed in the x, y coordinate system by random Y

X

vector; the

same error is expressed in the x, y coordinate system by random vector ´Y

´X.The orthogonal

(rotational) transformation which relates the two vectors is

Fig. 8-8.


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Fig. 8-9.

,Y

X

cossin

sincos

Ý

´X

(8-50)

where is the angle of rotation.

Now the covariance matrices for random vectors

Y

X and

Ý

´X

are

2

yxy

xy

2

x and ,

0

02

y

2

x

respectively. The off-diagonal terms in the covariance matrix for

Ý

´X are zero because X' and Y' are uncorrelated (x' and y' are the

principal axes of the ellipse).

Applying the general law of propagation of variances andcovariances, Eq. (6-19), to the vector relationship given by Eq. (8-

50), we get:

.cossinsincos

cossinsincos

00

2

yxy

xy

2

x

2

ý

2

´x

(8-51)


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Solution

,m0246.014.022.080.0 2yxxy

,m0340.02

14.022.0

2

2222y2x

.m0285.0)0246.0(

4

)14.0()22.0(

4

2

21

2

22221

2

xy

22

y

2

x

Thus,

21

2

xy

22

y

2

x

2

y

2

x2

´x42

22

´x m0625.00285.00340.0 and

21

2

xy

22

y

2

x

2

y

2

x2

´y42

.m0055.00285.00340.0 22

´x

Thus the semimajor axis is

m25.00625.0´x

and the semiminor axis is

.m074.00055.0´y

now

.711.1

14.022.0

)0246.0(222tan

222

y

2

x

xy

Since xy , and 2

y

2

x are both positive, 2 lies in the first quadrant.

Thus, º7.59)711.1(1

tan2

, and the orientation of the error ellipse is º8.29 .

To determine the probability associated with an error ellipse, it is most convenient toconsider independent (uncorrelated) random errors X and Y. (If the errors are correlated,

they can always be transformed into uncorrelated errors by rotation through angle .)


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For uncorrelated random errors, p = 0 and Eq. (8-49) reduces to

.c

yx 22

y

2

2

x

2

(8-58)

Now consider the position of a point defined by the two random errors X and Y. This pointwill lie on or within the error ellipse if

.cyx 22

y

2

2

x

2

(8-59)

Since X and Y are two independent normal random variables with zero means, the random

variable

2

y

2

2

x

2

YXU

(8-60)

has a chi-square distribution with two degrees of freedom. The probability density functionof U can be easily derived from the general chi-square density function, Eq. (8-3), noting

that for two degrees of freedom, 12n .Thus, the probability density function of U is

2ue2

1)u(f

for .0u (8-61)

The probability that the position given by values of X and Y lies on cr within the error

ellipse is

222

y

2

2

x

2

cUPcyx

P

2

0

2udue

2

1

.e1 2u (8-62)

The probability 2cUP is represented by the volume under the bivariate normal densitysurface within the region defined by the error ellipse.

2cUP for various values of c is

given in Table 8-5. Since for the standard error ellipse 1c , we see from Table 8-5 that the

probability is 0.394 that the position of a point plotted from the two random errors will lieon or within the standard error ellipse.


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Table 8-5

c p[U

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8-3 Given a random variable T which has a t distribution with 10 degrees offreedom. Determine from Table III of Appendix B the probability that T takes on a value:(a) less than 0.70; (b) greater than 0.70; (c) between 0.26 and 0.70; (d) between -3.17 and

3.17; (e) between -1.81 and 0,26. Note: Since the t distribution is symmetric about zero, tTP1tTP .8-4 A distance is measured 25 times. All measurements are independent and have thesame precision (i.e., the measurements constitute a random sample of size 25). Followingare the measurements:

231.354 m 231.361 m 231.384 m 23 1. 34 7 m 231.335 m

231.31 2 m 231.3 55 m 231.34 7 m 231.3 66 m 231.361 m

231.320 m 231.348 m 231.34 1 m 231.3 38 m 231.337 m

231.361 m 231.3 41 m 231.3 50 m 231.3 33 m 231.355 m

231.32 2 m 231.3 31 m 231.37 6 m 231.3 35 m 231.344 m

Evaluate the sample mean, sample median, sample midrange, sample range, sample meandeviation, sample variance, and sample standard deviation.

8-5 The following 20 observations of a pair of angles, and , are obtained:

OBSERVATION

1 31°14'16.2" 42º08'24.0"

2 31°14'15.2" 42°08'24.4"

3 31°14'15.6" 42°08'24.5'4 31º14'14.5" 42º08'23.8"

5 31°14'14.0" 42º08'25.7"

6 31'14'15.8" 42°08'26.1"

7 31°14'16.0" 42º08'21.8"

8 31°14'14.1" 42º08'23.3"9 31°14'16.4" 42°08'24.8"

10 31°14'13.6" 42°08'23.2"

11 31°14'16.7" 42°08'25.7"

12 31º14'14.1" 42°08'22.7"

13 31°14'15.3" 42º08'26.2"14 31°14'15.2" 42°08'25.8"

15 31°14'12.9" 42°08'25.3"

16 31°14'17.9" 42°08'24.0"17 31°14'16.2" 42°08'27.4"

18 31°14'14.2" 42°08'25.8"

19 31º14'14.1" 42°08'23.7"

20 31°14'14.8" 42º08'24.0"

Compute the sample variances,2

S and 2S , and the sample covariance, S . From these

sample statistics compute the sample correlation coefficient for and


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.SS

S

8-6 Angles and in Problem 8-5 are added to form a new angle .Compute

a sample of 20 values for directly from the 20 pairs of and values in Problem

8-5. Then compute from these data the sample variance 2s , the sample covariance

2s and the sample correlation coefficient . Check the computed values

for 2s , and

2s by appl yin g vari ance-covariance propagati on to the vector

function

,10

01

using the sample variances and covariance for and , computed in Problem 8-5, as

elements of the covariance matrix for .

8-7 A distance is measured 10 times. All measurements are independent and have thesome precision. The standard deviation of each measurement is known to be 0.025 m. Thefollowing values are obtained: 307.532 m, 307.500 m, 307.474 m, 307.549 m, 307.490 m,307.527 m, 307.556 m, 307.502 m, 307.489 m, and 307.514 m. Evaluate the sample mean

for the distance and the standard deviation of this sample mean. Construct 50% and 95%confidence intervals for the population mean of t distance.8-8 if in Problem S-7 the standard deviation of each measurement is not known,construct 50% and 95% confidence intervals for the unknown standard deviation, under

the assumption that the measurements are normally distributed. Is there any significantdifference between the sample standard deviation and the standard deviation (0.025 m)given in Problem 8-7?8-9 An angle is measured six times with the following results:

40°10'15.6" 40°10'16.4"40º10'10.8" 40º10' 13.5"40°10'08.9" 40°10'11.0"

The measurements are assumed to be normally distributed and independent. Construct90% and 99% confidence intervals for the population mean, variance,and standard deviation.8-10 An angle is measured 16 times with a theodolite . The measured values are:

1. 52°35'24" 9. 52°35'24"2. 52°35'28" 10. 52°35'29"

3. 52º35'22" 11. 52°35'35"

4, 52°35'20" 12. 52°35'31"

5. 52°35'25" 13. 52°35'29"

6. 52°35'29" 14. 52°35'26"7. 52°35'18" 15. 52°35'30"

3. 52°35'26" 16. 52°35'31"


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It is suspected that the theodolite was disturbed between the eighth and ninthmeasurements. Construct 90% confidence intervals for the means of the first eight and lasteight measurements. Is there evidence the theodolite was disturbed?

8-11 An angle is independently measured 10 times with the same precision. Theobserved values are 90°00'05", 90°00'10", 90°00'00", 90°00'07", 89°59'54", 89°59'58",90°00'06", 90°00'03", 89°59'57", 90°00'10".

(a)Test the hypothesis that the mean of the measurement equals 90°00'00" against the

alternative that the mean does not equal 90°00'00". Use a 5% level of significance.(b)Test the hypothesis that the standard deviation of the measurement is 4" against the

alternative that it is not 4". Use a 5% level of significance.8-10 A level rod is observed 15 times with a precise level that is equipped with a

micrometer. Following are the rod readings (assumed to be a random sample from a

normal population):

1412.80 mm 1412.85 mm 1412.87 mm1413.09 mm 1412.50 mm 1412.80 mm1412.86 mm 1412.84 mm 1412.66 mm

1412.80 mm 1412.84 mm 1412.84 mm

1412.78 mm 1413.02 mm 1412.72 mm

Test the following hypotheses at the 5% level of significance:

(a) 0H : 1413.00 mm against 1H : 1413.00 mm

(b) 0H : 1412.75 mm against 1H : 1412.75 mm

(c) 0H : 0.08 mm against 1H : 0.08 mm

(d) 0H : 0.20 mm against 1H : 0.20 mm

8-13 Two independent calibrations of a 50 m steel tape yield two different values forits length: 50.0026 m, and 50.0008 m. The standard deviation of a single tape calibration isknown to be 0.7 mm.

(a) Test at the 2% level of significance for any significant difference between the twocalibration values.

(b) Assuming there is no significant difference between the two calibration values,construct a 99% confidence interval for the length of the tape based upon the mean of thetwo calibration values.8-14 Plane coordinates X and Y of a survey station have a bivariate normal distribu-

tion. The mean and standard deviation of X are 1700.50 m and 0.20 m, respectively; themean and standard deviation of Y are 810.65 m and 0.10 m, respectively, The coefficientof correlation between X and Y is 0.60. Evaluate the principal dimensions (semimajor and

semiminor axes) and orientation of the standard error ellipse associated with this survey

station position.8-15 The following covariance matrix is associated with the random error in thehorizontal (x, y) position of a point:

,m160.0096.0

096.0090.02


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Under the assumption that the random error has a bivariate normal distribution, evaluatethe principal dimensions and orientation of its standard error ellipse. Sketch the standarderror ellipse, showing pertinent dimensions.

8-16 If X and Y have a bivariate normal distribution with 0yx , 0yx

and 0xy , it can be shown that the radial distance 2

Y2

XR has the following

density function:

2

2

22

r exp

r )r (f for .0r

This is the density function of the Rayleigh distribution. Evaluate 0R P and comparewith the corresponding entry in Table 8-5.8-17 The computation of a closed traverse results in the following misclosure in the xand y coordinates:

cm3.3xx 0c

,cm9.6yy 0c

where xo and 0y are the given coordinates of the origin of the survey and cx and cy are the

computed coordinates. The covariance matrix of the computed coordinates, referenced tothe given coordinates, is

.cm83.887.0

87.063.42

Is the misclosure acceptable in the sense that it lies within the 0.95 probability errorellipse?8-18 The x, y position of a survey station is computed by the method of least squares.

The initial (approximate) position of the station is given by m60.10400x , m50.2143

0y ,

and the normal equations in the least squares solution are

.20.0

40.0

y

x

500.0250.0

250.0125.1

The reference variance is 2m040.0 ,Assume the solution requires only one iteration.

Evaluate the least squares position of the survey station and the principal dimensions andorientation angle of its 99% confidence region (an ellipse identical in size, shape andorientation to the corresponding 0.99 probability error ellipse but centered on the least

squares position).8-19 The principal axes of a standard error ellipse coincide with the x and y axes

(Fig.8-10). The standard deviations x , and y are as shown in the figure. Let of

OP0 be the standard deviation in any direction . In general, P does not lie on the

ellipse; instead, its locus is the so-called pedal curve, shown as a broken line in Fig. 8-10.

(a) Show that


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.sencos 22y22

x

2

0

( Hint : Put the x' axis in the direction of OP.)(b) Then show that the equation of the pedal curve in the x, y coordinate system is

.0yxyx 22y22x222

8-20 A steel tape of length 1 is used to measure the distance between two survey

stations. A total of n full tapelengths are required to measure the distance. Random errors,

nE,2E,1E in aligning the tape introduce a positive error V in the observed distance.

Assume these alignment errors are independent and normally distributed with zero mean

and standard deviation , and that the effect of each error iE , in the direction of taping can

be approximated by 22

i

E .

(a) Show that Y22V , where Y is distributed as chi-sguare with n degrees offreedom. (b) Derive expressions for the mean and standard deviation of V. (c) If a 50 m

tape is used to measure a distance of 800 m, and m50.0 , evaluate v such

that 95.0vV p .

Fig. 8-10


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General Least Squares

Adjustment

9.1. INTRODUCTION

In Examples 3-5 and 4-6 a straight line is fitted through three points. In these examples,the y-coordinates were assumed to be the observations, while the x-coordinates were

considered as constants. Based on the straight-line equation

0 baxy (9-1)

the condition equations took the form for the adjustment of indirect observations

.f Bv (9-2)

The elements of the unknown parameters vector are the slope a and the y-intercept b, and

the residuals are those associated with the observed y-coordinates.Let us now consider the case in which not only the y-coordinate but also the x-coordinateof each point is an observed quantity. The straight-line equation (9-1), written for any

point, will then contain two observations, x and y, and two parameters, a and b. In this

form, it does not fit the condition equations form of either one of the two techniques ofleast squares adjustment discussed in Chapter 4. In adjustment of indirect observations,

where the form of the condition equations is given by Eq. (9-2), each condition equationcontains only one observation. In adjustment of observations only, where the conditionequations are of the form

,f Av (9-3)

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no parameters are included in the conditions. Consequently, for the case at hand, there isneed for a more general least squares technique that can handle combined observations and

parameters in the condition equations without the restriction of having only one

observation in each equation. Such a technique is the subject of this chapter.To permit as much generality as possible, the technique will place no restriction on thestructure of the covariance, cofactor, or weight matrices of the observations, i.e., it willaccept measurements that may be correlated and/or of unequal precision.

Before proceeding with the derivation of the general least squares technique, we shallrework the straight-line problem of Examples 3-5 and 4-6, now taking the x-coordinates aswell as the y-coordinates as observations.

EXAMPLE 9-1

A straight line, baxy , must be fitted through three points. The following data are

given:

POINT x(cm) y(cm)

)

2

cm(

2

x )

2

cm(

2

y 1 2.00 3.20 0.04 0.10

2 4.00 4.00 0.04 0.08

3 6.00 5.00 0.04 0.08

These are precisely the same data as given for Example 4-6, Part (2), with variances for the

x-coordinates added. All measured coordinates are assumed to be uncorrelated. It isrequired, under these new conditions, to find least squares estimates for the two parametersa and b.

Solution

For any one of the three points, the equation of the straight line is

.0 bvxavyFxy

Since a and xv , are both unknown, this equation is nonlinear in the unknowns. It cannot

therefore be used directly, as was the case in Example 4-6, but must be linearized asfollows:

0 b b

Fa

a

Fv

y

Fv

x

F bxay

yx00

or

, bxay b1axv1va 00yx0

General Least Squares Adjustment 237

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.

0

0

20.0

00.2)00.6(5.000.5

00.2)00.4(5.000.4

00.2)00.2(5.020.3

f

Since both the x and y coordinates are treated as observed variables, and since there are sixcoordinate values in total, the covariance matrix is 6x6. Furthermore, since all theobservations are uncorrelated, the covariance matrix is diagonal, i.e.,

,cm

08.0

04.0

08.0

04.0

10.0

04.0

2

2

3y

2

3x

2

2y

2

2x

2

1y

2

1x

The observation vector and the residual vector v are

.

v

v

v

v

v

v

vand

y

x

y

x

y

x

3y

3x

2y

2x

1y

1x

3

3

2

2

1

1

The constant term vector f can be writing as

,

b

b

b

y

x

y

x

y

x

1a0000

001a00

00001a

f

0

0

0

3

3

2

2

1

1

0

0

0

which, symbolically, is

,dAf

where d is obviously the constant vector

0 b

0 b

0 b

.

Now, the vector of observations can be transformed into a vector of equivalent

observations, c , as follows:

.Ac


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The same transformation applies to the vector of residuals, v, i.e.,

.Avv0 Thus,

df c and the linearized condition equations become

,dBv c0 which is the form of Eq. (4-7) for the technique of adjustment of indirect observations.From the general law of propagation of variances and covariances, expressed by

Eq. (6- 19), we can obtain the covariance matrix of c .

.cm

09.000

009.00

0011.0

AA 2t

Letting 2

cm09.02

0 , the weight matrix of the equivalent observations is [see Eq. (4-19)]:

.w 12

0

1

1

818.0

09.01

09.01

01.01

)09.0(

and the squares solution, according to Eqs. (4-28), (4-29), and (4-38), is

818.2636.11

636.11272.55

WBB N t

1636.0

3272.0Wf Bt

t

7147.25715.0

5715.01384.0 N 1

.2571.0

0482.0t N

1

The corrected parameter values are

4518.00482.05.0aaa 01 .2571.22571.00.2 b b b 01


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The corrected values must now be used as new approximations in the solution for a newcorrection vector. Thus

14518.00000

0014518.000

000014518.0

A1

16

14

12

B1 and as before,

0321.0

0643.0

0393.0

f 1

0882.0

0882.0

1082.0

AA t

1

t

11

1

1

8151.0

W 1

1c

2

01c

815.2630.11

630.11260.55BWB N 11c

t

11

0002.0

0005.0f WBt11c

t

11

7219.25729.0

5729.01387.0 N

1

1

.00026.0

00005.0t N 11

11

Since the values in 1 , are sufficiently small, the iterative procedure is terminated, and the

final estimates of the parameters arc

.cm257.2 b̂,452.0â

We have seen in the preceding example that in some problems the condition equations aremore readily written in the form f BAv than in either of the two simpler forms

discussed in Chapter 4. While it is possible in principle to solve any problem by any

technique of Least squares, it is quite often


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more convenient, and more efficient, to solve a given problem by a particular technique.Among others, curve-fitting problems and coordinate-transformation problems are bestsolved by the general technique of this chapter when all the coordinates in the condition

equations are observed variables.

9.2. DERIVATION

Given a mathematical model, the minimum number of observations necessary for its

unique determination is 0n . When n measurements which are consistent with the model

are acquired, such that0nn , the redundancy is

.nnr 0 (9-4)

This means that among the n observational variables there exist r independent conditionequations (see Sections 3.3, 4.1, and 4.4). If we wish to carry unknown parameters into the

adjustment, then we must write an additional condition equation for each parameter.Hence, if u parameters are carried, the number of condition equations will be

.ur c (9-5)

The lower limit for the value of u is obviously zero, in which case no parameters are

carried and there are r c conditions among the observations only. On the other hand, the

upper limit for the value of u is 0n , for which n0nr c , so that we may not write any

more conditions than the total number of observations. Thus,

0nu0 (9-6)

,ncr (9-7)

the lower limits of which define the case for adjustment of observations only, and theupper limits define the case for adjustment of indirect observations, the two special cases

treated in Chapter 4 (see Section 9.4). In many problems, the number of parameters carried

in the adjustment is neither zero nor 0n . In such cases, neither one of the two special

techniques of Chapter 4 would be directly suitable. Instead, c condition equations existwhich, when they are originally linear, take the form

,dBvA (9-8)

where


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A is a nc rectangular coefficient matrix nc ;

B is a uc rectangular coefficient matrix nc ; is the 1n given observational vector;v is the corresponding 1n vector of residuals;

is the 1u vector of parameters;

d is a 1c vector of constants.

The precision of the n given observations may be expressed by either a co-variance

matrix , a cofactor matrix Q, or a weight matrix W. In the subsequent derivation, the

cofactor matrix will be used.There are two possible procedures for deriving the least squares estimate of the parametervector . The first was essentially followed in Example 9-1 in which the condition

equations combining observations and parameters are transformed into the form of indirectobservation adjustment. For the sake of completeness, this procedure is summarized here.

Let

Ac (9-9)

represent a vector of c equivalent observations, each of which is a linear combination of

the n original observations; and let

Avvc (9-10)

be the corresponding c residuals. If Q is the cofactor matrix of the given observations, then

cQ , the cofactor matrix of the equivalent observations, is, according to Eq. (6-27), given

by

c,nn,nn,cc,c

AQAQ t

c (9-11)

In terms of the equivalent observations, the condition equations given by Eq. (9 -8) become

dBvcc or (9-12)

.f dBv cc

If the weight matrix of the equivalent observations is

1t1cc AQAQW (9-13)


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then the normal equation matrices for the conditions expressed by Eq. (9-12) are,according to Eqs. (4-28) and (4-29),

BAQABBWB N 1ttct

(9-14)

and

f AQABf WBt 1ttct

(9-15)

and the least squares estimates of the parameters are, from Eq. (4-38),

.t N 1 (9-16)

The reader should recognize that Eq. (9-13) is identical to Eq. (4-53) in Chapter 4, and

should understand that the use of the symbols cQ and cW in Chapter 4 was justified since

they, indeed, represent cofactor and weight matrices, respectively, of the set of equivalent observations.In the second derivation procedure we shall apply the minimum criterion directly. Recallfrom Chapter 4 that the minimum criterion for observations with a weight matrix W is

.imunminWvv t (9-17)

Although in Chapter 4 we restricted ourselves to uncorrelated observations for the sake ofsimplicity, this expression applies to all cases regardless of the structure of the weight

matrix. Thus, W may be a full matrix, for which case the observations are correlated with

equal or unequal precision; or it may be a diagonal matrix for uncorrelated observationswith unequal precision; or it may be a scalar or identity matrix reflecting uncorrelatedobservations with equal precision.

The linear, or linearized, condition equations are

.f BAv (9-18)

When they arc originally linear, then from Eq. (9-8)

.Adf (9-19)

The more common case is to have the conditions nonlinear with c functions of the generalform

0x,FF (9-20)

in which is the vector or n observations and x the vector of u parameters. Thus, the

linearization of Eq. (9-20) leads to Eq. (9-18), with


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,0x,Ff ,

x

FB,

FA

(9-21)

where the three matrices A, B, and f are evaluated at the numerical values for the

observations and a set of u approximate values 0x for the unknown parameters.

In a manner similar to the technique of adjustment of observations only (see Section 4.4), avector k of c Lagrange multipliers is used. The minimum criterion then becomes

.imunminf BAvk 2Wvv' tt (9-22)

To achieve a minimum, the partial derivatives of ' with respect to both v and must be

equated to zero, or

0Ak 2Wv2

v

' tt

,0Bk 2' t

which after transposition and rearrangement become

k AWv t (9-23).0k Bt (9-24)

Solving Eq. (9-23) for v, we get

,k QAk AWv tt1 (9-25)

and substituting into Eq. (9-18), we get

,f Bk AQAt

which, in view of Eq. (9-11), becomes

.Bf k Qc (9-26)

Solving Eq. (9-26) for k yields

,Bf WBf Qk c

1

c (9-27)

Finally, substituting into Eq. (9-24) gives

0Bf WB ct


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or

,f WBBWB ctct (9-28)

i.e., ,t N (9-29)where

BAQABBWB N 1ttct

and

,f AQABf WBt 1ttct

which are precisely the relationships given by Eqs. (9-14) and (9-15). Equation (9-16) can

then be used to solve for .

If the condition equations are nonlinear, the vector estimate of the parameters is

.xx̂ 0 (9-30)

The vector of residuals. v, is obtained by substituting the righthand side of Eq. (9-27) for kin Eq. (9-25):

.Bf WQAv ct (9-31)

The vector of adjusted observations, ̂ , is obtained by adding v to the observation vector,

, i.e.,

vˆ (9-32)

EXAMPLE 9-2

Using the data of Example 9-1, calculate the adjusted coordinates of the three points.

Solution

.cm

08.004.0

08.0

04.0

10.0

04.0

andcm09.0 222

0


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Thus

.

889.0

444.0

889.0

444.0111.1

04.0

1Q

2

0

Values for the final iteration are

14518.00000

0014518.000

000014518.0

A

0321.0

0643.00393.0

f ,

16

1412

B

0

0

0

,

100

010

008151.0

Wc

Thus

cm

03.0

01.0

06.001.0

04.0

01.0

f QABf WQAv t

c

t

and

,cm

97.401.6

06.4

99.3

16.3

01.2

03.001.0

06.0

01.0

04.0

01.0

00.500.6

00.4

00.4

20.3

00.2

vˆ

i.e., the adjusted coordinates of the three points are


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POINT x(cm) y(cm)

1 2.01 3.16

2 3.99 4.06

3 6.01 4.97

The reader can verify that these adjusted positions lie on the line .26.2x452.0y

In addition to the estimates themselves, it is equally important to calculate the precision ofthe estimates. This is discussed in the following section.

9.3. PRECISION ESTIMATION

To derive Q the cofactor matrix of the parameter estimates, , we first substitute Eq. (9-

15) into Eq. (9-16) to get

.f WB Nc

t1 (9-33)

The vector f is then replaced by Ad , according to Eq. (9-19), to give

.AdWB N ct1

(9-34)

The only random vector in Eq. (9-34) is . Its cofactor matrix is Q. Thus, the Jacobian of

with respect to is

AWB NJ ct1

(9-35)

and from the propagation law expressed by Eq. (6 -28), we get

tQJJQ

tct1ct1 AWB NQAWB N 1

c

t

c

t1 BNW)AQA(WB N

, N 1 (9-36)

since cW , and N are symmetric and, from Eqs. (9-13) and (9-14), tAQA

1

cW

, and

BcWt

B N , respectively.

If the condition equations are originally linear, then is the vector of estimated parameters, and Q , given by Eq. (9-36), is the corresponding cofactor matrix. If,

however, the condition equations are nonlinear, they are linearized by a series expansion


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(see Chapter 2), and is a vector of parameter corrections to be added to the

approximations0x . It is important, then, to derive the cofactor matrix xxQ for the final

vector of estimates x. Equation (9-30) indicates that the final estimate is the sum of 0x ,

and . If 0x , is taken as the parameter approximation at the beginning of the last iteration

of the least squares solution and is the final correction, then

, NQQ 1xx

(9-37)

because 0x , can be regarded as a vector of numerical constants the components of which

have been determined by all iterations preceding the last.

Equation (9-37) shows that the cofactor matrix of the parameters estimated by leastsquares turns out to be simply the inverse of the normal equations coefficient matrix N.Thus, the precision of the parameter estimates is a byproduct of the calculation of the

parameter estimates themselves.While Eq. (9-37) gives the cofactor matrix, or relative covariance matrix, of the estimated

parameters, it is more often necessary to find the absolute precision of the parameter

estimates, i.e., their covariance matrix, . If the reference variance, 20 , is known

beforehand (a priori), then it is a straightforward matter to calculate or xx :

. NQ 12

0

2

0xx

(9-38)

EXAMPLE 9-3

With reference to Example 9-1, calculate the covariance matrix for the two parameter

estimates â and b̂ .

Solution

From Example 9-1, the cofactor matrix of the parameter estimates is

.7219.25729.0

5729.01387.0 NQQ

1

1xx

The a priori reference variance 2

0 is 0.09 2cm .Thus

.2450.00516.0

0516.00125.0 N

12

0xx

If, however,2

0 is not known beforehand, an estimate for it, 2

0̂ , can be calculated a

posteriori from the results of the adjustment:

,r

Wvv

uc

Wvvˆ

tt2

0 (9-39)


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where

v is the vector of observational residuals;

W is the a priori weight matrix of the observations;c is the number of condition equations;u is the number of parameters;

r is the redundancy (degrees of freedom) which is equal to 0nn , 0n , being the total

number of observations requiered to define the underlying model uniquely.*

The a posteriori covariance matrix of the parameter estimates is then

. NˆQˆ 1202

0xx

(9-40)

The a priori value of 2

0 , if given, can be statistically tested by using 2

0̂ in place of 2S and r

in place of 1n in Section 8.11, Chapter 8. When 2

0̂ is consistent with 2

0 , the latter should

always be used in calculating covariance matrices such as xx .This is because the value

of 2

0̂ ; is only one estimate from one data set with limited redundancy, while 2

0 presumed

to be far better known. Should 2

0̂ turn out to he inconsistent with 2

0 , then several steps are

taken to determine the reason. [This topic is outside the scope of this book; for thoseinterested, see Mikhail (1976)].

To evaluate the quadratic form, Wvtv , we use the relationship Bf cW

tQAv of Eq.

(9-31). Thus,

Bf WQAWBf WQAWvv ctt

c

tt

which reduces to

tc

tt f f Wf Wvv (9-41)

if we note that Q and cW are symmetric, that1

WQ

, and that

tAQA

1

cW

, BcWt

B N , f cWt

Bt , and t1

N

, from Eqs. (9-13) through (9-16), re-

spectively.When the condition equations are nonlinear, the iterative procedure is usually carried outuntil the final value of is so small as to be essentially zero. Hence, for the nonlinear case,

Eq. (9-41) reduces to

.f Wf Wvvc

tt (9-42)

*It is interesting to note that when 10n n,= 1 and 1W , En. (9-39) reduces to

1n

2v

1n

vtv2

0ˆ

the familiar expression for sample variance.


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Two other cofactor matrices are of interest — vvQ , the cofactor matrix of the residuals;

and, more important,ˆˆ

Q the cofactor matrix of the adjusted observations.

To derive vvQ we note from Eqs. (9-27), (9-15), and (9-19) that

tBNf Wk 1c f WBBNf W ct1c

,AdWBBNIW ct1c (9-43) and from Eq. (9-25) that

.k QAv t Applying the general law of cofactor propagation to Eq. (9-43), and noting that d is avector of constants, we obtain

tct1cct1ckk AWBBNIWQAWBBNIWQ which reduces to ct1ckk WBBNIWQ (9-44) since cWtB1BNIcW

is idempotent.* Applying the general law of cofactor propagation

to Eq. (9-25), we obtain

ttkk tvv QAQQAQ AQWBBNIWQA ct1ct (9-45)

To deriveˆˆ

Q , we note from Eqs. (9-32), (9-25) and (9-43) that

vˆ

AdWBBNIWQA ct1ct ,cWBBNIWQAI ct1ct (9-46)

where dcWtB1BNIcWtQAc , a vector of constants. Applying the general law ofcofactor propagation to Eq. (9-46) we get

,AWBBNIWQAIQAWBBNIWQAIQ tct1ctct1ctˆˆ which reduces to

AQWBBNIWQAQQ ct1ctˆˆ (9-47)

*An idempotent matrix has the property that it is equal to its square.


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It is important to recognize from Eqs. (9-45) and (9-47) that

,QQQvvˆˆ

(9-48)

which is the same relationship as given by Eqs. (6-46) and (6-61).

EXAMPLE 9-4

With reference to the problem of Examples 9-1 and 9-2, calculate the covariance matrixfor the adjusted coordinates.

Solution

From Examples 9-1 and

889.0

444.0

889.0

444.0

111.1

04.0

Q

14518.00000

0014518.000

000014518.0

A

.7219.25729.0

5729.01387.0 N,

100

010

008151.0

W,

16

14

12

B 1

c

Thus,

8403.03217.01605.0

3217.03579.03212.0

1969.03941.08030.0

WBBNc

t1

and

1597.03217.01605.0

3217.06421.03212.0

1969.03941.01970.0

WBBNI ct1


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and

.

889.000

00.200

0889.00

000.2000111.1

0000.2

QAt

Using Eq. (9-45), the cofactor matrix of the residual vector v can be calculated:

AQWBBNIWQAQ ct1ctvv

.

126.0Symmetric

028.0006.0

254.0057.0507.0

057.0013.0114.0026.0

159.0036.0317.0071.0198.0

029.0006.0057.0013.0036.0006.0

From Eq. (9-48) we can get the cofactor matrix of the adjusted coordinates, ̂ :

,QQQ vvˆˆ

.

763.0Symmetric

028.0438.0

254.0057.0382.0

057.0013.0114.0418.0

159.0036.0317.0071.0913.0

029.0006.0057.0013.0036.0438.0

Finally, since 2

cm09.02

0 the covariance matrix of the adjusted coordinates is

.cm

0687.0

0025.00394.0

0229.00051.00344.0

0051.00012.00103.00376.0

0143.00082.00285.00064.00822.0

0026.00005.00051.00012.00032.00394.0

Q 2ˆˆ

2

0ˆˆ


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9.4. SPECIAL CASES

In Chapter 4, two specific techniques of least squares adjustment were presented. These

techniques are:

1. Adjustment of indirect observations.2. Adjustment of observations only.

After studying the general case covered in this chapter, it should be clear that the twotechniques of Chapter 4 arc comparatively simpler. They are, in fact, two special cases ofthe general technique, as was indicated in Section 9.2.

Adjustment of Indirect Observations

With reference to Section 9.2, this special case is achieved when u (the number of

parameters carried in the adjustment) is equal to its upper limit, 0n (the minimum number

of observations necessary for a unique determination), making c (the number of condition

equations) equal to its upper limit, n (the number of observations).According to Eq. (9-18), the initial condition equations can be expressed in general linearor linearized form as

,f BvA 000 (9-49)

where 0A , 0B , and 0f arc the initial matrix and vector inputs.

When cn , the 0A matrix becomes a square nn matrix. Furthermore, since the condition

equations are independent, 0A must be nonsingular, i.e., 0A , must have an inverse. Thus,

if Eq. (9-49) is premultiplied through by 1

0A

, and B and f are set equal to

0B1

0a

and

0f 1

0a , respectively, we get

,f Bv (9-50)

which is identical to Eq. (9-2), the form of the condition equations for adjustment ofindirect observations. The distinctive feature of this equation is that the coefficient matrixof v is an identity matrix.

With A replaced by the identity matrix, we have, from Eqs. (9-11), (9-13), (9-14), and (9-

15)

QIQIQc (9-51)

WQQW 11

cc

(9-52)

WBBBWB N tc

t (9-53)


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and

.Wf Bf WBt tc

t (9-54)

Equations (9-53) and (9-54) are identical to Eqs. (4-28) and (4-29), respectively,developed in Chapter 4.

When IA and 1QcW

, as they are in this special case, Eq. (9-31) reduces to

,Bf v (9-55)

which is identical to Eq. (4-32), an obvious rearrangement of the condition equation.

If N is calculated using Eq. (9-53), 1 NQ

, as given by Eq. (9-36), is identical to the

cofactor matrix for obtained in Eq. (6-43).

Finally, with IA and 1QcW

, Eq. (9-45) reduces to

t1

vv BBNQQ

(9-56)

which is identical to Eq. (6-44); and Eq. (9-47) reduces to

t1

ˆˆ BBNQ

(9-57)

which is identical to Eq. (6-45).Thus, in every respect, the method of least squares adjustment of indirect observations is a

special case of the general procedure.

Adjustment of Observations Only

With reference once more to Section 9.2, this special case is achieved when u is equal toits lower limit, zero, making c equal to its lower limit, r (the redundancy). With no

parameters carried in the adjustment, the B term in Eq. (9-18) vanishes, leaving

,f Av (9-58)

which is identical to Eq. (9-3), the form of the condition equations for adjustment ofobservations only.

To make the B term vanish we can simply set 0B . If this is done, Eq. (9-27) reduces to

f Wk c (9-59)

which is identical to Eq. (4-52). Equation (9-25), which is

k QAv t


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needs no reduction since it is already identical to Eq. (4-48).

As for the cofactor matrices, when 0B , Eq. (9-45) reduces to

AQWQAQ ctvv (9-60)

which is identical to Eq. (6-58), and Eq. (9-47) reduces to

AQWQAQQ ct

ˆˆ

(9-61)

which is identical to Eq. (6-60). Obviously, the relationship expressed by

Eq. (9-48).

,QQQvvˆˆ

is consistent with Eqs. (9-60) and (9-61).

This should make it clear that the technique of adjustment of observations only is also aspecial case of the general procedure.In principle, a problem that can be solved by the general method of adjustment can also be

solved by using the special procedures, provided appropriate transformations are made.The necessary transformations may be relatively complicated, however, and it is notadvocated that they be attempted. Instead, it is suggested that the most appropriate of thethree available techniques be selected to solve a specific adjustment problem.

Selection of the most appropriate technique is based on experience.

9.5. SUMMARY OF SYMBOLS AND EQUATIONS

The basic symbols and equations of the least squares method are summarized for quick

reference when solving adjustment problems.

Symbols

is the vector of observations,

̂ is the vector of adjusted observations,

v is the vector of residuals.x is the vector of parameters.

0x is the vector of approximate values for x,

is the vector of parameter estimates (linear case) or of parameter corrections

(nonlinear case).

x̂ is the vector of parameter estimates (nonlinear case),d is a vector of constants.

f is the condition equations constant terms vector.


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c is the vector of equivalent observations.

n is the number of given measurements, and thus the number of elements in , ̂ ,

and v.

0n is the number of observations necessary to specify uniquely the model that

underlies the adjustment problem.r is the redundancy, or the number of statistical degrees of freedom.u is the number of parameters carried in the adjustment, and thus the number of

elements in A, x, 0x , and x̂ .

c is the number of independent condition equations, and the number of elements

in d, f, and c .

A is a nc coefficient matrix for the observations.

B is a uc coefficient matrix for the parameters.

Q is the nn a priori cofactor matrix of .

W is the nn a priori weight matrix of .

cQ is the cc cofactor matrix of c .

cW is the cc weight matrix of c .

N is the uu coefficient matrix of the normal equations.

t is the 1u vector of "constants" in the normal equations.

k is the 1c vector of Lagrange multipliers.

Q , xxQ , vvQ and ˆˆQ are the cofactor matrices of , x, v, and ̂ , respectively.

2

0 is the reference variance (a priori value).

2

0̂ is the least squares estimate of the reference variance (a posteriori value).

, xx

, vv

and ˆˆ

are the covariance matrices of , x, v, and , respectively.

Equations, General Case

0nnr (9-4) ur c (9-5)

0nu0 (9-6)ncr (9-7)

The general linear or linearized form of the condition equations is

.f BAv (9-18)

where for the linear case

.Adf (9-19)


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and for the nonlinear case

,0x,Ff ,xF

B,

F

A

(9-21)in which

0x,FF (9-20)

The least squares solution is

t

c AQAQ (9-11)

1t1cc AQAQW (9-13)

BAQABBWB N 1ttct

(9-14)

f AQABf WBt 1ttct (9-15).t N

1 (9-16).xx̂ 0 (9-30)

.Bf WQAvc

t (9-31)

.vˆ (9-32)

The cofactor matrices are

1 NQ

(9-36)

1

xx NQ (9-37)

AQWBBNIWQAQQ ct1ctˆˆ (9-47) .QQQ

vvˆˆ

(9-48)

The estimate of the reference variance is

,r

Wvv

uc

Wvvˆ

tt2

0 (9-39)

where

.f f Wf Wvv t

c

tt (9-41)

The covariance matrices are

12

0

2

0xx NQ

, if 20 is know a priori (9-38)


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12

0

2

0xx NˆQˆ

, if 20 is know a priori; (9-40) ˆˆ

2

0ˆˆ Q

, if 20 is know a priori

ˆˆ

2

0Q̂ , if2

0 is know a priori.

Equations, Special Case-Adjustment of Indirect Observations

0nu Upper limit (9-6).nc Upper limit (9-7)

Then condition equations are of the form

,f Bv (9-50)


WQQW 11

cc

(9-52)

WBBBWB N tc

t (9-53)

Wf Bf WBt tc

t (9-54)

.t N 1 (9-16)

)casenonlinear (.xx̂ 0 (9-30)

,Bf v (9-55)

vˆ (9-32)


1 NQ

(9-36)

, NQQ 1xx

(9-37)t1

vv BBNQQ (9-56)

t1

ˆˆ BBNQ

(9-57)

Equations, Special Case-Adjustment of Observations Only

0u Lower limit (9-6).r c Lower limit (9-7)


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The r condition equations are of the form

.f Av (9-58)


t

c AQAQ (9-11)

1

cc QW (9-13)

f Wk c (9-59)

,k QAv t (9-25)

.vˆ (9-32)


AQWQAQc

tvv (9-60)

.QQQvvˆˆ

(9-48)

PROBLEMS

9-1 Figure 9-1 depicts a plane isosceles triangle ABC. The sides and height of the

triangle are measured; the observations are 1 , 2 , and 3 , as shown. If the base x is the

only parameter to be carried in the adjustment, give the model elements

Fig. 9-1.


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( ,u,r ,0n,n and c) and write the condition equations in the linearized form, f BAv .

9-2 In Fig. 9-2, distances OA, AB, BC, and CO are observed. The observed values

are 1 , 2 , 3 , and 4 , respectively. All angles are held fixed. It is required to find leastsquares estimates for the coordinates of C ( 0x and 0y ). (a) Write suitable condition

equations for this problem in the form f Av . (b) Write the condition equations in the

form f BAv , carrying x, and y, as the parameters.

9-3 Two sides and the three angles of the triangle in Fig. 9-3 are measured. The

measured values are 1 , 2 , 1 and 2 , as shown. Least squares estimates for side x andaltitude h are required. Write suitable condition equations for this problem in the form

f BAv , carrying x and h as the parameters.

9-4 If in Problem 9-1 m00.10001 , m10.10002 and m25.8003 , all uncorrelated

and with equal precision, find the least squares estimate for x

Fig. 9-2.

Fig. 9-3.


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9-5 If in Problem 9-2 m20.10001 , m55.5002 , m75.7073 ,and m60.11184 ,

and the covariance matrix for t4321 is

,cm

1005000

5010000

00200100

00100200

2

find the least squares estimates for 0x and 0y Also determine the principal dimensions and

orientation of the standard error ellipse for the computed position of C.

9-6 With reference to the triangle in Problem 9-3, 00'00"401 , 00'00"952 ,

00'30"45

3 45°00'30", m00.1000

1

and m00.15502

. The observations are uncor-

related, the standard deviation of each observed angle is 15", and the standard deviation of

each observed side is 0.10 m. Find least squares estimates for x and h. Evaluate also thecovariance matrix for x and h, and construct 95% confidence intervals for x and h.9-7 With reference to Fig. 9-4, the following angles are measured:

00'00"10AOBANGLE1

00'00"8BOCANGLE2

00'07"18AOCANGLE3

00'00"5CODANGLE4

00'00"12DOCANGLE5

00'12"25BOEANGLE6

Fig. 9-4.


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The covariance matrix for the observation vector is

Symmetric

.arcof ondssec

4

24

024

0024

00024

000024

2

Find the least squares estimate for angle COD. Construct also a 99% confidence interval

for the angle COD.9-8 The following data are observed:

x(m) y(m)

1.00 2.952.05 3.054.10 3.604.85 4.30

8.00 4.809.10 5.25

All observations (x and y) are independent and have the same precision. Find least squares

estimates for the parameters of the straight line baxy fitted to the data. Find also thea posteriori estimate for the reference variance and use this value to estimate the standarddeviation of each observation and the standard deviations of the estimates for a and b. Test

at the 2% level of significance the hypothesis that b= 2.70 m.9-9 Rework Example 48, Chapter 4, using the time as an observed variable as well as

the altitude, with standard deviations 1.0s and 20 ", respectively.9-10 Calculate the covariance matrix for the least squares estimates of the maximum

altitude and the time at which maximum altitude occurs, as determined in Problem 9-9.Evaluate also the standard deviations and coefficient of correlation for these least squaresestimates.9-11 The angles shown in Fig. 9-5 are measured with a theodolite. The observed

values and their weights are

ANGLE OBSERVED VALUE WEIGHT

a 70°14'30" 1

b 62°27'14" 2

c 103°38'26" 2

d 61°52'04" 1e 109°10'04" 1

f 76°56'36" 1


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Fig. 9-5.

(a) Use the method of least squares to determine adjusted values for these angles. (b)

Construct 95% confidence intervals for angles b and f.

9-12 The following data are given for the level net in Fig. 9-6

LINE FROM TO OBSERVED

ELEVATIONDIFFERENCE(m)

DISTANCE

(Km)

1 A B +34.090 2.90

2 B C +15.608 6.153 C A -49.679 17.324 A D +16.010 15.845 D B +18.125 9.22

6 D C +33.704 10.50

Fig. 9-6.


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The elevation of A is fixed at 324.120 m above mean sea level. The variance of eachelevation difference is directly proportional to the distance.(a) Determine least squares estimates for the elevations of B, C, and D and evaluate the

cofactor matrix for these estimates. (b) Evaluate the a posteriori reference variance andconstruct 95% confidence intervals for the elevations of B, C, and D.


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Applications

in Plane Coordinate

Surveys10.1. INTRODUCTION

Many survey projects are based upon two-dimensional positioning within a plane

rectangular coordinate system. This chapter addresses the application of least squaresadjustment in such plane coordinate surveys. It includes formulation and linearization ofthe three basic condition equations (distance, azimuth, and angle) encountered in theadjustment of plane coordinates by the method of indirect observations (also known as the

method of variation of coordinates), least squares position adjustment for a typical procedure employed in plane coordinate surveying, and least squares transformation ofcoordinates from one plane rectangular system to another_

10.2. THE DISTANCE CONDITION AND ITS LINEARIZATION

The adjusted distance, ijŜ , between two points i and j is given by

,

212

iY

jY

2

iX

jX

ijŜ

(10-1)

where )iY,iX( and ) jY, jX( are the rectangular plane coordinates of i and j, respectively.

This is the distance condition.Linearization of Eq. (10-1) according to Eq. (2-18) is

jY

jY

ijS

jX

jX

ijS

iY

iY

ijS

iX

iX

ijS

0ijS

ijŜ

(10-2)

where

,212

0iY0

jY

20i

X0 jX

0

ij

S

(10-3)

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noting that )0

iY,

0

iX( and )

0

jY,

0

jX( are approximate values for the coordinates of i and j,

respectively.

Equation (10-2) assusmes that all four coordinates are unknown variables for whichapproximations are necessary. If one of the two points is a control point, the coordinates ofthis point Would be known and would thus be taken as constants. In this case, Eq. (10-2)would include only two partial derivatives; namely, the derivatives with respect to the two

coordinates of the unknown point, For example, if point j is a control point, Eq. (10-2)reduces to

iY

iY

ijS

iX

iX

ijS

0ijS

ijŜ

(10-4)

in which

,

212

0

i

Y

j

Y2

0

i

X

j

X0

ij

S

(10-5)

noting that ) jY, jX( are the known coordinates of the control point j, and )0

iY,

0

iX( are

approximate values for the coordinates or the unknown point i. For the remainder of thissection we will consider the general case of having all four coordinates as unknowns.

Now, according to Eq. (9-32), the adjusted distance is

,vSŜijijij

(10-6)

whereij

S is the observed value of the distance andij

v is the corresponding residual. Thus, it

follows from Eqs. (10-2) and (10-6) that

.ijS

0

ij

S jY

jY

ijS

jX

jX

ijS

iY

iY

ijS

iX

iX

ijS

ijv

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