Verify the normal distribution of residuals v; of check 
and control point coordinates for each model 3 Vj =1,2,... 
obtained by block adjustment. The condition is satisfied if 
the null hypothesis is not rejected: 
H, : v; y N(0, 0°) (7) 
To verify the null hypothesis apply the x? test against the 
theoretical normal distribution 
k 
H, : Y (f; - npjy![np; S x14-, (8) 
j=1 
where k is the number of subsets in which the sampled resid- 
uals are subdivided 
f; is the number of sampled residuals in the subset j 
n is the total number of sampled residuals 
p; is the theoretical probability (according to a Binomial 
distribution for an outcome inside the subset j 
r are the degrees of freedom 
a is the probability for an error of first kind 
In case of a normal distribution, f; will have a Binomial 
distribution with the theoretical mean np; and the variance 
np;(1 — pi) 
If the null hypothesis is not rejected apply the data snoop- 
ing test already proposed by F.Crosilla/G. Garlatti (1991) 
for digital cartography. Let z;, and x; be the x-coordinates 
of point P; under control coming from photogrammetric (p) 
and cartographic (c) procedures respectively: 
Tip = N(Hizpy ap) 
Lie 7 N (Hisce, 02.) (9) 
Define the random variable 
Vi = Tip — Tic (10) 
following a Normal Distribution 
yi m N(Misp — Hines 02, + 02, — 20zpnc) (11) 
Once the probability « for a first kind error is accepted and 
the parametric space S is partitioned in the subspace of ac- 
ceptance (A) and rejection (R) 
A = [zp,fi € 5 : —za/2 € yifoy « za/2} 
R = (zpt0€S:y/oi € -zaf2 or yifoiy > za/2} 
(12) 
where 
za[2 : P[z « —zo/2] z p[z » za/2] 2 a/2. Vze S (13) 
than holds 
Hj x30 if rat. € À 
H, p if Vip; Tic € R a 
3.2 Non parametric hypothesis tests 
In case the normal distribution of f; is not accepted non 
parametric tests should be applied. The first question which 
arises is the symmetry behaviour therefore we have to verify 
the symmetry of the distribution of the sampled y; Vi = 
1,2,...,m (m number of points). The null hypothesis is for- 
mulated such that the mean value of distribution corresponds 
with the median value. 
576 
Let be the mean value of y; = ÿ Vi = 1,2,...,m and 
the median value of y; = yso Vi — 1,2,...,m Verify that 
ÿ = Yso. This can be done by the two-tailed Quantile Test 
for which the following null hypothesis is introduced: 
H,:the .50 population quantileis ÿ (15) 
or equivalently 
H,: P(Y € Y) > .50quantile and P(y < ÿ) € .50quantile 
(16) 
in which Y has the same distribution as the sampled y;. 
For a decision rule let us introduce the quantities: 
T5 number of observations j 9) 
T, number of observations € j and 
T, — T if none of the observations m 
The critical region corresponds to values of T? which are 
too large, and to values of T, which are too small. This region 
is found by entering a table of the Binomial distribution with 
the sample size m and the hypothesized probability .50. We 
now have to solve the following problem: find the number f; 
such that 
Pla<H)=ar (17) 
where z has the binomial distribution with parameters m 
and .50, and where ay is about half of the desired level of 
significance. Then find the number tz such that 
PC > ta) = a 
or. Pla <a) = 1— 0 (18) 
where a, is chosen such that a; + az is about equal the 
desired level of significance (Note: o4 and az are not integer 
numbers). The final decision rule is given by 
EL, =f if FA; <H or T» 2 (19) 
otherwise accept H, with a significance level equals o, 4 a2. 
The symmetry of the distribution can also be verified by the 
Wilcozon Signed Rank Test which is reported in the follow- 
ing. In this context we will differentiate in two cases: 
Case 1: the condition of symmetry is accepted 
Verify that the median y so = 0. Then apply the two-tailed 
Wilcoxon signed rank test 
H, : y.so = 0 
Hi:yso = p 
If H, = 0 then it follows for the symmetry condition 
(20) 
ÿ=0 (21) 
and therefore systematic and gross errors are not present 
within the population with significance level œ. The test 
statistic T equals the sum of the ranks assigned to those 
values of y, > 0 - Vi 1,2, ..., m. 
Tu m (22) 
i=1 
where R; = 0 if y; < 0 and R; equals the rank asigned to 
positive values of yj. 
As a decision rule we have to reject H, at a level of sig- 
nificance o if T' exceeds w,_a/2 Or if T' is less than wa/2- If 
T is between wa/2 and w,_a/2 OF equal to either quantile,