THE SIGN TEST
Where an analysis of data does not meet the basic
parametric assumptions of the "t" test we can use the sign test. The
sign test merely counts the number of cases in one group who exceed
their matched partners and compares this with the number of persons in
the second group who exceed their matched partners.
EXAMPLE
Cottage values in two suburbs were subdivided into 10
different architectural styles. The average value for each style is
matched between the 2 suburbs as follows. Therefore, after adjusting
each suburb's housing stock according to style, is there a difference
in value between the 2 suburbs?
AVERAGE VALUE PER ARCHITECTURAL STYLE
SUBURB

SUBURB 
SIGN 
A

B 
(AB) 
21

16

+ 
10

14

 
14 
8

+ 
21

13 
+ 
28

10

+ 
19

19

0 
14 
17 
 
12 
11 
+ 
11

13 
 
18

18

0 
If the groups had changed about equally, the pluses
and minuses will be randomly distributed around a median of 0. The null
hypothesis is therefore, that the median difference = 0. If there are
considerably more of one sign than the other, the distribution of
differences is clearly not random and the hypothesis of equal change in
the 2 groups must be rejected.
TESTING WITH <=10 PAIRS
H0 is tested when there are 10 or fewer cases by use
of the binomial expansion at probability of 0.5 and n = to the number
of pairs observed. In the above table there are 10 matched pairs, 5 of
which are pluses, 3 minuses and 2 zeros. The zeros are disregarded and
n=8. By chance we would expect 4 pluses and 4 minuses from the 8 non
zero pairs. We find the probability of getting 5 pluses in a binomial
expansion by reading the values from Pascal's triangle.
See pascal's triangle
The line in Pascal's triangle where n=8 reads
1,8,28,56,70,56,28,8,1 which sums to 256 individuals. Since the median
(70) is the point where p=.50 that is, 4 pluses out of 8 we move to the
right and find that 56 would represent the times out of 256 we would
expect to get 5 pluses out of 8, 28 times would give 618, 8 times would
get 718 and 1 time would 818.
Therefore, to determine the probability of getting 5 or more out of 8 = (56+28+8+1)/1256 = 0.36.
At .05, 0.36 is not significant and therefore, we
accept the null hypothesis and the 5 pluses out of 8 could have easily
been by chance alone.
PASCAL'S TRIANGLE
n FREQUENCIES OF COMBINATIONS sum
1 1 1 2
2 1 2 1 4
3 1 3 3 1 8
4 1 4 6 4 1 16
5 1 5 10 10 5 1 32
6 1 6 15 20 15 6 1 64
7 1 7 21 35 35 21 7 1 128
8 1 8 28 56 70 56 28 8 1 256
TESTING WITH >10 PAIRS
For more than 10 pairs the normal curve can be used
as an approximation of the probabilities. The necessary Z scores are
found with a mean of 0.5n and a STDEV of square root of N90.25). The Z
score for a given number of pluses(X) is:
Z = ((X+I#0.5)#0.5n)hIN(o.25)
If the number of pluses > 0.5n then use X0.5 and
if the number of pluses <0.5n then use X+0.5 in computing Z. This
procedure corrects for the discontinuity of the data. Having computed
the Z score, the valuer can consult the table of areas under the normal
curve (see Z value table) to determine the proportion of samples that
would have more pluses than our obtained number. Either a 1 or 2 tailed
test may be applied depending on the nature of the hypothesis.
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