Statistical Analysis for Education and Psychology Researchers

Table 7.2: Original and rank scores for ten schools

SCHOOL per
cent

%FSME Rank %CAT Rank

A 24.0 7 1.5 2.25 24.1 7 1.5 2.25
B 24.3 8 2.5 6.25 21.8 5 −0.5 0.25
C 15.3 5 −0.5 0.25 23.3 6 0.5 0.25
D 40.8 9 3.5 12.25 36.4 10 4.5 20.25
E 10.7 2 −3.5 12.25 8.4 1 −4.5 20.25
F 6.3 1 −4.5 20.25 13.1 2 −3.5 12.25
G 23.1 6 0.5 0.25 35.1 8 2.5 6.25
H 45.0 10 4.5 20.25 36.0 9 3.5 12.25
I 12.9 3 −2.5 6.25 17.7 3 −2.5 6.25
J 13.9 4 −1.5 2.25 18.5 4 −1.5 2.25
Σ=82.5 Σ=82.5

3 The denominator for equation 7.1 is given by (82.5^2 )0.5=82.5 and the numerator,

(the products summed) is given by:

SCHOOL

A 2.25

B −1.25

C −0.25

D 15.75

E 15.75

F 15.75

G 1.25

H 15.75

I 6.25

J 2.25

Σ=73.5

4 Using formula 7.1, rs=73.5/82.5=0.89.

If there are tied ranks, the rank value given to each member of the tied group is the
average of the ranks which would have been assigned if there were no ties. For example,
if schools E and F both had FSME scores of 10.7, the rank value assigned to each would
be 1.5 (the average of the 1st and 2nd ranks ie (1+2)/2).

Interpretation

If the ten schools represent a random sample from a population of schools in an education
authority, the null hypothesis of no relationship between FSE and CAT scores in the
population can be tested. Looking at the scatterplot shown in Figure 7.2 a one-tailed
alternative hypothesis would seem reasonable. That is FSE and CAT scores are positively
related. The probability associated with the observed rs 0.89 is obtained by treating the
value, ((n−2)0.5×rs) /(1−rs^2 )0.5 as coming from a t distribution with n−2 df where rs is the

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