RESEARCH METHOD COHEN ok

512 QUANTITATIVE DATA ANALYSIS
Box 24.10
Distribution of test scores
Test scores
Frequency
Valid 2 1 0.1
3 223 22.3
4 276 27.6
5 32 3.2
6 69 6.9
7 149 14.9
8 185 18.5
9 39 3.9
10 26 2.6
Total 1,000 100.0
Valid per cent
If we have interval and ratio data then, in
addition to the modal scores and cross-tabulations,
we can calculate the mean (the average) and the
standard deviation. Let us imagine that we have
the test scores for 1,000 students, on a test that
was marked out of 10 (Box 24.10).
Here we can calculate that the average
score was 5.48. We can also calculate the
standard deviation, which is a standardized
measure of the dispersal of the scores, i.e.
how far away from the mean/average each
score is (see http://www.routledge.com/textbooks/
9780415368780 – Chapter 24, file 24.6.ppt). It is
calculated, in its most simplified form (there being
more than one way of calculating it), as:
indicates that the scores are widely dispersed. This
is calculated automatically by software packages
such as SPSS at the simple click of a single button.
In the example here the standard deviation in
the example of scores was 2.134. What does this
tell us First, it suggests that the marks were not
very high (an average of 5.48). Second, it tells
us that there was quite a variation in the scores.
Third, one can see that the scores were unevenly
spread, indeed there was a high cluster of scores
around the categories of 3 and 4, and another
high cluster of scores around the categories 7 and
8. This is where a line graph could be useful
in representing the scores, as it shows two peaks
clearly (Box 24.11).
It is important to report the standard deviation.
For example, let us consider the following. Look
at these three sets of numbers:
(1) 1 2 3 4 20 mean = 6
(2) 1 2 6 10 11 mean = 6
(3) 5 6 6 6 7 mean = 6
If we were to plot these points onto three
separate graphs we would see very different results
(Boxes 24.12 to 24.14) (see http://www.routledge.
com/textbooks/9780415368780 – Chapter 24, file
24.7.ppt).
Box 24.11
Alinegraphoftestscores
√ ∑ d 2
300
SD =
where
N − 1
200
d 2 = the deviation of the score from the mean
(average), squared
Count
100
= the sum of
N = the number of cases
Alowstandarddeviationindicatesthatthescores
cluster together, while a high standard deviation
0
2
3
4
5 6 7
Test scores
8
9
10