WASON DATA ANALYSIS
The raw data shows how many correct and incorrect responses there were in each version of the Wason
card sorting task (i.e. Abstract, Concrete) for the two conditions involving task order (i.e., AbstractConcrete, Concrete-Abstract). Each condition consists of two tasks and the responses are shown in
relation to task.
Remember that we have not provided the values for the concrete task in the abstract-concrete condition,
as these values were unrelated to our hypotheses
Main Table
Task Order
Abstract - Concrete
Concrete - Abstract
Hypothesis 1: Table
Observed values
First Problem
Abstract
Concrete
TOTAL
Hypothesis 2: Table
Observed values
Abstract Problem
First (Before Concrete)
Second (After Concrete)
TOTAL
N
Abstract Task
Concrete Task
Correct
Incorrect Correct Incorrect
90
8
82
-90
11
79
33
-57
# Correct # Incorrect
TOTAL
First Problem
Abstract
Concrete
% Correct
% Incorrect
# Correct # Incorrect
TOTAL
Abstract Problem
% Correct
First (Before Concrete)
Second (After Concrete)
% Incorrect
For each hypothesis there are two tables to fill out which can then be used for conducting a chi square
analysis.
HYPOTHESIS ONE
Is performance better on some versions of the Wason card sorting task than others the first time it is
performed?
Fill out the first table (under ‘observed values’) with how people responded to the first problem they
were faced with. E.g. In total 90 people had the abstract task first: 8 responded correctly and 82
incorrectly.
Put these values in the table and fill out the remaining conditions and totals. This table can now be used
to fill out the percentage table.
E.g. 8 out of total of 90 responded incorrectly- (8/90) x 100 = 8.8%
You should be able to see from your table that performance was better in the concrete version of the task
compared to the abstract version. Is this difference significant? A chi square test can be used to find out.
This is the Chi Square formula:

χ
∑
O
E
2
O  E 
2
2


E
= simply the symbol for chi-square.
= the sum of…
= observed values i.e. the actual responses
= expected values i.e. what we would expect if all things equal.
Each cell in the observed values table will have its own O and its own E. For each cell the observed
value is taken from the expected value, and then squared, this is then divided by the expected value. The
results from all the cells are then simply added together to give us a Chi square value.
It is useful to think of the cells in this way, and use a table like the one below to work out the formula:
First Problem
Abstract
Concrete
COLUMN TOTAL
O
# Correct # Incorrect
Cell 1
Cell 2
Cell 3
Cell 4
ROW
TOTAL
CELL 1
CELL 2
CELL 3
CELL 4
8
82
33
57
E
O-E
(O-E)2
(O-E)2/E
You must keep a record of your workings, and these must be included as an appendix in your lab report.
The best way to do this is to create a written version of the above table.
The observed values can be filled in, but the expected values must be calculated using the following
simple formula:
Row total x Column total
Overall total
E.g. Cell 1 = (90 x 41)
180
= 20.5
Calculate the expected value for each cell. Again you must keep a record of your workings either written
or in Excel. Fill in the table with these expected values and start working out the formula. When the
table is complete, simply add up the bottom row to reach your value of chi square.
We now must find the critical value of χ2 to compare our own to. If our value is smaller than the critical
value, then the results are due to chance and not significant. To get the critical value, you need to work
out the degrees of freedom based on our original table.
d.f. = (number of rows -1) x (number of columns -1)
Look up your degrees of freedom on the ‘table of critical values for Chi Square’ (on the Graham’s web
page). You should be able to see that your value exceeds the critical value at p = 0.05, p = 0.01, and
p = 0.001. It can be reported in one of these ways:
χ2 = 19.74, p<.001, df=1
χ2 (1) = 19.74, p<.001
This tells us that the difference in performance between tasks is significant, and that probability that the
results occurred by chance is 0.1%.
HYPOTHESIS TWO
Is performance on the abstract task affected if it follows a concrete scenario?
Follow the same steps for hypothesis two. Hypothesis two is concerned with the abstract task only and
whether individuals were helped in providing the correct answers for the abstract task if they first had to
complete a concrete task. Fill out the first table, convert these to percentages in the second table, and use
the second table to construct a graph.
Construct another table to work out the chi square formula, work out the degrees of freedom and look up
the critical value. Remember only if our value exceeds the critical value, are any differences in
performance (as shown by the graph) significant.
Abstract Problem
First (before concrete)
Second (after concrete)
COLUMN TOTAL
CELL 1
# Correct
Cell 1
Cell 3
CELL 2
O
E
O-E
(O-E)2
(O-E)2/E
χ2 (.....) = ........., p ..........
#
ROW
Incorrect TOTAL
Cell 2
Cell 4
CELL 3
CELL 4