P 1 >P 2 , H 1 : P 1 <P 2 and H 1 : P 1 ≠P 2. In most situations the two-tailed test is the appropriate
test. In this example a two-tailed alternative hypothesis is considered; Does the younger
age cohort (≤6 years) have a significantly larger or smaller proportion of left-handed
subjects with upper limb injury? Fishers’s exact test determines the probability under the
null hypothesis of obtaining the observed frequencies, or more extreme distributions of
cell frequencies, with the same (fixed) marginal totals.
1 The first step is to identify all possible frequency distributions with the same fixed
marginal totals as the observed frequency counts (A+B=4; C+D=6; A+C=6; and
B+D=4). The number of possible tables is equal to the smallest marginal frequency
plus 1 if none of the cells are 0. In this example there are 5 possible tables (4+1).
These are numbered (i) to (v) according to the frequency of cell A:
Table (i) Table (ii) Table (iii)
0 4 |4 1 3 |4 2 2 |4
6 0 |6 5 1 |6 4 2 |6
6 4 |10 6 4 |10 6 4 |10
Table (iv) Table (v)
31|4 40|4
33|6 24|6
6 4 |10 6 4 |10
2 Select an appropriate alpha level, for example, 0.05 (two-tailed).
3 The probability of each possible frequency distribution, with the same fixed marginal
totals is then evaluated using equation 6.5.
As a check, the total probability of p(i) to p(v) should sum to 1.
p(i)=(4!×6!×6!×4!)/ (10!×0!×4!×6!×0!)=0.0048
p(ii)=(4!×6!×6!×4!)/ (10!×1!×3!×5!×1!)=0.1143
p(iii)=(4!×6!×6!×4!)/ (10!×2!×2!×4!×2!)=0.4286
p(iv)=(4!×6!×6!×4!)/ (10!×3!×1!×3!×3!)=0.3810
p(v)=(4!×6!×6!×4!)/ (10!×4!×0!×2!×4!)=0.0714
Total=p(i)+p(ii)+p(iii)+p(iv)+p(v)=1.0001
4 For a two-tailed test, the probability is the sum of the probabilities of all tables with a
probability less than or equal to the probability of the observed table. That is (p(i)
0.0048+p(ii) 0.1143+p(iv) 0.3810+p(v) 0.0714) p=0.5715.
For one-tailed tests the right tail probability is the sum of the probabilities of all
tables more extreme and including the probability of the observed table (when
tables are arranged in order of magnitude from cell A minimum frequency to cell
A maximum frequency). The right-tail probability is therefore (p(iv) 0.3810+p(v)
0.0714)=0.4524. The left-tail probability is the sum of the probabilities of all
tables less extreme and including the probability of the observed table, that is
(p(iv) 0.3810+p(iii) 0.4286+p(ii) 0.1143+p(i) 0.0048)=0.929.
Statistical analysis for education and psychology researchers 182