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Fisher’s
Exact Test
·
Does not rely on Normality assumption.
·
Uses “exact” distribution instead of a Normal approximation.
·
Use in place of χ^{2} test when any expected cell
frequency is less than 5.
Example: caries incidence

Caries 



yes 
no 
total 
control 
10 
26 
36 
intervention 
6 
62 
68 
total 
16 
88 
104 
The nullhypothesis of Fisher’s Exact test is that there is no relationship between the two categories. Thus, every possible arrangement of observations in the respective cells is equally likely (we assume the row and column totals are fixed).
The
pvalue is computed by calculating the number of possible arrangements of
observations that produce tables that are more extreme than the observed and
then dividing this by the total number of possible arrangements of the
observations.
Example: Caries incidence
observed
table 
Caries 


yes 
no 
total 
p̂_{c} p̂_{i}= 
0.190 

control 
10 
26 
36 

intervention 
6 
62 
68 
probability
under H_{0} 

total 
16 
88 
104 
p_{H0}= 
0.01048 
Tables more extreme (result in greater
difference in proportions) 

11 
25 
36 
p̂_{c} p̂_{i}= 
0.23 
15 
21 
36 
p̂_{c} p̂_{i}= 
0.40 

5 
63 
68 
1 
67 
68 

16 
88 
104 
p_{H0}= 
0.00236 
16 
88 
104 
p_{H0}= 
0.00000 

12 
24 
36 
p̂_{c} p̂_{i}= 
0.27 
16 
20 
36 
p̂_{c} p̂_{i}= 
0.44 

4 
64 
68 
0 
68 
68 

16 
88 
104 
p_{H0}= 
0.00038 
16 
88 
104 
p_{H0}= 
0.00000 

13 
23 
36 
p̂_{c} p̂_{i}= 
0.32 
0 
36 
36 
p̂_{c} p̂_{i}= 
0.24 

3 
65 
68 
16 
52 
68 

16 
88 
104 
p_{H0}= 
0.00004 
16 
88 
104 
p_{H0}= 
0.00055 

14 
22 
36 
p̂_{c} p̂_{i}= 
0.36 
1 
35 
36 
p̂_{c} p̂_{i}= 
0.19 

2 
66 
68 
15 
53 
68 

16 
88 
104 
p_{H0}= 
0.00000 
16 
88 
104 
p_{H0}= 
0.00602 

Total probability of all as or more
extreme tables = 
0.01985 
Formula for Probability of
Table in Fisher’s Exact Test

Probability _{} 
SPSS
output
McNemar’s Test for Proportions (Paired
Data)
Example: Change in plaque index
Fiftythree
study participants assessed twice for plaque index (PI), at baseline and 4
weeks later. We wish to assess whether
the proportion of patients with high PI changes.

PI at 4 weeks 

PI at baseline 
low 
high 
low 
29 
1 
high 
13 
10 
Incorrect methods:
1. Comparing _{} with _{} using the Ztest: This will not give a valid pvalue because
it does not compare independent samples.
The same 53 people are used in each proportion.
2. Performing a Chisquare or
Fisher’s Exact test on the above 2×2 table: These would test whether the proportion of
high’s at baseline is related to the proportion of high’s at 4 weeks. They would not test whether or not the
proportions are different.

PI at 4 weeks 

PI at baseline 
low 
high 
low 
29 
1 
high 
13 
10 
McNemar’s
Test assesses the null hypothesis
H_{0}: P(PI
high at baseline) = P(PI high at 4 week),
by
noting that it is equivalent to:
H_{0}: 
P(PI changes high to low) = P(PI changes low to high), 

for all discordant pairs. 
The discordant pairs
are those that have different values for the two observations. Note that each entry in the table is the
number of pairs.
The latter H_{0} can
be evaluated using a onesample test for proportions with,
where p = proportion of discordant pairs that increase.

PI at 4 weeks 

PI at baseline 
low 
high 
low 
29 
1 
high 
13 
10 
·
If n > 20 (where n
is # of discordant pairs) can use Ztest for proportions (chapter 9.3).
·
If n < 20 (as in
the current example, n = 14) use the
binomial distribution to compute the exact pvalue.
Let X = number of discordant
pairs that increase, which, under H_{0}, is binomial(n
= 14, p = 0.5).
The twosided pvalue is the
probability that we would see a more unbalanced sample of the discordant pairs
than 13 vs 1, which is
P(X
< 1) + P(X > 13)
= P(X=0) + P(X=1) + P(X=13) +
P(X=14)
= 0.0001 + 0.0009 + 0.0009 + 0.0001
= 0.0020
SPSS output
Analysis of Categorical Data
Summary
Ø Proportions from two
independent samples
Þ
Large samples – Ztest for proportions
Þ
Small samples – Fisher’s Exact Test
Ø Proportions from > 2 independent
samples
Þ
Chisquare test
Ø Proportions from paired data
Þ
McNemar’s Test