Methods for Two Categorical Variables – Fisher’s Exact Test
Example: Suppose there are a total of 30 pine trees on a specific island; 15 are on the windward shore
and 15 on the lee shore. Of these trees, four on the windward shore are damaged, while all the trees on
the lee shore are healthy.
The lee shore line is shown in red (the
wind moves toward the shore), and the
windward shore is shown in light green.
Research Question – Is there evidence that there are more damaged trees on the windward
shore (which may indicate the wind or some other factor on the windward
shore has a negative impact) or is the observed difference simply due to chance?
The data are summarized in the table given below.
Windward shore
Lee shore
Total
Damaged
4
0
4
No Damage
11
15
26
Total
15
15
30
When we analyze data on two variables, the first step is to determine which is the __________________
Variable and which is the ___________________ (or predictor) variable.

Response variable: The __________________ variable on which comparisons are made.

Explanatory (or predictor) variable: This defines the ____________ to be compared.
Questions:
1. What are the two variables used in this study?
2. Which is the response variable? Which is the explanatory variable?
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Descriptive Methods for Two Categorical Variables

A ______________________________ shows the joint frequencies of the two categorical
variables. The rows of the table denote the categories for the first variable and the columns
denote the categories of the second variable.

A _________________________ is a visual representation of the relationship between two
categorical variables. A mosaic plot graphically represents the information given in the
contingency table.
To obtain these descriptive summaries we need to enter the data into JMP as follows.
Next, click Analyze Fit Y by X. Put the response variable (Damage) in the Y, Response box , the
explanatory variable (Shoreline) in the X, Factor box , and Count in the Freq box as shown below.
Click OK and JMP will return the following output.
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Recall, the research question for this study was…
Is there evidence that there are more damaged trees on the windward shore (which may
indicate the wind or some other factor on the windward shore has a negative impact) or is the
observed difference simply due to chance?
As always, to find the p-value we will find the probability of obtaining a sample at least as extreme as
was observed, assuming there is no difference between the two groups. That is, assuming that the
shore has no effect on whether trees are damaged.
To calculate the probability, we need only consider that we have 4 damaged and 26 healthy trees
overall. Furthermore, we assume that a damaged tree is equally likely to be from either shoreline (that
is, there is no difference in the risk of damage between the lee and windward shorelines). If we
randomly divide these 30 trees into two groups of size 15, what is the probability that all four damaged
trees will end up in the windward group?
Note that the number of damaged trees in the windward group can assume the following values: 0, 1, 2,
3, or 4. The probability of observing each of these outcomes is calculated under the assumption of no
difference in risk using a distribution known as the _____________________________ distribution and
is shown below.
Questions:
3. Assuming there is no difference in the risk of damage between the two shorelines, what is the
probability that all four damaged trees would come from the windward shoreline?
4. Can you find this probability in the JMP output below?
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Fisher’s Exact Test
Fisher’s Exact test is based on the probability of observing a table at least as extreme as the original
data. The hypothesis test is carried out as outlined below.
Step 0: Define the research question
Is there evidence that there are more damaged trees on the windward shore (which may
indicate the wind or some other factor on the windward shore has a negative impact) or is the
observed difference simply due to chance?
Step 1: Determine the null and alternative hypotheses
H0: There is no difference in the proportion of damaged trees on the windward vs. lee shorelines
Ha: The proportion of damaged trees on the windward shoreline is greater than the proportion
of damaged trees on the lee shoreline
We can also re-write the hypotheses using symbol notation.
H0: pwindward ≤ plee
or
pwindward = plee
Ha: pwindward > plee
Step 2: Finding the test statistic and p-value

There is _____ test statistic for this hypothesis test.
Step 3: Report the conclusion in context of the research question
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Example: The following table shows a sample of patients categorized with respect to two categorical
variables, congenital heart defect (present or absent) and karyotype (trisomy 21, also called Down
syndrome, or trisomy 13, also called Patau syndrome).
Karotype
Down syndrome
Patau syndrome
Total
Congenital Heart Defect
Present
Absent
Total
24
36
60
20
5
25
44
41
85
Research Question – Is there evidence that a congenital heart defect is found more commonly in
patients with one of the two karyotypes examined?
Step 0: Define the research question
Is there evidence that the proportion of Down syndrome patients with a congenital heart defect
is different from the proportion of Patau syndrome patients with a congenital heart defect?
Step 1: Determine the null and alternative hypotheses
H0:
Ha:
Step 2: Finding the test statistic and p-value
Step 3: Report the conclusion in context of the research question
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Example: An advertisement by the Schering Corporation in 1999 for the allergy drug Claritin mentioned
that in a pediatric randomized clinical trial, symptoms of nervousness were shown by 4 of 188 patients
on Claritin and 2 of 262 patients taking a placebo.
Step 0: Define the research question
Is there evidence that the proportion who experience nervousness greater for those taking
Claritin than for those who take the placebo?
Questions:
5. What is the response variable?
6. What is the explanatory variable?
7. Fill in the contingency table below using the above scenario.
Nervousness?
Drug
Claritin
Placebo
Total
Yes
No
Total
450
Step 1: Determine the null and alternative hypotheses
H0:
Ha:
Step 2: Finding the test statistic and p-value
Step 3: Report the conclusion in context of the research question
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Example: Prevention of deep vein thrombosis (DVT) is a critical issue in patients undergoing total hip
replacement surgery. Orthopedic surgeons recognize the importance of prophylaxis in the management
of their patients but do not agree on an optimal method. In this study, two different prophylaxis
methods are to be compared for the prevention of proximal DVT after total hip replacement surgery.
Patients undergoing total hip replacement were randomly assigned to one of the two prophylactics.
After surgery, it was noted whether patients had complications from proximal DVT or not. The results
are presented in the following contingency table.
Treatment 1 Treatment 2 Total
No Complications
72
68
140
DVT Complications
3
12
15
Total
75
80
155
Step 0: Define the research question
Is there evidence that that risk of DVT complications differs between the two prophylactic
treatments?
Step 1: Determine the null and alternative hypotheses
H0:
Ha:
Step 2: Finding the test statistic and p-value
Step 3: Report the conclusion in context of the research question
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Observational Studies vs. Designed Experiments
Reconsider the above example. Fisher’s exact test provided evidence that the risk of complications does
differ between the two treatments (p-value = 0.0281). Now, the question is this: can we conclude that it
really is something with the two treatments that causes the risk to be higher for Treatment group 2?
The answer to this question lies in whether the experiment itself was a designed experiment or an
observational study.
 Observational study  Involves collecting and analyzing data ___________________
randomly assigning treatments to experimental units.
 Designed Experiment  A treatment is ____________________imposed on individual subjects
in order to observe whether the treatment causes a change in the
response.
Key statistical idea:
The random assignment of treatments used by researchers in a designed experiment should balance out
between the treatment groups any other factors that might be related to the response variable.
Therefore, designed experiments can be used to establish a cause-and-effect relationship (as long as the
p-value is small).
On the other hand, observational studies establish only that an association exists between the predictor
and response variable. With observational studies, it is always possible that there are other lurking
variables not controlled for in the study that may be impacting the response. Since we can’t be certain
these other factors are balanced out between treatment groups, it is possible that these other factors
could explain the difference between treatment groups.
Note that the “DVT complications” study is an example of a designed experiment since participants were
randomly assigned to the two groups. We were trying to show that there was a difference in risk of
complications between the two groups. The small p-value rules out observing the difference in risk
between these two groups (4% vs. 15%) simply by chance, and the randomization of subjects to
treatment groups should have balanced out any other factors that might explain the difference. So, the
only explanation left is that Treatment 1 is truly better than Treatment 2.
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Example: Past research has suggested a high rate of alcoholism among patients with primary unipolar
depression. A study of 210 families of females with primary unipolar depression found that 89 had
alcoholism present. A set of 299 control families found 94 present. The data can be entered into JPM as
shown below.
Questions:
8. What is the response variable?
9. What is the explanatory variable?
Step 0: Define the research question
Is the alcoholism rate in females different among patients with primary unipolar depression
versus the control group? That is, is the proportion of the Depression group with Alcoholism
different from the proportion of the Control group with alcoholism?
Step 1: Determine the null and alternative hypotheses
H0:
Ha:
Step 2: Finding the test statistic and p-value
Step 3: Report the conclusion in context of the research question
Question:
10. Can we say that having unipolar depression causes alcoholism? Explain.
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An Alternative to Fisher’s Exact Test
The Chi-square test can be used to carry out the analysis for testing a difference between two
proportions. Let’s again look at the congenital heart failure example.
Step 0: Define the research question
Is there evidence that the proportion of Down syndrome patients with a congenital heart defect
different from the proportion of Patau syndrome patients with a congenital heart defect?
Step 1: Determine the null and alternative hypotheses
H0: pDown = pPatau
Ha: pDown ≠ pPatau
Step 2: Finding the test statistic and p-value
We will again use the Chi-square test statistic =

 observed - expected 
expected
2
. Therefore, we must again
find expected counts. However, this time we don’t have a set of hypothesized proportions. Thus, we’ll
have to use some information from our data.
Karotype
Down syndrome
Patau syndrome
Total
Congenital Heart Defect
Present
Absent
Total
24
36
60
20
5
25
44
41
85
Recall, under the null hypothesis we’re assuming that there is no difference between the two groups.
That is, it would be like taking the 44 people with a congenital heart defect and randomly assigning them
to the Down syndrome or Patau syndrome groups. Therefore, it would be like expecting 52% (44/85) of
the 60 Down syndrome patients to have a congenital heart defect and the remaining 42% (41/85) to not.
Similarly, it would be like expecting 52% of the 25 Patau syndrome patients to have a congenital heart
defect and the remaining 48% to not. We can summarize the expected counts in the table below.
Karotype
Congenital Heart Defect Down syndrome Patau syndrome Total
Present
44
Absent
Total
41
60
25
85
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We can again use JMP to compute the test statistic and p-value.
Step 3: Report the conclusion in context of the research question
Example: A study was carried out to examine the feeding habits of vampire bats. The main question is
whether or not cows in estrus or not in estrus have the same chance of being attacked by vampire bats
or not.
Bitten by vampire bat?
Yes
No
Cow in estrus
5
3
Cow not in estrus
1
7
Step 0: Define the research question
Is there evidence that the proportion of cows in estrus attacked by vampire bats is the same as
the proportion of cows not in estrus attacked by vampire bats?
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Step 1: Determine the null and alternative hypotheses
H0: The proportion of cows in estrus that have been bitten is equal to the proportion of
cows not in estrus that have been bitten.
Ha: The proportion of cows in estrus that have been bitten is different from the
proportion of cows not in estrus that have been bitten.
H0:
Ha:
Step 2: Finding the test statistic and p-value
Step 3: Report the conclusion in context of the research question
Questions:
11. Look closely at the output. Should you trust the results of the Chi-square test? Explain.
Assumptions behind the Chi-Square Test:
The chi-square test may be inappropriate for tables with very small expected cell frequencies. One rule
of thumb suggests that most of the expected cell frequencies in the table should be 5 or more;
otherwise, the chi-square approximation may not be reliable.
JMP and most other statistical software packages will warn you when the results of the chi-square test
are suspect.
Note that the chi-square test gives an approximate p-value; on the other hand, Fisher’s exact test gives
the exact p-value. Therefore, when the assumptions for the chi-square test are not met for a 2 x 2
table, you should use Fisher’s exact test. If the assumptions are met, the two p-values should be
approximately equal.
12. What is your conclusion based on Fisher’s Exact test?
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