Qualitative - Binary, Nominal and Ordinal Data Analysis in Medical Science

Abstract

The outcome of any medical research is belonged to the human beings. The correct application of statistical test has its paramount importance. This article provides the details of categorical data analysis test with example and with its interpretation. It is included when to use Chi-square test (2*2 & R*C), Yate`s Correction, Fisher Exact test, Mc Nemartest manually as well as using Open Epi and R codes.

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Date of Submission
:
2
7
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0
6
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202
2
Date of Acceptance
:
2
8
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0
8
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2022
Date of
Publication
:
3
0
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0
9
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2022
Correspondence:
Dr. Swati Patel (E
-
mail:
Swati84patel@gmail.com
)
Copy Right:
The
Authors
retain the copyrights of this article
, with first publication rights granted to Medsci Publications.
National Journal of Community Medicine│Volume 13│Issue 09│September 2022 Page 663
How to cite this article:
Patel S. Qualitative
-
Binary, Nominal and Ordinal Data Analysis in Medical Science. Natl J Community Med
2022;13(9):663-668. DOI: 10.55489/njcm.130920222200
COUNTINUOUS MEDICAL EDUCATION
pISSN09763325│eISSN22296816
Open Access Article (CC BY-SA)
www.njcmindia.com
DOI: 10.55489/njcm.130920222200
Qualitative - Binary, Nominal and Ordinal
Data Analysis in Medical Science
Swati Patel1
1Surat Municipal Institute of Medical Education & Research, Surat
INTRODUCTION
In study the categorical /Binary data represented by
frequency table whereas to know the Association be-
tween either binary or categorical data Chi- Square
test is used. The chi square analysis determines as-
sociation between categorical responses between
two or more independent groups. Chi square tests
can only be used on actual numbers and not on per-
centages, proportions and means. There are two type
of chi- square test (i) Good ness of fit chi-square test
(II) Chi-square test of Independence. The Chi-square
goodness of fit test is a statistical hypothesis test
used to determine whether a variable is likely to
come from a specified distribution or not. It is often
used to evaluate whether sample data is representa-
tive of the full population. Chi-square test of inde-
pendence is a statistical hypothesis test used to de-
termine whether two Binary / Categorical variables
are likely to be related/associated or not. It can be
applying when two qualitative variables i.e bina-
ry/categorical (nominal or ordinal) are independent.
In medical research when researcher mainly inter-
ested to know the association between two binary or
dichotomous variable, different statistical test ap-
plies for that 2* 2 table to know the association be-
tween them according to the frequency of each cell.
This article mainly describes the appropriate usage
of chi-square test, Yate`s correction of chi-square
test, Fisher Exact test and Mc-Nemar test.
Chi – Square goodness of fit
The chi-square test of goodness-of-fit use when you
have one nominal variable within two or more cate-
gories (such as <18 ,18-30, 31-45,...) You compare
the observed counts of observations in each category
with the expected counts, which you calculate using
some kind of theoretical expectation (such as a 1:1
Male: Female ratio or a 1:2:1 ratio in a genetic cross).
In other words when you draw random samples and
to know the proportion of outcome is equal or not to
check it chi- square good ness of fit is useful.
If the expected number of observations in any cate-
gory is too small, the chi-square test may give inac-
curate results, and you should use an exact test.
Pearson’s Chi- Square test (Chi- Square test of in-
dependence)
Chi-square test applied for actual frequen-
cies/numbers as cross tabulated in a contingency
table. It is not appropriate to use for percent-
ages/proportion. It can be applied on 2* 2 contin-
ABSTRACT
The outcome of any medical research is belonged to the human beings. The correct application of statisti-
cal test has its paramount importance. This article provides the details of categorical data analysis test
with example and with its interpretation. It is included when to use Chi-square test (2*2 & R*C), Yate`s
Correction, Fisher Exact test, Mc Nemartest manually as well as using Open Epi and R codes.
Keywords: Qualitative analysis, nominal data, ordinal data

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National Journal of Community Medicine│Volume 13│Issue 09│September 2022 Page 664
gency table or R*C table (R is number of rows and C
is Number of Columns).
When a researcher interested to analyze his/her data
using a chi-square test of independence, he/she need
to make sure that the data should follow the follow-
ing assumptions. If the data doesn’t meet the as-
sumption of chi-square test, you can`t apply the chi-
square test for independence.
Assumption 1: Two variables should be measured
at binary or ordinal or nominal level.
Assumption 2: Your two variables should consist of
two or more categorical, independent groups. Ex-
ample independent variables that meet this criterion
include Outcome of Patients (2 groups: Death, Sur-
vive), Socioeconomic level (e.g., 3groups: Upper,
Middle, lower), Pain Score (4 group: No Pain, Mild,
Moderate, Severe), profession (e.g., 5 groups: sur-
geon, doctor, nurse, dentist, therapist).
Assumption: 3 Cells in the contingency table should
be mutually exclusive.
It’s assumed that individuals can only belong to one
cell in the contingency table. That is, cells in the table
are mutually exclusive — an individual can not be-
long to more than one cell.
We can verify that this assumption is met by check-
ing that no individual has been counted in more than
one cell.
Assuming each individual in the dataset was only
surveyed once, this assumption should be met be-
cause it’s not possible for an individual to be, say, a
Male Republican & Female Democrat simultaneously.
Assumption 4: Expected value of cells should be 5 or
greater in at least 80% of cells.
It’s assumed that the expected value of cells in the
contingency table should be 5 or greater in at least
80% of cells and that no cell should have an expected
value less than 1.
Methods:
Researcher can enter the data in Excel. It has also fa-
cility to prepare cross table. Using the pivot table in
excel cross table can be prepared. (2,3)
Excel → Insert → Pivot table
1. The above table.1 full filed the all assumptions of
Pearson’s chi- square test, two categori-
cal/binary independent variables are given, Cells
in the contingency table are mutually exclusive
2. Calculate expected value of each cell. (no more
than 20% of cells expected value is < 5)
Expected cell = row total (rij)*Colum total(cij)/N
3. Calculate 𝜒
=∑()

oij = observed frequency of ith row and jth Column
eij = expected frequency of ith row and j th Column
Post hoc test
When chi-squared test is significant, the next step is
to perform a post hoc test to find out which cells
from the contingency table are different from their
expected values. Without any prior knowledge of
each cell, we are interested in testing all cells in a
contingency table at once. Three test statistics are of-
ten calculated for each cell: Raw Residual (RawR),
Standardized Residual (StdR), and Adjusted Residual
(AdjR). The larger these residuals are, the greater the
contribution of these residuals to the overall chi-
squared test. (5)
When the chi-square test of a more than 2×2 is sig-
nificant (and sometimes when it isn't), it is desirable
to investigate the data further. MacDonald and Gard-
ner (2000) use simulated data to test several post-
hoc tests for a test of independence, and they found
that pairwise comparisons with Bonferroni Correc-
tion P values work well.
Raw Data application
Researcher planned study to assess the prevalence
and associated factors of depression in breast cancer
patients. Total 152 patients included with Brest can-
cer, depression measure using PHQ -2 Scales. The re-
searcher hypothesised that
Null Hypothesis: There is no association between
depression and stages of cancer.
Alternative Hypothesis: There is an association be-
tween depression and stages of cancer.
How to prepare table using raw data
Table 1: Details of Depression and Stages of Can-
cer in Brest cancer patients
Stages of Cancer Depression
Present Absent
I 8 18
II 13 30
III 26 41
IV 11 5
Table 1.1: Details of observed and expected value
Stages of Cancer Depression Total
Yes No
I Observed 8(o11) 18(o12) 26
expected 12(e11) 16(e12)
II Observed 13(o21) 30(o22) 43
expected 16(e21) 27(e22)
III Observed 26(o31) 41(o32) 67
expected 26(e31) 41(e32)
IV
Observed
11(o41)
5(o42)
16
expected 6(e41) 10(e42)
Total 58 94 152
𝜒
=(𝑜11 − 𝑒11)
𝑒11 + ⋯ … . . + (𝑜42 − 𝑒42)
𝑒42
= (8 − 12)
12 +(18 − 16)
16 + ⋯ … … + (5 − 10)
10
= 8.104

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4. Compare 𝜒
 value with tabulated value according
to degree of freedom
(df =(r-1) *(c-1) = (4-1) *(2-1) =3).
5. Conclusion: - 𝜒
>𝜒.,
 , reject the null hypothe-
sis. i.e there is association in between stages of can-
cer and depression among the breast cancer patients.
Manually Calculation part is very laborious for chi-
square test, especially for R*C table. Now days many
software’s/ online statistical calculators are available
using this anyone can do calculation very easily but
important thing is that researcher need to know the
assumption of it, once you prepare the cross table
(R*C), to calculate chi square value you can use OPEN
EPI & EPI TOOLS software, which are free of cost.
Figure 1: chi-Square test using Open EPI
1. https://www.openepi.com/SampleSize/SSCohort.htm use this website.
2. Click on R*C, you will get new window.
3. Click on enter in new window, it will ask your number of rows and column of cross table for chi- square
calculation.
4. Click on calculate.
5. In Result you will get three values i.c chi square =8.104, df =3 and P-value = 0.04392.
6. This software will also mention the assumption of chi- square test (i.e % of expected cell`s value < 5)
Interpretation
Here 4×2 table has a chi-square value is 8.104 with 3
degrees of freedom, P value 0.04392, which only in-
dicating that there is an association between Depres-
sion and different stages of cancer among the Brest
Cancer patients. But to know which stages of cancer
is significantly associated with the depression to
know that pairwise comparison researcher needs to
apply Post hoc test.
Post hoc chi-square: -
There are six possible pairwise comparisons, so you
can do a 2×2 chi-square test for each one. Post hoc
Chi-square computed p-values for each cell in this
contingency table of this example. All Pairs P value is
> 0.008 (Adjusted Bonferroni level of significance
0.05/6 =0.008). No pairwise association has been
observed between stages of cancer and presence of
depression using R software.
2* 2 contingency table data analysis
When in Study researcher has 2*2 contingency table
is easy to calculate chi-square than more than 2*2
table. According to the objective of study researcher
can apply different test for 2*2 table.
Chi-square test –
For 2 * 2 cross table assumptions are same as more
than 2*2 chi-square test. When 2*2 cross table is giv-
en to know the association between two binary vari-
ables, it will be applying when all cell`s frequency is
>5.
Case study
During the Pandemic of COVID19 a team of doctors
recorded that the severity has been observed among
the patients who admitted in COVID ward with hy-
pertension. Using this data, they assumed that sever-
ity of COVID 19 is related or associated with hyper-

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National Journal of Community Medicine│Volume 13│Issue 09│September 2022 Page 666
tension. Total 276 patients recorded for this study,
divided them two parts hypertensive and non-
hypertensive according to severity and non-severity.
Table 1.2 Details of COVID 19 admitted patient’s
severity wise
Hypertensive Non-Hypertensive Total
Sever 52 89 141
Non-Sever 29 106 135
Total 81 195 276
Null Hypothesis: Severity of disease is not associated
with hypertension.
Alternative hypothesis: Severity of disease is asso-
ciated with hypertension
Calculation of Chi square:
𝜒
=(𝑎𝑑 − 𝑏𝑐)
(𝑎 + 𝑏)(𝑐 + 𝑑)(𝑎 + 𝑐)(𝑏 + 𝑑)
=((∗)(∗))
()()()() =7.886
Interpretation: Here in the given 2*2 cross table 𝜒

> 𝜒.,
 (3.84 will get from statistical tables) , accept
the alternative hypothesis. i.e., Severity of disease is
associated with hypertension.
Using Open EPI software, the calculation will become
easy.
Figure1.2 Calculation using OPEN EPI
Interpretation – The given 2×2 table has a chi-square
value is 7.886 with 2 degrees of freedom, P value(2-
tail) 0.004981, which only indicating that severity of
disease is associated with hypertension and
2.136(OR) times more chance of severity among hy-
pertensive patients of COVID 19 as compared to non-
hypertensive COVID 19 Patients.
Yate Correction –
Yate`s correction is the method of modification of
chi-square for only for 2’*2 contingency table, to re-
duce the error in approximation, Frank Yates, an
English statistician, suggested a correction for conti-
nuity which adjusts the formula for Pearson's chi-
square test by subtracting 0.5 from the difference be-
tween each observed value and its expected value in
a 2 × 2 contingency table (Yates, 1934). This reduces
the chi-square value obtained and thus increases its
p-value. (6)
The effect of Yates' correction is to prevent overes-
timation of statistical significance for small data. This
formula is chiefly used when at least one cell of the
table has an expected count smaller than 5. Unfortu-
nately, Yates' correction may tend to overcorrect.
This can result in an overly conservative result that
fails to reject the null hypothesis when it should. So it
is suggested that Yates' correction is unnecessary
even with quite low sample sizes (Sokal and Rohlf,
1981), such as total sample sizes less than or equal to
20.
Fisher Exact test
Fisher's exact test assesses the null hypothesis of in-
dependence applying hypergeometric distribution of
the numbers in the cells of the table instead of ap-
proximation. Fisher's exact test is practically useful
only in analysis of small samples but actually it is val-
id for all sample sizes. While the chi-squared test re-
lies on an approximation, Fisher's exact test is one of
exact tests. Especially when more than 20% of cells
have expected frequencies < 5, we need to use Fish-
er's exact test because applying approximation
method is inadequate. Manually calculation of Fisher
Exact test is very difficult but today many packages

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National Journal of Community Medicine│Volume 13│Issue 09│September 2022 Page 667
provide the calculation of Fisher's exact test for 2 × 2
contingency tables but not for bigger contingency ta-
bles with more rows or columns. For example, the
SPSS statistical package and OPEN EPI automatically
provides an analytical result of Fisher's exact test as
well as chi-squared test only for 2 × 2 contingency
tables. But When Fisher’s exact test of bigger contin-
gency tables, we can use web pages providing such
analyses. SAS and R can be also use for to calculate
Fisher Exact test for more than 2 rows and 2 col-
umns.
A researcher studied to know the association be-
tween age group and severity of CIVID19 hospital-
ised cases. This study included total 99 patients of
COVID 19.
Table 1.3: Details of Age group and Severity of
COVID 19 Case
Age Mild Moderate Severe
<18 yrs 6 4 0
18-30 yrs 12 7 3
31-50yrs 12 6 5
>50 yrs 4 23 17
There are 4 rows and 3 columns in the above table,
given data has 25% of 12 cells have expected values
< 5 which doesn`t meet the Cochran’s criteria, so to
know the proportions are independent or not In-
stead of Chi- Square test Fisher Exact test should be
use.
Interpretation: Here P< 0.05, so the proportion of
age group and severity of COVID19 are Associate
(Not Independent). To know the age wise severity
comparison post hoc pairwise comparison.
For 4*3 contingency table if P< 0.05 (by Fisher Exact
test), you could do a 2×3 Fisher's exact test for each
of these pairwise comparisons, but there are 6 possi-
ble pairs, so you need to do correct multiple compar-
ison. One way to do this is with a modification of the
Bonferroni-corrected pairwise technique suggested
by MacDonald and Gardner (2000), replacing Fish-
er's exact test for the chi-square test they used. You
do a Fisher's exact test on each of the 6 possible
pairwise comparison (i.e <18 yrs & 18-30 Yrs, <1
8yrs &31-50 Yrs ,…… etc) then apply the Bonferroni
Correction multiple tests. With 6 pairwise compari-
sons, the P value must be less than 0.05/6, or 0.008,
to be significant at the P<0.05 level.
Age group P- Value adjusted Fisher
<18 Yrs & 18-30 Yrs 0.8480
<18 Yrs & 31-50 Yrs 0.6510
<18 Yrs & >50 Yrs 0.0017*
18-30 Yrs & 31-50 Yrs 0.8480
18-30 Yrs & > 50 Yrs 0.0017*
31-50 Yrs & >50 Yrs 0.017
Interpretation: - In this pairwise comparison, two
pairs of age group <18 yrs. & >50 yrs & 18-30 yrs & >
50 yrs are significantly associated with severity of
COVID19.
Mc Nemarc test
It is not showing the association between two varia-
bles. It is non-parametric (distribution-free) test as-
sesses if a statistically significant change in propor-
tions have occurred on a dichotomous attribute at
two time points on the same population. It is ap-
plied using only a 2×2 contingency table with the di-
chotomous variable at time 1 and time 2. In medical
research, if a researcher wants to determine whether
or not a particular therapy has an effect on a disease
(e.g., yes vs. no), then a count of the individuals is
recorded (as + and – sign, or 0 and 1) in a table be-
fore and after being given the Theraphy. Then,
McNemar’s test is applied to make statistical deci-
sions as to whether or not a drug has an effect on the
disease.
Example
A researcher planed study to know the effect of two
different treatments on for breast cancer after mas-
tectomy. The two group of treatment group com-
pared on the basis of other prognostics factors. Total
621 patients included in this study. Patients are as-
signed to pairs matched on age (within 5 yrs.) and
clinical condition
Outcome of
Treatment B
Outcome of Treatment A
Survive for 5
yrs
Die within 5
yrs
Survive for 5 yrs 510 (a) 16(b)
Die within 5 yrs 5 (c ) 90(d)
Ho: No Difference in proportion of survival/(5 yrs)
by treatment A and B.
𝜒
 𝑜𝑟 𝑄 = ((𝑏 − 𝑐)− 1)
(𝑏 + 𝑐) = (16 − 5)− 1)
16 + 5 = 4.76
In order to test if the treatment is helpful, we use on-
ly the number discordant pairs of twins, b and c,
since the other pairs of twins tell us nothing about
whether the treatment is helpful or not.
𝜒
 𝑜𝑟 𝑄 = ( )
() or
which for large samples is distributed like a chi-
squared distribution with 1 degree of freedom. A
closer approximation to the chi-squared distributing
uses a continuity correction.
𝜒
𝑜𝑟 𝑄 = ((𝑏 − 𝑐)− 1)
(𝑏 + 𝑐)
CONCLUSION
Calculated value (4.76) > 𝜒.,
(3.84) ,reject the
null hypothesis. i.e treatment give different result
from subjects of matched pairs, then the treatment

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A subjects of pair is significantly more likely to sur-
vive for 5 yrs than the treatment subjects.’
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Annexure
Name of Test R code
Post hoc Chi-square test
R1 = c(8,18)
R2 = c(13,30)
R3 = c(26,41)
R4 = c(11,5)
rows = 4
x = matrix(c(R1,R2,R3,R4),
nrow=rows,
byrow=TRUE)
rownames(x) = c("One", "Two", "Three", "Four")
colnames(Matriz) = c("Yes", "No")
x
library(rcompanion)
pairwiseNominalIndependence(Matriz,
fisher = FALSE,
gtest = FALSE,
chisq = TRUE,
method = "fdr")
Fisher Exact test x <- matrix(c(6,4,0,12,7,3,12,6,5,4,23,17), byrow = TRUE, nrow = 4, ncol = 3);
fisher.test(x);
Outcome:
p-value = 8.491e-05 or 0.00008491
library(rcompanion)
pairwiseNominalIndependence(x,
fisher = TRUE,
gtest = FALSE,
chisq = FALSE,
method = "fdr")
McNemar's Chi-squared test > x <- matrix(c(510,5,16,90) , nrow =2 );
> x
[,1] [,2]
[1,] 510 16
[2,] 5 90
> mcnemar.test(x)
data: x
McNemar's chi-squared = 4.7619, df = 1, p-value = 0.029