THE CHI-SQUARE TEST IN DEPTH AND WITH EXAMPLES

THE CHI-SQUARE TEST
THE CHI-SQUARE TEST
BACKGROUND AND NEED OF THE TEST
BACKGROUND AND NEED OF THE TEST
Data collected in the field of
Data collected in the field of
medicine is often qualitative.
medicine is often qualitative.
--- For example, the presence or
--- For example, the presence or
absence of a symptom, classification
absence of a symptom, classification
of pregnancy as ‘high risk’ or ‘non-
of pregnancy as ‘high risk’ or ‘non-
high risk’, the degree of severity of a
high risk’, the degree of severity of a
disease (mild, moderate, severe)
disease (mild, moderate, severe)

The measure computed in each
The measure computed in each
instance is a proportion,
instance is a proportion,
corresponding to the mean in the case
corresponding to the mean in the case
of quantitative data such as height,
of quantitative data such as height,
weight, BMI, serum cholesterol.
weight, BMI, serum cholesterol.
Comparison between two or more
Comparison between two or more
proportions, and the test of
proportions, and the test of
significance employed for such
significance employed for such
purposes is called the “Chi-square
purposes is called the “Chi-square
test”
test”

---- KARL PEARSON IN 1889, DEVISED
---- KARL PEARSON IN 1889, DEVISED
AN INDEX OF DISPERSION OR TEST
AN INDEX OF DISPERSION OR TEST
CRITERIOR DENOTED AS “CHI-
CRITERIOR DENOTED AS “CHI-
SQUARE “. (
SQUARE “. (X
X 2 ).
).
The formula for
The formula for X
X 2 –test is,
–test is,

Chi- Square
Chi- Square
 
( o - e ) 2
e
X
X 2
=
=

reject
reject H
Ho
o if
if 
2
2
>
> 
2
2
.
.
,
,df
df
where df = (
where df = (r
r-1)(
-1)(c
c-1)
-1)

2
2
= ∑
= ∑
(
(O
O -
- E
E)
)2
2
E
E
3.
3. E is the expected frequency
E is the expected frequency
^
^
E
E =
=
^
^
(total of all cells)
(total of all cells)
total of row in
total of row in
which the cell lies
which the cell lies
total of column in
total of column in
which the cell lies
which the cell lies
•
•
1.
1. The summation is over all cells of the contingency
The summation is over all cells of the contingency
table consisting of r rows and c columns
table consisting of r rows and c columns
2.
2. O is the observed frequency
O is the observed frequency
4.
4. The degrees of freedom are df = (r-1)(c-1)
The degrees of freedom are df = (r-1)(c-1)

APPLICATION OF CHI-SQUARE
APPLICATION OF CHI-SQUARE
TEST
TEST
 TESTING INDEPENDCNE (OR
TESTING INDEPENDCNE (OR
ASSOCATION)
ASSOCATION)
 TESTING FOR HOMOGENEITY
TESTING FOR HOMOGENEITY
 TESTING OF GOODNESS-OF-
TESTING OF GOODNESS-OF-
FIT
FIT

Chi-square test
Chi-square test
Purpose
Purpose
To find out whether the association between two
To find out whether the association between two
categorical variables are statistically significant
categorical variables are statistically significant
Null Hypothesis
Null Hypothesis
There is no association between two variables
There is no association between two variables

Chi-square test
Chi-square test
 
  df
c
b
c
b
d
c
b
a
d
b
c
a
N
bc
ad
N
df
b
a
d
c
d
b
c
a
bc
ad
n
f
d
e
e
o
corrected
c
r
1
2
2
2
2
2
2
)
1
)(
1
(
2
2
~
1
~
)
)(
)(
)(
(
2
~
)
)(
)(
)(
(
)
(
.
~
1

























 

1.
2.
3.
4.
Test statistics
Test statistics

Chi-square test
Chi-square test
 Objective
Objective :
: Smoking is a risk factor for MI
Smoking is a risk factor for MI
 Null Hypothesis:
Null Hypothesis: Smoking does not cause MI
Smoking does not cause MI
D (MI) D-(No MI) Total
Smokers 29 21 50
Non-smokers 16 34 50
Total 45 55 100

Chi-Square
Chi-Square
E
O
29
E
O
21
E
O
16
E
O
34
MI Non-MI
Smoker
Non-Smoker

Chi-Square
Chi-Square
E
O
29
E
O
21
E
O
16
E
O
34
MI Non-MI
Smoker
Non-smoker
50
50
55
45 100

Estimating the Expected Frequencies
Estimating the Expected Frequencies
E
E =
=
^
^ (row total for this cell)•(column total for this cell)
(row total for this cell)•(column total for this cell)
n
n
Classification
Classification 1
1
Classification
Classification 2
2
1
1
2
2
3
3
4
4 c
c
r
r
1
1 2
2 3
3
C
C1
1 C
C2
2 C
C3
3 C
C4
4 C
Cc
c
R
R1
1
R
R2
2
R
R3
3
R
Rr
r
E
E =
=
^
^ R
R2
2C
C3
3
n
n

Chi-Square
Chi-Square
E
O
29
E
O
21
E
O
16
E
O
34
MI Non-MI
Smoker
Non-smoker
50
50
55
45 100
50 X 45
100
22.5 =
22.5

Chi-Square
Chi-Square
E
O
29
E
O
21
E
O
16
E
O
34
Alone Others
Males
Females
50
50
55
45 100
22.5 27.5
22.5 27.5

Chi-Square
Chi-Square
 Degrees of Freedom
Degrees of Freedom
df
df = (R-1) (C-1)
= (R-1) (C-1)
 Critical Value (Table A.6) = 3.84
Critical Value (Table A.6) = 3.84
 X
X2
2
= 6.84
= 6.84

THE CHI-SQUARE TEST IN DEPTH AND WITH EXAMPLES

Age
Age
Gender
Gender <30
<30 30-45
30-45 >45
>45 Total
Total
Male
Male 60 (60)
60 (60) 20 (30)
20 (30) 40 (30)
40 (30) 120
120
Female
Female 40 (40)
40 (40) 30 (20)
30 (20) 10 (20)
10 (20) 80
80
Total
Total 100
100 50
50 50
50 200
200
Chi- square test
Find out whether the sex is equally
distributed among each age group

Test for Homogeneity (Similarity)
Test for Homogeneity (Similarity)
To test similarity between frequency distribution or
To test similarity between frequency distribution or
group. It is used in assessing the similarity between
group. It is used in assessing the similarity between
non-responders and responders in any survey
non-responders and responders in any survey
Age (yrs) Responders Non-responders Total
<20 76 (82) 20 (14) 96
20 – 29 288 (289) 50 (49) 338
30-39 312 (310) 51 (53) 363
40-49 187 (185) 30 (32) 217
>50 77 (73) 9 (13) 86
Total 940 160 1100

THE CHI-SQUARE TEST IN DEPTH AND WITH EXAMPLES

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