The chi-square test is used to compare observed and expected frequencies to determine if two characteristics are related, based on a null hypothesis. Key requirements include frequency data organized into categories, with a total observations count greater than 20 and expected frequencies above 5. In a sample analysis, the null hypothesis was rejected, indicating an association between gender and hypertensive status at the significance level of 0.05.

1. The Chi Square Test
Muhammad Shahid
Sr. lecturer IIN

2. Introduction
 We often have occasions to make
comparisons between two characteristics
of something to see if they are linked or
related to each other.
 One way to do this is to work out what we
would expect to find if there was no
relationship between them (the usual null
hypothesis) and what we actually observe.

3. Introduction
 The test we use to measure the
differences between what is observed
and what is expected according to an
assumed hypothesis is called the chi-
square test.

4. For Example
 Some null hypotheses may be:
– ‘there is no relationship between the height
of the post RN and Generic BSN students ’.
– ‘there is no connection between the size of
farm and the type of farm’

5. Important
 The chi square test can only be used on
data that has the following
characteristics:
The data must be in the form
of frequencies
The frequency data must have a
precise numerical value and must be
organised into categories or groups.
The total number of observations must be
greater than 20.
The expected frequency in any one cell
of the table must be greater than 5.

6. Formula
χ 2 = ∑ (O – E)2
E
χ2 = The value of chi square
O = The observed value
E = The expected value
∑ (O – E)2 = all the values of (O – E) squared then added
together

7.  Write down the NULL HYPOTHESIS
and ALTERNATIVE HYPOTHESIS
and set the LEVEL OF
SIGNIFICANCE.
 NH ‘ there is no association between gender and
hypertensive status
 AH ‘there is association between gender and
hypertensive status
 We will set the level of significance at 0.05.
Step 01

8. Construct a table with the information you have observed
or obtained.
Observed Frequencies (O)
HTN
Status
Males Females
N 50 50
PH 20 40
H 10 30

9.  The number of degrees of freedom to use is: the
number of rows in the table minus 1, multiplied by the
number of columns minus 1. This is (2-1) x (3-1) = 1 x 2 =
2 degrees of freedom.
Critical value X2=5.991
 Look up the significance tables. These will
tell you whether to accept the null
hypothesis or reject it.
Step 02 & 03 significance level

10.  Step 04 calculation of calculated value
HTN
Status
MALES FEMALES
N E1=40 E2=60
PH E3=24 E4=36
H E5=16 E5=24
Expected frequency = row total x column total
Grand total
Eg: expected frequency for Normal male E1 = (100 x 80) / 200 = 40

11. HTN
Status
Males Females
Row Total
N 50 50 100
PH 20 40 60
H 10 30 40
Column
total
80 120 200

12. O E O-E (O-E)2 (O-E)2
E
50 40 10 100 100/40 =2.5
50 60 -10 100 100/60 =1.66
20 24 -4 16 16/24 =0.66
40 36 4 16 16/36 =0.44
10 16 -6 36 36/16 =2.25
30 24 6 36 36/24 =1.5
(X2)=9.01
Add up all of the above numbers to obtain
the value for chi square: χ2 = 9.01

13.  Comparison of critical and calculated
value
We find that our answer of 9.01 is greater than the
critical value of 5.991 (for 2 degrees of freedom and a
significance level of 0.05) and so we reject the null
hypothesis.

14. Step Decision
 We reject the null hypotheses because
the calculated value fall into rejection
area

15. Step 06 conclusion
 There is an association between gender
and hypertensive status at alpha 0.05