2. It is a non parametric test , not based on any
assumption or distribution of any variable.
It is very useful in research and it is most
commonly used when the data are given in
frequencies .
It can be used with any data which can be
reduced to proportion or percentages.
This test involves calculation of quantity
called chi- square , which derived from Greek
letter ( χ)² and pounced as ‘kye’. Cont…
3. It is an alternate test to find the
significance of difference in two or more
than two proportions.
Chi –square test is yet another useful test
which can be applied to find the
significance in the same type of data with
the following advantages .
cont…
4. To compare the values of two binomial
samples even if they are small such as
incidence of diabetes in 20 obese persons
with 20 non obese persons.
To compare the frequencies of two
multinomial samples such as number of
diabetes and non diabetes in groups
weighing 40-50, 50-60, 60-70, and > 70 kg
s of weight.
5. Test of association between events in
binomial or multinomial samples .
Two events often can be tested for their
association such as cancer and smoking .
Treatment and outcome of disease .
Vaccination and immunity .
Nutrition and intelligence .
Cholesterol and heart disease .
Weight and diabetes, B P and heart disease.
To find they are independent of each other or
dependent on each other i.e. associated .
6. Three essentials to apply chi - square test
are
A random sample
Qualitative data
Lowest frequency not less than 5
Steps :-
Assumption of Null Hypothesis (HO)
Prepare a contingency table and note down
the observed frequencies or data (O)
cont…
7. Determine the expected number (E) by
multiplying CT × RT /GT (column total ,row
total and grand total )
Find the difference between observed and
expected frequencies in each cell (O-E)
Calculate chi- square value for each cell with
(O-E)²/E
Sum up all chi –square values to get the total
chi-square value (χ)² d.f. (degrees of freedom)
= χ²= ∑(O-E)²/E and d.f. is (c-1) (r-1)
8. Chi- square test applied in a four fold table
will not give reliable result , with one degree
of freedom.
If the observed value of any cell is < 5 in
such cases Yates correction can be applied
by subtracting ½ (χ)² = ∑ (O-E-1/2)²/E
Even with Yates correction the test may be
misleading if any expected value is much
below 5 in such cases Yates correction can
not be applied.
cont…
9. In tables larger than 2 × 2 Yates correction can not be
applied .
The highest value of chi- square χ² obtainable by
chance or worked out and given in (χ)² table at different
degrees of freedom under probabilities (P) such as
0.05, 0.01, 0.001 .
If calculated value of chi-square of the sample is found
to be higher than the expected value of the table at
critical level of significance i.e. probability of 0.05 the
H O of no difference between two proportions or the H
O of independence of two characters is rejected.
If the calculated value is lower the hypothesis of no
difference is accepted.
10. Groups Died Survived Total
A (a) 10 (b) 25 35
B (c) 5 (d) 60 65
Total 15 85 100
Expected value can be computed with CT×RT/GT
for (a) 15×35/100= 5.25
(b) 85×35/100= 29.75
(c) 15×65/100= 9.75
(d) 85×65/100= 55.25
11. χ²= ∑(O-E)²/E d.f. =( c-1) (r-1)
(A) =(10-5.25)²/5.25 = (4.75)²/5 = 4.30
(B) = (25-29.75)²/29.75 =(5)²/29.75 = 0 .76
(C) = (5-9.75)²/9.75 = (5)²/9.75 = 2.31
(D) = (60-55.25)²/55.25 = (5)²/55.25 = 0.40
7.77
The calculated chi- square value is 7.77 is more than
Chi- square table value with 1 d.f. At 0.01, 6.64
which significant. It can be said that there is
significant difference between two groups.
12. Chi- square can be calculated in the following
way also.
(χ)² = ( ad-bc)²×N
(a b × c d ×ac ×b d) = (600-125 )²
( 35× 65× 15×
85)
= (475)² x100 225625 00 /2900625
(2900625) = 7.77
Here the calculated chi-square value7.77 is >
6.64 at
0.01 with 1 d.f. shows significant difference b/n