CHI SQUARE biostat easy explained

1. Chi-square Test

2. Introduction
• The Chi-square test is one of the most commonly used non-
parametric test.
• Chi-square test is a useful measure of comparing
experimentally obtained result with those expected
theoretically and based on the hypothesis.
• It was introduced by Karl Pearson as a test of association. The
Greek Letter χ2 is used to denote this test.
• It can be applied when there are few or no assumptions about
the population parameter.
• It can be applied on categorical data or qualitative data using
a contingency table.

3. • Chi square: it describe the magnitude of the
discrepancy between theory and observation.
• χ2 = ∑ (0 - E)2
E
• O = observed frequency
• E = expected frequency

4. • Expected frequency E= RT x CT
N
• RT = The row total for the row containing cell
• CT = The column total for the row containing
cell
• N = Total number of observation.
• Degree of Freedom(υ): (C-1)x(R-1)

5. Condition for Applying Chi-square test
• Each cell should contain minimum 5 observations.
• All individual observations should be independent
and completely random.
• The total sample size should be at least 50
observations.
• The data should be expressed in original units.

6. Various Applications
• Test for independence of attributes: we can find out
whether two or more attributes are associated or not.
• Chi-square test as Goodness of fit: On several occasions
the decision maker needs to understand whether an
actual sample distribution matches or coincide with a
known probability distribution such as poisson,
binominal or normal.
• Chi-square test for goodness of fit enables us to
determine the extent to which the theoretical probability
distribution coincide with empirical sample distribution.

7. • Chi-square for Yate’s correction for continuity: the
distribution of chi-square test statistics is continuous
but the data under the test is categorical which is
discrete.
• Test for homogenicity: This is useful in case when we
intend to verify whether several population are
homogenous with respect to some characteristics of
interest.

8. • In an antimalarial campaign in India, quinine was
administered to 500 persons out of total population
of 2000. The number of fever cases is shown below.
• Discuss the usefulness of quinine in checking malaria
• H0= Quinine is not effective in checking malaria.
• Ha= Quinine is effective in checking malaria.
Treatment Fever No Fever Total
Quinine 20 480 500
No Quinine 100 1400 1500
Total 120 1880 2000

9. • E= RT x CT
N
E= 500 x 120 = 30
2000
E= 1500 x 120 = 90
2000
E= 500 x 1880 = 470
2000

10. Treatment Fever Expected
Value
No Fever Expected
Value
Total
Quinine 20 30 480 470 500
No Quinine 100 90 1400 1410 1500
Total 120 1880 2000

11. • Calculation Chi-square
• DF= (C-1)(R-1)
(2-1)(2-1) = 1
• Chi-square (0.05) = 3.84
• H0 is failed and rejected. Hence quinine is useful
in checking malaria.
Observation
(O)
Expected value
(E)
(O-E) (O-E)2 (O-E)2 /E
20 30 -10 100 100/30 = 3.33
100 90 10 100 100/90 = 1.11
480 470 10 100 100/470 = 0.21
1400 1410 -10 100 100/1410 = 0.07
Total= 4.72

12. • A drug x claimed to be effective in curing cold. In an
experiment on 500 persons with cold, half of them
were given the drug x and half were given placebo.
The patients reaction to the treatment and recorded
in the table.
• On the basis of data can it be calculated that there is
a significant difference in the effect of drug x and
placebo.
Treatment Helped Reacted No Effect Total
Drug 150 30 70 250
Placebo 130 40 80 250
Total 280 70 150 500

13. • H0= Drug is as effective as placebo
• Ha= Drug is more effective than placebo
Treatment Helped Expected
value
Reaction Expected
value
No effect Expected
value
Total
Drug 150 140 30 35 70 75 250
Placebo 130 140 40 35 80 75 250
Total 280 70 150 500

14. • E= RT x CT
N
E= 250 x 280 = 140
500
E= 250 x 70 = 35
500
E= 250 x 150 = 75
500

15. Observed
Value (O)
Expected Value
(E)
(O-E) (O-E)2 (O-E)2 /E
150 140 10 100 100/140= 0.71
130 140 -10 100 100/140= 0.71
30 35 -5 25 25/35 = 0.71
40 35 5 25 25/35 = 0.71
70 75 -5 25 25/75 = 0.33
80 75 5 25 25/75 = 0.33
Total = 3.5

16. • DF = (C-1) x (R-1)
(3-1) x (2-1) = 2
• Chi-square (0.05) = 5.99
• Null hypothesis is passed and accepted.
• There is no significant in effect of drug x and
placebo.

17. THANK YOU