The document summarizes the Chi-Square test. It begins with an introduction explaining that the Chi-Square test is a non-parametric statistical test that uses Chi-Square distribution and compares observed versus expected frequencies in categories. It then discusses the characteristics, applications, conditions for use, calculation method, provides an example, and concludes with the limitations of the test.
2. ACKNOWLEDGMENT
I wish to acknowledge our
respected HOD Madam
Dr Madhurima Chattopadhyay, Our
seminar
faculty-Samik Chakraborty Sir,
Soumik Das Sir, and the entire
AEIE
department of HERITAGE
INSTITUTE OF
TECHNOLOGY,KOLKATA who has
been a
4. INTRODUCTION
CHI-SQUARE TEST is a non-parametric test, not based
on any assumption or distribution of any variable.
This statistical test follows a specific distribution
known as CHI-SQUARE DISTRIBUTION.
To measure the differences between what is observed
and what is expected according to a hypothesis is
called CHI SQUARE TEST.
5. CHARACTERISTICS
This test is based on frequencies such as the number of
responses in two or more categories, not on the parameters
like mean and standard deviation.
The test is not useful for estimation and used for testing the
hypothesis.
As no rigid assumptions are necessary ,relatively less
mathematical details are involved .
This test can also be applied to a complex contingency table
with several classes and a very useful test in research work.
The three essential requirements for Chi-square test are:
A random sample
Qualitative data
Lowest expected frequency not less than 5.
6. APPLICATION
Useful in test of Goodness of Fit of
Distribution.
Useful in test of Independence of Attributes.
Useful in test of Homogeneity.
7. CONDITIONS FOR
APPLICATION
Data must be in the form of frequencies.
The frequency data must have a precise numerical
value and must be organized into categories or
groups.
Observations recorded and used, are collected on a
random basis
All items in the sample must be independent.
The overall numbers of items must also be
reasonably large(at least 50).
9. EXAMPLE
E
EO 2
2 )(
= 9. (calculated value)
o Degrees of freedom of this given problem is: (n-1)=(6-1)=5.
o The table value of for 5 degrees of freedom at 5% level of
significance is 11.071. Calculated value is less than table value.
o Thus the result supports the hypothesis and it can be concluded
the die is unbiased.
The probability of obtaining any one of the 6 number is 1/6.so the
expected frequency of any 1 number coming upward is 132*(1/6)=22.
10. CONCLUSION
This test tells the presence or absence of an
association between the events but does not
measure the strength of association.
This test doesn’t indicate the cause and effect,
only tells the probability of occurrence of
association by any chance.
The occurrence of one individual observation
has no effect upon the occurrence of any other
observation in the sample under consideration.
11. BIBLIOGRAPHY
Sawhney A K : A course in Electrica
l & Electronic Measurements & Instr
uments
E. A. Doebelin, Measurement Syste
ms: Application and Design
www.wikipedia.com
www.google.com
