Chi Square Test…..
This topic comes under Biostatistics…….
This is useful for Maths students, B.Pharm Students ,M.Pharm Students who studying Biostatistics.
This Presentation Contain following...
#History and Introduction
#Conditions
#Formula
#Classification
#Types of Non-Parametric Chi Square Test
#Test of Independence
#Steps for Test of Independence
#Problem and Solution for Test of Independence
#Test of Goodness of Fit
#Problem and Solution for Test of Goodness of Fit
#Applications of Chi Square Test
Thanks for the Help and Guidance of Dr. M. S. Bhatia Sir
March 2024 Directors Meeting, Division of Student Affairs and Academic Support
Chi sqaure test
1. Prepared by
Mr. Pritam Pravin Kolge
M. Pharm (P’ceutical Quality Assurance)
Bharati Vidyapeeth College of Pharmacy,
Kolhapur
Guided by
Dr. M. S. Bhatia
Professor & Head
Dept. of P’ceutical Chemistry
2. ❑ Karl Pearson introduced Chi Square Test in in the year 1900.
❑ χ2 test enables us to explain whether or not two attributes are associated.
❑ It is used to distinguish whether an observed set of frequencies differs
from a Expected (specified) frequency Distribution
❑ The Chi Square statistic is commonly used for testing relationships between
categorical variables. Eg, Gender, Colour, Education
❑ It is denoted by the Greek Letter χ2
Chi Square Test
3. ❑ All items in the sample must be independent.
❑ No group should contain very few items, say less than 10. Some statisticians take
this number as 5. But 10 is regarded as better by most statisticians.
❑ Total number of items (Grand Total) should be large, say at least 50.
Conditions for the Applications of 2
6. Chi-Square Test as a Non-Parametric
Test
1) Test of Independence.
2) Test of Goodness of Fit
7. ❑ Testing independence determines whether two or more observations across
two populations are dependent on each other (that is, whether one variable
helps to estimate the other.
❑ If the calculated value is less than the table value at certain level of
significance for a given degree of freedom, indicate that two attributes are
independent or not associated.
❑ If calculated value is greater than the table value, indicate that two attributes
are dependent or associated, and reject the null hypotheses.
Test of Independence
8. Step1-
Note the frequencies observed (O) in each class of one
event, row-wise and the frequencies in each group of the other
event, column-wise with the Sum.
Step2-
Calculate Expected frequency (E)
Steps for Test of Independence
(Row Total) (Column Total)
E=
Grand Total
9. Step4-
Determine Degrees of Freedom
df = (R-1)(C-1)
Where,
R= No. of Rows
C= No. of Columns
Step3-
Calculate test statistic
10. Step5- Compare Calculated Value against a tabled/critical value. With use
of degree of freedom and Probability
Decision:- If calculated 2 is greater than table value, reject Ho……Vice
Versa
11. • Suppose a researcher is interested in voting preferences on gun
control issues.
• A questionnaire was developed and sent to a random sample of 90
voters.
• The researcher also collects information about the political party
membership of the sample of 90 respondents.
• At the 0.05 level of significance, is there a difference in Democrat
and Republican?
Favour Neutral Oppose
Democrat 10 10 30
Republican 15 15 10
Problem
12. Favor Neutral Oppose f row
Democrat 10 10 30 50
Republican 15 15 10 40
f column 25 25 40 N = 90
Row Frequency
Column Frequency
Observed Frequencies
Solution Step1-
Note the frequencies observed (O) in each class of one event, row-wise and the
frequencies in each group of the other event, column-wise with the Sum.
14. Favor Neutral Oppose f row
Democrat fo =10
fe =13.9
fo =10
fe =13.9
fo =30
fe=22.2
50
Republican fo =15
fe =11.1
fo =15
fe =11.1
fo =10
fe =17.8
40
f column 25 25 40 N = 90
Eg. Of Calculation of Expected Frequency
E= 50*25/90
=13.9
Step3-
Calculate test statistic
16. Step4-
Determine Degrees of Freedom
df = (R-1)(C-1)
df = (R-1)(C-1) =
(2-1)(3-1) = 1*2= 2
df = 2
Step5-
Compare Calculated Value against a tabled/critical value
Probability (α) = 0.05
Calculated Value = 11.3
Tabled/Critical Value = 5.99
18. Decision-
• Calculated Value 11.03 exceeds as compared to critical value 5.99
• So there is no Independence
• Null hypothesis (Ho) is rejected
• Chi Square Value is more than zero, So Democrats & Republicans differ
significantly in their opinions on gun control issues.
19. ❑ It enables us to see how well does the assumed theoretical distribution fit to
the observed data.
❑ When the calculated value of χ2 is less than the table value at certain level of
significance, the fit is considered to be good one and if the calculated value
is greater than the table value, the fit is not considered to be good.
Test of Goodness of Fit
20. As personnel director want to test the perception of fairness of three
methods of performance evaluation of 180 employees, 63 rated Method 1
as fair, 45 rated Method 2 as fair, 72 rated Method 3 as fair. At the 0.05
level of significance, is there a difference in perceptions?
Method Observed
frequency
Expected
frequency
(O-E) (O-E)2 (O-E)2/E
1 63 60 3 9 0.15
2 45 60 -15 225 3.75
3 72 60 12 144 2.4
Avg = 60 χ2 = 6.3
Problem
Solution
21. Step1-
Calculate Expected frequency
Take the Average of Observed Frequency and apply to every Method.
That is
63+45+72= 180
Average= 60
Step2-
Calculate test statistic
χ2 = 6.3
23. Step4-
Compare Calculated Value against a tabled/critical value
Probability (α) = 0.05
Calculated Value = 6.3
Tabled/Critical Value = 5.99
24. Decision-
• Calculated Value 6.3 exceeds as compared to critical value 5.99
• So there is no Independence
• Null hypothesis (Ho) is rejected
• Chi Square Value is more than zero, So there is difference in Perceptions
of Every Employee for Method Selection.
25. Applications of Chi Square Test
❑ For finding difference in proportions
❑ For finding association between variables