The Chi-square test is used to test the 'Goodness of fit' of the data. it helps to understand the difference between observed values and expected observations.
10. Biostatistics test of significance, chi square test
1. skhot1976@gmail.com B.Sc.-III Paper- XIV (DSE –F26) Bioinformatics, Biostatistics and Economic Botany
skhot1976@gmail.com
Paper XIV
Unit. 2. Biostatistics
2.9 Test of Significance
Chi-Square Test (x2)
Dr. Sudhakar Sambhaji Khot
M.Sc., Ph.D., SET
Assistant Professor in Botany
Y. C. Warana Mahavidyalaya, Warananagar
2. skhot1976@gmail.com B.Sc.-III Paper- XIV (DSE –F26) Bioinformatics, Biostatistics and Economic Botany
Dr. S. S. KHOT
2.4 Statistical methods for testing significance:
Chi-square test
‘Test of significance’: The process to access the significance of differences between
two samples drawn from same population or from two closely related populations’
e.g. Difference between yield of variety A and B
difference between productivity of crops with and without use of fertilizers
Helps to understand whether the observed difference between two samples is
actually due to chance and insignificant or it is really significant and influenced by
some factor.
The process is part of inferential statistics
3. skhot1976@gmail.com B.Sc.-III Paper- XIV (DSE –F26) Bioinformatics, Biostatistics and Economic Botany
Dr. S. S. KHOT
2.4 Statistical methods for testing significance:
Chi-square test
Chi-square test (x2) : (kye square)
• It is sum of deviation square in observed and expected frequencies divided
by expected frequency.
• 𝑥2=Σ[
(𝑂−𝐸)2
𝐸
]
Where, O= observed value, E= Expected value
• The test was proposed by Fisher and developed by Pearson.
• Used to test the goodness of fit
• Calculate the x2 value
• Compare it with tabulated value at df=n-1 and p=0.05
If calculated x2 < tabular x2 , the difference is non-significant, hypothesis accepted
If calculated x2 > tabular x2 , the difference is significant, hypothesis rejected.
4. skhot1976@gmail.com B.Sc.-III Paper- XIV (DSE –F26) Bioinformatics, Biostatistics and Economic Botany
Dr. S. S. KHOT
2.4 Statistical methods for testing significance:
Chi-square test
Example: In dihybrid experiment, in F2 generation following plants appeared:
Yellow, Round = 940; Yellow wrinkled= 260
Green, Round = 340; Green, wrinkled= 60
With x2 test, verfy that the plant follows 9:3:3:1 ratio.
Solution:
• Null Hypothesis: the experiment follows 9:3:3:1 ratio
• Degree of Freedom (df)= n-1= 4-1=3
• Total Plants = 940+260+340+60 = 1600
• Expected Ratio = 9:3:3:1 (out of total 16)
• Expected freq of Yellow, Round = 9/16 x 1600= 900
• Expected freq of Yellow, wrinkled = 3/16 x 1600= 300
• Expected freq of Green, Round = 3/16 x 1600= 300
• Expected freq of Green, Wrinkled = 1/16 x 1600= 100
5. skhot1976@gmail.com B.Sc.-III Paper- XIV (DSE –F26) Bioinformatics, Biostatistics and Economic Botany
Dr. S. S. KHOT
2.4 Statistical methods for testing significance:
Chi-square test
Calculated x2:
𝒙𝟐=𝜮[
(𝑶−𝑬)𝟐
𝑬
] = 28.44
Tabulated x2 at df=3 and
p=0.05 is 7.81
F2 plants O E (O-E) (O-E)2 (O-E) 2
/ E
Yellow, Round 940 900 40 1600 1.78
Yellow, Wrinkled 260 300 -40 1600 5.33
Green, Round 340 300 40 1600 5.33
Green, Wrinkled 600 100 -40 1600 16.00
𝜮[
(𝑶−𝑬)𝟐
𝑬
] = 28.44
Inference: The calculated x2 (28.44) is greater than tabulated x2 (7.81)
Therefore, the is significant and the hypothesis is rejected.
The plant does not follows 9:3:3:1 ratio
6. skhot1976@gmail.com B.Sc.-III Paper- XIV (DSE –F26) Bioinformatics, Biostatistics and Economic Botany
Dr. S. S. KHOT
Questions, if any ?
Thank You
