![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.](https://image.slidesharecdn.com/10-210907062751/85/10-Biostatistics-test-of-significance-chi-square-test-3-320.jpg)
![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](https://image.slidesharecdn.com/10-210907062751/85/10-Biostatistics-test-of-significance-chi-square-test-5-320.jpg)
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