F test

I M.Com.

1. F - test Dr. R. MUTHUKRISHNAVENI SAIVA BHANU KSHATRIYA COLLEGE, ARUPPUKOTTAI

2. F – test The F-test(Variance Ratio Test) is developed by Statistician R.A. Fisher The object of the test is find out whether the two random independent samples variance differ significantly or not Formula for F-test , F = ; Table value for degrees of freedom(v1,v2) at 5% or 1 % level of significance  

3. F- Test F = = Estimate of Variance of sample1 = = Estimate of Variance of sample1 = Degrees of freedom v1 = ; v2=  

4. Assumptions Normality – Data are normally distributed Homogeneity – variance within each group are equal Independence of error –the error is independent for each group

5. Illustration - 1 In a sample of 8 observations, the sum of squared deviations of items from the mean was 84.4. In another samples of 10 observations, the value was found to be 102.6. Test whether the difference is significant at 5% level. V1 = 7 & V2 = 9 degrees of freedom at 5% level of significance F 0.05 = 3.29

6. Solution Let us take hypothesis that the difference in the variance of two samples is not significant H : n1= 8 n2 = 10 = = = = 12.06 = = = = 11.4 F = = = 1.06  

7. Solution - continue Calculated value of F = 1.06 Table value of F = 3.29 (v1 = 7 and v2 = 9 at 5% level of significance) Inference CV < TV, hence we accept the null hypothesis and conclude that the difference in the variance of two samples is not significant at 5%

8. Illustration 2 The following data present the yields in Kilogram of common 10 subdivision of equal area of two agricultural plots Test whether two samples taken from two random populations have same variance.(5% point of F for V1=9 and V2=9 is 3.18) Plot A 620 570 650 600 630 580 570 600 600 580 Plot B 560 590 560 570 580 570 600 550 570 550

9. Solution Let us take the null hypothesis that the samples come from populations having the same variance H : applying F-test F = = ; =  

10. Calculation of sample variance ( ( 2 ( ( 2 620 20 400 560 -10 100 570 -30 900 590 20 400 650 50 2500 560 -10 100 600 O 570 630 30 900 580 10 100 580 -20 400 570 570 -30 900 600 30 900 600 550 -20 400 600 570 580 -20 400 550 -20 400 6000 6400 5700 2400 620 20 400 560 -10 100 570 -30 900 590 20 400 650 50 2500 560 -10 100 600 O 570 630 30 900 580 10 100 580 -20 400 570 570 -30 900 600 30 900 600 550 -20 400 600 570 580 -20 400 550 -20 400 6000 6400 5700 2400 X 1 = = 60   X 2 = = 57    

11. Solution - Continue = = = 711.11 = = = 266.67 F = = = 2.67  

12. Solution - continue Calculated value of F = 2. 67 Table value of F = 3.18 (v1 = 9 and v2 = 9 at 5% level of significance) Inference CV < TV, hence we accept the null hypothesis and conclude that the samples come from populations having the same variance at 5%