chisquare-GoF

1. 9/22/23 The Chi-Square Distribution Goodness of Fit 1 Chi-Square Distribution 1 ! !!!∗!!!! ! ! = ∗ ! 2!/!Γ! 2 2 1

2. 9/22/23 Chi-Square Distribution * R code * Produces graph to the right 3 Chi-Square Table 4 2

3. 9/22/23 Chi-square: Goodness of fit test • We can test to see if our observed frequencies “fit” our expectations • This is the c2 Goodness-of-Fit test • This converts the difference between the frequencies we observe and the frequencies we expect to a distribution with known probabilities 2 ( ) O−E E ∑ χ2= ;df = #categories−1 5 P-value for χ2 • Upper tail (only) • P-value = pchisq(χ2 calc,df,lower.tail=FALSE); 6 3

4. 9/22/23 Critical Value for χ2 • Upper tail (only) • χ2crit = qchisq(1- α,df); 7 P-Value Rules for χ2 • If the P-value is greater than α, you will fail to reject H0 • If the P-value is less than α, you will reject H0 8 4

5. 9/22/23 Critical Value Rules for χ2 • Upper tail (only) • If χ2calc is greater than χ2crit , you will reject H0 • If χ2calc is less than χ2crit , you will fail to reject H0 9 Goodness-of-Fit: An Example • It has been reported that • 10.3% of U.S. farmer do not own a tractor, • 34.2% owning 1 tractor, • 38.4% owning 2 tractors, and • 17.1% owning 3 or more tractors. 10 5

6. 9/22/23 Goodness-of-Fit • The data for a random sample of 100 farmers in Florida are summarized below. Can we reject the possibility that the tractor-ownership distribution in this state differs from that of the nation as a whole? (α=0.05) # Tractors Owned 0 1 2 3 or more # Farmers 20 35 23 22 11 Goodness-of-Fit H0: p0 = 0.103, p1 = 0.342, p2 = 0.384, p3+ = 0.171 tractor-ownership distribution in this state is the same as it is in the nation as a whole. HA: At least one of the proportions does not equal the stated value. Tractor- ownership distribution in this state is not the same as it is in the nation as a whole. 12 6

7. 9/22/23 Goodness-of-Fit # Tractors Owned Observed Expected 0 20 10.3 1 35 34.2 2 23 38.4 3 or more 22 17.1 13 Goodness-of-Fit (O-E)2 (O-E)2/E # Tractors Owned 0 Observed Expected (O-E) 20 10.3 9.7 94.09 9.13 1 35 34.2 0.8 0.64 0.02 2 23 38.4 -15.4 237.16 6.18 3 or more 22 17.1 4.9 24.01 1.40 Σ = 16.73 d.f. = 3 14 7

8. 9/22/23 P-value for χ2 • Upper tail (only) • P-value = pchisq(χ2 calc,df,lower.tail=FALSE); 15 Critical Value for χ2 • Upper tail (only) • χ2crit = qchisq(1- α,df); 16 8

9. 9/22/23 P-Value Rules for χ2 • If the P-value is greater than α, you will fail to reject H0 • If the P-value is less than α, you will reject H0 17 Critical Value Rules for χ2 • Upper tail (only) • If χ2calc is greater than χ2crit , you will reject H0 • If χ2calc is less than χ2crit , you will fail to reject H0 18 9

10. 9/22/23 P- value pchisq(16.73,3,lower.tail=FALSE) = 0.0008030961 19 critical value qchisq(.95,3) = 7.814728 20 10

11. 9/22/23 Goodness-of-Fit a = 0.05 df = k – 1 = 4 – 1 = 3 II. Test Statistic: c2 = 16.73 III. Conclusion: Since the test statistic of c2 = 16.73 is greater than the critical value of c2 = 7.815, we reject H0 with at least 95% confidence. Implications: There is enough evidence to show that tractor ownership in this state differs from that in the nation as a whole. 21 Goodness-of-Fit: An Example •On a large dairy farm, the manager used two sources of semen for artificial insemination: conventional and sexed. • Of the 521 calves resulting from conventional semen, 47% were female. • Of the 514 calves resulting from sexed semen, 84% were female. 22 11

12. 9/22/23 Goodness-of-Fit: An Example •We want to know if: 1. Does sex of calf from the conventional semen agree with biological expectations (i.e. 1:1 ratio)? 2. Does sex of calf from the sexed semen agree with biological expectations (i.e. 1:1 ratio)? 3. Using the conventional semen as a control (expectation), does sex of calf agree with internal control? • α=0.05 23 Goodness-of-Fit: Conventional Semen H0: pf = 0.50, pm = 0.50 sex ratio agrees with biological expectations. Ha: sex ratio does NOT agree with biological expectations. 24 12

13. 9/22/23 Goodness-of-Fit: Conventional Semen (O-E)2 (O-E)2/E Sex of Calf Observed Expected (O-E) Heifer 521×0.47 = 244.87 521×0.53 = 276.13 521×0.50 = 260.5 521×0.50 = 260.5 -15.63 244.30 0.94 Bull 15.63 244.30 0.94 χ2 = 1.88 d.f. = 1 25 P-value for χ2 • Upper tail (only) • P-value = pchisq(χ2calc,df,lower.tail=FALSE); 26 13

14. 9/22/23 Critical Value for χ2 • Upper tail (only) • χ2crit = qchisq(1- α,df); 27 P-Value Rules for χ2 • If the P-value is greater than α, you will fail to reject H0 • If the P-value is less than α, you will reject H0 28 14

15. 9/22/23 Critical Value Rules for χ2 • Upper tail (only) • If χ2calc is greater than χ2crit , you will reject H0 • If χ2calc is less than χ2crit , you will fail to reject H0 29 P- value pchisq(1.88,1,lower.tail=FALSE) = 0.1703341 30 15

16. 9/22/23 critical value qchisq(.95,1) = 3.841459 31 Goodness-of-Fit: Conventional Semen • Conclusion: • Because the p-value of 0.1703341 is greater than 0.05 (α), we fail to reject the null hypothesis. • Because the test statistic of 1.88 is less than the critical value of 3.841459, we fail to reject the null hypothesis. • Implications: • for conventional semen, sex ratio agrees with biological expectations. 32 16

17. 9/22/23 Goodness-of-Fit: Sexed Semen H0: pf = 0.50, pm = 0.50 sex ratio agrees with biological expectations. Ha: sex ratio does NOT agree with biological expectations. 33 Goodness-of-Fit: Sexed Semen (O-E)2 (O-E)2/E Sex of Calf Observed Expected (O-E) Heifer 514×0.84 = 431.8 514×0.16 = 82.2 514×0.50 = 257.0 514×0.50 = 257.0 174.8 30,555 118.9 Bull -174.8 30,555 118.9 χ2 = 237.8 d.f. = 1 34 17

18. 9/22/23 P-value for χ2 • Upper tail (only) • P-value = pchisq(χ2calc,df,lower.tail=FALSE); 35 Critical Value for χ2 • Upper tail (only) • χ2crit = qchisq(1- α,df); 36 18

19. 9/22/23 P-Value Rules for χ2 • If the P-value is greater than α, you will fail to reject H0 • If the P-value is less than α, you will reject H0 37 Critical Value Rules for χ2 • Upper tail (only) • If χ2calc is greater than χ2crit , you will reject H0 • If χ2calc is less than χ2crit , you will fail to reject H0 38 19

20. 9/22/23 P- value pchisq(237.8,1,lower.tail=FALSE) = 1.186896e-53 39 critical value qchisq(.95,1) = 3.841459 40 20

21. 9/22/23 Goodness-of-Fit: Conventional Semen • Conclusion: • Because the p-value of 1.186896e-53 is less than 0.05 (α), we reject the null hypothesis. • Because the test statistic of 237.8 is greater than the critical value of 3.841459, we reject the null hypothesis. • Implications: • for sexed semen, sex ratio does NOT agree with biological expectations; predominately heifer calves. 41 Goodness-of-Fit: Sexed Semen, internal control H0: pf = 0.47, pm = 0.53 sex ratio agrees with internal control (conventional semen) expectations. Ha: sex ratio does NOT agree with internal control (conventional semen) expectations. 42 21

22. 9/22/23 Goodness-of-Fit: Sexed Semen, internal control (O-E)2 (O-E)2/E Sex of Calf Observed Expected (O-E) Heifer 514×0.84 = 431.8 514×0.16 = 82.2 514×0.47 = 241.6 514×0.53 = 272.4 190.2 36,176 149.7 Bull -190.2 36,176 132.8 χ2 = 282.5 d.f. = 1 43 P-value for χ2 • Upper tail (only) • P-value = pchisq(χ2calc,df,lower.tail=FALSE); 44 22

23. 9/22/23 Critical Value for χ2 • Upper tail (only) • χ2crit = qchisq(1- α,df); 45 P-Value Rules for χ2 • If the P-value is greater than α, you will fail to reject H0 • If the P-value is less than α, you will reject H0 46 23

24. 9/22/23 Critical Value Rules for χ2 • Upper tail (only) • If χ2calc is greater than χ2crit , you will reject H0 • If χ2calc is less than χ2crit , you will fail to reject H0 47 P- value pchisq(282.5,1,lower.tail=FALSE) = 2.14196e-63 48 24

25. 9/22/23 critical value qchisq(.95,1) = 3.841459 49 25