B how many degrees of freedom df are there for each x2 test

Problem 3.18 full pods 845 constricted pods 336 violet flowers 687 white flowers Probability (p) 0,90 0.50 0.20 0.05 0,01 0.001 1 0.02 0.46 64 3.84 6.64 10.83 2 0.21 1.39, 3.22 5.99 9.21 13.82 3 0.58 2.37 4:64. 7.82 11.35 16.27 4 1.06 3.36 5.99 9.49 13.28 18.47 5 1.61 4:35 7.29 11.07 15.09 20.52 6 2.20 5.35 8.56 12.59 16.81 22.46 df 7 2.83 6.35 9:80 14.07 1848 24.32 8 3.49 7.34 9 4.17 8.34 12.24 16.92 20.09 26.13 10 4.87 9.34 27.88 13.44 18.31 23.21 29.59 15 8.S5 14,34 19.31 25.00 30.58 37.30 25 16.47 24.34 30 68 37 65 44.31 52.62 50 67 76.15 86.60 X2 values Fails to reject the null hypothesis Rejects the null hypothesis e Pearson Education, Inc.

Solution

Chi-Square value= 7.49

Degrees of freedom= number of phenotypes - 1

= 2-1=1

Reject null hypothesis.

Phenotype Observed (O) Expected(O) O-E (O-E)^2 (O-E)^2/E

Violet flowers 687 696.75 -9.75 95.06 0.14

White flowers 242 232.25 9.75 95.06 0.41

Chi-Square value= 0.55

Degrees of freedom= number of phenotypes - 1

= 2-1=1

Fails to reject null hypothesis.

Phenotype	Observed (O)	Expected(O)	O-E	(O-E)^2	(O-E)^2/E
Full pods	845	885.75	-40.75	1660.56	1.87
Constricted pods	336	295.25	40.75	1660.56	5.62