Contingency X
P(aB) = P(a) P(B) = 0.25(0.857) = 0.21425.
The expected number of aaB_ progeny is 0.21425(280) = 59.99.
P(aabb) = P(a) P(b) = 0.25(0.143) = 0.03575.
The expected number of aabb progeny is 0.03575(280) = 10.5
| |
B |
b |
| A |
180
(179.97) |
30
(30.03) |
| a |
60
(59.99) |
10
(10.5) |
We can now test Ho:A and B are independent factors using
a Contingency X
test. We will use Yate’s correction, due to small numbers
in some classes. If O-E is less than 0.5, then we set O-E
= O.
X
= S(| O-E|
-1/2)
E
= (|180 - 179.97|
- 1/2)
+ (|30 - 30.03| - 1/2)
179.97
30.03
+ (|60 - 59.99|
- 1/2)
+ (|10 - 10.5| - 1/2)
59.99
10.5
= 0.00 + 0.00 + 0.00 + 0 = 0.0
A contingency X
test of 0.0 with1df is not significant. When a
= 0.05 the critical X
value must be 3.84 or larger. We could use the formula for
a 2 x 2 contingency X
test with Yate’s correction term.
X
= (ad-bc - 1/2N)
N
(a+b)(a+c)(c+d)(b+d)
| |
B |
b |
|
| A |
180(a) |
30(b) |
a+b |
| a |
60(C)
a+c |
10(d)
b+d |
c+d |
X
= [180(10) - 30(60) - 1/2(280)]
280
(210)(240)(70)(40)
= (1800-1800-140)
280
141120000
= 5488000
141120000
= 0.039 ~ 0
Which is a non-significant X
when testing Ho:Factors A and B are independent. The explanation
for the significant deviation from the 9:3:3:1 ratio is that
there is a distorted segregation ratio at the B locus.
The
rows show how the proportions vary for the B locus for the
same level of the A factor. The columns show how the proportions
vary for the A locus and a fixed level of factor B.
