X
Tests Part IV
Contingency X
Test
Row multiplied by respective column multiplied by total N.
Remember, when events are independent we can multiply the
probability of each separate event to get the joint probability
of both events. We then multiply this joint probability by
the total number of observations to get the expected number
of observations.
X
= S(|
O - E |- 1/2)
E
X
= (|80 - 83.17| -
1/2)
+ (|30 - 26.85| - 1/2)
83.17 26.85
X
+ (|90 - 86.92| - 1/2)
+ (|25 - 28.05| - 1/2)
86.92 28.05
X
= 0.0857 + 0.2615 + 0.0766 + 0.2318
X
= 0.66
Example
128 individuals with two experiments and 64 individuals per
experiment.
Observed:
| Environment |
Dominant |
Recessive |
Totals |
| A |
48 |
16 |
64 |
| B |
48 |
16 |
64 |
| |
96 |
32 |
128 |
Expected Relative Frequencies:
genotDominant: Recessive = 3:1
genotExperiment A: Experiment
B = 1:1
Expected Relative Frequencies:
| Environment |
Dominant |
Recessive
|
Relative
Frequency |
| A |
1
2 |
x |
3
4 |
1
2 |
x |
1
4 |
1
2 |
| B |
1
2 |
x |
3
4 |
1
2 |
x |
1
4 |
1
2 |
| |
3
4 |
1
4 |
|
| Environment |
Dominant |
Recessive |
| A |
3
x 128
8 x 128 |
1
x 128
8 x 128 |
| B |
3
x 128
8 x 128 |
1
x 128
8 x 128 |
X
= S(|O-E|
- 1/2)2
E
X
= (|48 - 48| - 0.5)2
+ (|16 - 16| - 0.5)2
48 16
X
= (|48 - 48| - 0.5)2
+ (|16 - 16| - 0.5)2
48 16
X
= 0 + 0 + 0 + 0 = 0 When |O-E| 
0.5 then set
X
= numerator to zero.
