X2
Tests Part IV
Contingency X2
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.
| X2 = S |
(| O - E |- 1/2)2 |
| E |
| = |
(|80 - 83.17| - 1/2)2 |
+ |
(|30 - 26.85| - 1/2)2 |
| 83.17 |
26.85 |
| + |
(|90 - 86.92| - 1/2)2 |
+ |
(|25 - 28.05| - 1/2)2 |
| 86.92 |
28.05 |
 = |
0.0857 + 0.2615 + 0.0766 + 0.2318 |
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:
Dominant: Recessive
= 3:1
Experiment 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 |
|
|
| B |
|
|
| = |
(|48 - 48| - 0.5)2 |
+ |
(|16 - 16| - 0.5)2 |
| 48 |
16 |
| = |
(|48 - 48| - 0.5)2 |
+ |
(|16 - 16| - 0.5)2 |
| 48 |
16 |
| |
= 0 + 0 + 0 + 0 = 0 When |O-E| < 0.5 then set |
| |
|
numerator to zero. |
