Hypothesis Testing Using Binomial Distribution Part
II
Example
n = 8, r = 7, x = 1/2, y = 1/2
cross Aa x aa 
1/2 Aa, 1/2 aa progeny expected
| expected |
4 |
4 |
|
4 |
4 |
4 |
4 |
4 |
|
4 |
4 |
| #
of Aa individuals |
8 |
7 |
6 |
5 |
4 |
3 |
2 |
1 |
0 |
| Probability |
1
256 |
8
256 |
28 256 |
56 256 |
70
256 |
56
256 |
28 256 |
8
256 |
1
256 |
| Deviation |
4 |
3 |
2 |
1 |
0 |
1 |
2 |
3 |
4 |
| We
observe |
7
Aa |
and |
1
aa |
| We
expect |
4
Aa |
and |
4
aa |
| Deviation |
3
Aa |
|
3
aa |
Deviations of 3 or more occur for families of 8:0,
7:1, 1:7, 0:8 and the associated probabilities: 1/256,
8/256, 8/256, 1/256. The sum of these probabilities
for classes that represent a deviation of three or more
from expected is:
| 1 + 8 + 8 + 1 |
= |
18 |
= 0.07 or 7% |
|
|
|
|
| 256 |
256 |
|
|
|
|
A deviation of 3 or more would occur in 7% of families
of size eight. We decided that we would fail to reject
Ho: 1 Aa: 1 aa with the probability of Type I error
= 5%. We will fail to reject Ho because our observed
ratio of 7 Aa: 1aa would occur in 7% of families. We
conclude that our observed sample of 7 Aa: 1 aa is not
so unusual to cause us to reject Ho.
