Homogeneity X
Part II
The sum of the five X
calculated from the five separate plants is 2.05 with 5df.
Now let us calculate the pooled X,
based on the data summed across experiments for each of the
two classes.
| Plants |
Round |
Wrinkled |
| 1 |
45 |
12 |
| 2 |
27 |
8 |
| 3 |
24 |
7 |
| 4 |
19 |
10 |
| 5 |
32 |
11 |
| Totals |
147 |
48 |
| |
|
|
df |
X |
| Observed |
147 |
48 |
|
|
| Expected |
146.25 |
48.75 |
|
|
| (O-E)/E |
0.0038 |
0.0115 |
1 |
0.015 |
Now let us summarize all our X
calculations and find the homogeneity X.
| Source |
X |
df |
Probability |
| Summed
Exp. |
2.050 |
5 |
|
| Overall
pooled |
0.015 |
1 |
0.95-0.90 |
| Homogeneity |
2.035 |
4 |
0.90-0.70 |
The homogeneity X
is less than 9.49 and so the data is homogeneous. All five
experiments are evidence of the same 3:1 ratio. The pooled
data fit a 3:1 ratio overall. If the homogeneity X
had been larger than 9.49, then this would have been evidence
of heterogeneity among plants. Perhaps there may have been
one gene segregating in some plants for round vs. wrinkled
seed and two genes segregating in other plants.
** The Yate’s correction term is not used for calculation
of homogeneity X.
X2 Test
The X2 test is an appropriate test and the binomial
test is an exact test. The X2 test is adequate
when the total number of scored individuals is large. For
a 1:1 segregation ratio, the total number of classified individuals
should be 30 or more to justify use of the X2 test.
Normal Distribution
As the number of individuals in each class increases and
p=q=1/2, the binomial distribution approaches a normal distribution.
When the number individuals in an experiment is large enough,
the normal distribution can be used to provide confidence
intervals for the observed frequency of a category. The normal
distribution is used as an approximation to the binomial distribution
when pqn is greater than 25.
