Another Example Of Hypothesis Testing With Binomial Distribution
Example
We cross two plants and evaluate a locus with complete dominance gene
action. We cannot distinguish BB individuals from Bb individuals based
on phenotype. The second plant has the bb phenotype.
genotHo: Bb x bb 
1/2 Bb: 1/2 bb
genotHA: BB x bb
1 Bb: o bb
In a family of 4 offspring, we observe 4 B_ phenotypes. Is this an
unlikely event under the Ho hypothesis? Let probability (Bb) zygote
= 1/2, which is a ‘successful’ event. What is the probability of 4 ‘successes’
and zero ‘failures’ in a sample size of n = 4?
Let n = 4, r = 4, p = 1/2, q = 1/2
dddddddddddddddddddd
The probability of 4 Bb plants and zero bb plants in a family of size
four is 1/16. If we observe one or more bb plants, this is proof that
the cross is Bb x bb. We can only observe bb offspring if both parents
carry the b allele.
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r = number of Bb plants or successes
The rejection region is one or more failures.
If we have four Bb plants out of a total sample
size of four, then we have observed zero bb plants.
We have no evidence that the cross is Bb x bb.
If we have 3, 2, or 1 Bb plants, then we have
1, 2, or 3 bb plants, respectively. We only reject
Ho when we have zero bb plants (4 Bb plants).
The probability of four Bb plants = 1/16 = 0.0625.
This is a one-tailed test with a
= 0.0625. Because the likelihood of observing
four Bb offspring is small (1/16) when Ho: Bb
x bb is true, we reject Ho. Four Bb plants in
a family size of four is unlikely when Ho is true
and we have no evidence to support Ho.
* Probability of at least one bb plant in family
of 4 progeny = 1 - probability (zero bb plants)
= 1 - 0.06 = 0.94. If Ho is true, we would expect
at least one bb plant 94% of trials.
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