Solution:
Part a
The null hypothesis is given as below:
Null hypothesis: H0: Ratio is equal to 3:1. (Trait is controlled by a single gene.)
Part b
The alternative hypothesis is given as below:
Alternative hypothesis: Ha: Ratio is not equal to 3:1. (Trait is not controlled by a single gene.)
Part c
Here, we have to use Chi square test for goodness of fit.
We assume 5% level of significance for this test. (? = 0.05)
Total number of categories = n = 2
Degrees of freedom = n – 1 = 2 – 1 = 1
Test statistic formula is given as below:
Chi square = ?[(O – E)^2/E]
Where O is observed frequencies and E is expected frequencies.
Calculation table is given as below:
Class |
O |
E |
(O - E)^2 |
(O - E)^2/E |
Round |
290 |
300 |
100 |
0.333333333 |
Shrunken2 |
110 |
100 |
100 |
1 |
Total |
1.333333333 |
Chi square = ?[(O – E)^2/E] = 1.333333333
DF = 1
P-value = 0.248213251
(By using Chi square table/Excel)
Part d
Here, P-value > ? = 0.05, so we do not reject the null hypothesis that Trait is controlled by a single gene.
There is sufficient evidence to conclude that trait is controlled by a single gene.
The Chi-square test is often used to test if different ratios are the same, for example...
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