In: Statistics and Probability
The article “Linkage Studies of the Tomato (Trans. Royal Canad.
Inst. (1931): 1-19) reported the following data on phenotypes
resulting from crossing tall cut-leaf tomatoes with dwarf
potato-leaf tomatoes. There are four possible phenotypes: (1) tall
cut-leaf, (2) tall potato-leaf, (3) dwarf cut-leaf, (4) dwarf
potato-leaf. Mendel’s laws of inheritance imply that π1= 9/16, π2=
3/16, π3= 3/16, π4= 1/16. (π means probability) Using the 0.01
significance level, test to determine if the reported data are
different from what we would have expected with Mendel’s
laws?
  
Find: Research question
b.) Hypothesis in Words (Ho, Ha)
c.Decision rule
d.Test and Results, name of test, results
a) Research question
Here, probabilities of four types of phenotypes are given. So,
we have observe sample data and collect observed
frequencies 
 of different types. Also using the probabilities given we
have to compute expected
frequencies 
 of all four types. We have to proceed as follows.
Suppose, we observed types for N trees. Then we get as follows.
| Type | Observed frequency
 
 | 
Expected frequency
 
 | 
| 1. Tall cut-leaf | ![]()  | 
![]()  | 
| 2. Tall potato-leaf | ![]()  | 
![]()  | 
| 3. Dwarf cut-leaf | ![]()  | 
![]()  | 
| 4. Dwarf potato-leaf | ![]()  | 
 
   | 
| Total | N | N | 
Now, we have to perform Chi-square test of goodness of fit.
We have to test for null
hypothesis 
against the alternative
hypothesis 
Number of groups 
Our test statistic is given by

Based on observed and expected frequencies we obtain calculated
value of test statistic as 
.
Level of significance 
Degrees of freedom 
So, critical value is given by
[Using R-code 'qchisq(1-0.01,3)']
Decision
rule- We reject null hypothesis if 
 or can not reject otherwise.