Q.3.1 - The Practice of Statistics for AP

Question

Biologists wish to mate pairs of fruit flies having genetic makeup RrCc, indicating that each has one dominant gene (R) and one recessive gene (r) for eye color, along with one dominant (C) and one recessive (c) gene for wing type. Each offspring will receive one gene for each of the two traits from each parent. The following Punnett square shows the possible combinations of genes received by the offspring: 

Any offspring receiving an R gene will have red eyes, and any offspring receiving a C gene will have straight wings. So based on this Punnett square, the biologists predict a ratio of 9 red-eyed, straight-winged (x):3 red-eyed, curly-winged (y):3 white-eyed, straight-winged (z):1 white-eyed, curly-winged (w) offspring. To test their hypothesis about the distribution of offspring, the biologists mate a random sample of pairs of fruit flies. Of 200 offspring, 99 had red eyes and straight wings, 42 had red eyes and curly wings, 49 had white eyes and straight wings, and 10 had white eyes and curly wings. Do these data differ significantly from what the biologists have predicted? Carry out a test at the A  0.01 significance level 

Step-by-Step Solution

Answer

Yes, these results are dramatically different from what the biologist projected.

Step 1: Given Information

Given in the question that the possible combinations of genes received by the offspring:  

Step 2: Calculate the expected frequencies

The following are the estimated frequencies for the four genetic fruit flies, based on the information provided:  

This is the weighted average of the four genetic fruits:$9: 3: 3: 1$

The following are the predicted frequencies for red-eyes with straight wings. 

$$\begin{gathered} =\frac{9}{16}\times 200 \\ =112.50 \end{gathered}$$

Then, for the red eyed curly winged and white eyed straight winged, compute the predicted frequencies:  

$$\begin{gathered} =\frac{3}{16}\times 200 \\ =37.50 \end{gathered}$$

Calculate the white eyed curly winged's predicted frequency:  

$$\begin{gathered} =\frac{1}{16}\times 200 \\ =12.50 \end{gathered}$$

As a result, here is how the contingency table will look: 

$\begin{array}{|lrr|}\hline & \text{ Observed }& \text{ Expected }\\ \text{ Red eyed straight winged }& 99& 112.5\\ \text{ Red eyed curly winged }& 42& 37.5\\ \text{ White eyed straight winged }& 49& 37.5\\ \text{ White eyed curly winged }& 10& 12.5\\ \hline\end{array}$

Step 3: Check the condition

Now, consider the following conditions: 

Random circumstance: It is satisfied since the genetic fruits flies data was chosen at random. It is satisfied because the total offspring are given to be 200, which is a large sample and greater than ten. 

Individual kids inherit their qualities independently from one another, satisfying the independent condition. As a result, all of the requirements are met. 

Assume the following hypothesis: 

$H_0$: The data are same

$H_1$: The data are different

Now compute the chi square as follows:

$$\begin{gathered} {\chi }^2=\frac{(99-112.5{)}^2}{112.5}+\frac{(42-37.5{)}^2}{37.5}+\frac{(49-37.5{)}^2}{37.5}+\frac{(10-12.5{)}^2}{12.5} \\ =6.187 \end{gathered}$$

Now, check the degree of freedom:

$$\begin{gathered} d f= Number of categories -1 \\ =4-1 \\ =3 \end{gathered}$$

Therefore, the $p$ value will be:

$p=0.0129$

We shall reject the null hypothesis if the P-value will be less than the significance level. 

As a result, we have  

$P>0.01\Rightarrow \text{ Not Reject }{H}_0$