Chapter 11.1, Problem 3.1CYU

To determine

To explain do these data differ significantly from what the biologist have predicted and carry out a test at the $\alpha =0.01$ significance level.

Answer to Problem 3.1CYU

Yes, these data differ significantly from what the biologist have predicted.

Explanation of Solution

In the question it is given that, biologists wishes to mate pairs of fruit flies having genetic makeup. The total offspring are given to be $200$ .Thus, according to the given data we will make the contingency table as:

First we will find the expected frequencies for the four genetic fruits flies as:

The ratio is given for the four genetic fruit, that is, $9:3:3:1$ , so the expected frequencies will be as,

For red-eyes straight winged:

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

For red eyed curly winged and white eyed straight winged:

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

For white eyed curly winged:

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

Thus, the contingency table will be as:

	Observed	Expected
Red eyed straight winged	99	112.5
Red eyed curly winged	42	37.5
White eyed straight winged	49	37.5
White eyed curly winged	10	12.5

Now, let us check the conditions as:

Random condition: It is satisfied as the data selected for the genetic fruits flies is randomly selected.

Large sample condition: It is satisfied as the total offspring are given to be $200$ , which is large sample and more than ten.

Independent condition: It is satisfied as individual offspring inherit their traits independently from each other.

Thus, all the conditions are met.

Let us define the hypothesis as:

  $\begin{array}{l}{H}_0:\text{The data do not differ}\\ H_1:\text{The data differ}\end{array}$

Now, let us calculate the chi-square as:

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

The degree of freedom will be as:

  $\begin{array}{l}df=\text{ Number of categories}-1\\ =4-1\\ =3\end{array}$

Thus, the P-value will be then as:

  $P=0.0129$

Thus, if the P-value is less than the significance level then we will reject the null hypothesis. Thus, we have,

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

Thus, we conclude that we do not have evidence to support the claim. Thus, these data differ significantly from what the biologist have predicted.