Chapter 11.1, Problem 11E

To determine

The test for the given data at the significance level of the proposed generic model.

Answer to Problem 11E

  $P$ is greater than the significance level which means that fail to reject $H_0$ . So, the phenotypes occurs in the given ratio $9:3:3:1$

Explanation of Solution

Given:

Phenotype	Red eyes and straight wings	Red eyes and curly wings	White eyes and straight wings	White eyes and curly wings
Frequency	$99$	$42$	$49$	$10$

Significance level: $5\%=0.05$

Ratio: $R_1:R_2:\text{}{W}_1:\text{}{W}_2=\text{}9:3:3:1$

  $n=4$

Concept Used:

Null hypothesis: $H_0$

The null hypothesis is rejected when the value of $P$ is less or equal to the significance level.

Calculation:

The null hypothesis: $H_0$

The phenotype occur in a ratio of $9:3:3:1$

Alternative hypothesis: $H_1$

The phenotype does not occur in a ratio of $9:3:3:1$

Now,

Take the sum: $S=9+3+3+1=16$

Draw a table consisting of observed value and expected value:

The expected value is calculated as:

  $\begin{array}{l}{E}_1=T\times \frac{R_1}{R}=200\times \frac{9}{16}=112.5\\ E_2=T\times \frac{R_2}{R}=200\times \frac{3}{16}=37.5\\ E_3=T\times \frac{W_1}{R}=200\times \frac{3}{16}=37.5\\ E_4=T\times \frac{W_2}{R}=200\times \frac{1}{16}=12.5\end{array}$

Phenotype	Red eyes and straight wings	Red eyes and curly wings	White eyes and straight wings	White eyes and curly wings	Total $T$
Observed value $O_i$	$99$	$42$	$49$	$10$	$200$
Expected value $E_i$	$112.5$	$37.5$	$37.5$	$12.5$	$200$

Now, use the test −statistics:

  $\begin{array}{l}{x}^2={\displaystyle \sum _{i=1}^4\frac{{\left(O_i-E_i\right)}^2}{E_i}}\\ x^2=\frac{{\left(99-112.5\right)}^2}{112.5}+\frac{{\left(42-37.5\right)}^2}{37.5}+\frac{{\left(49-37.5\right)}^2}{37.5}+\frac{{\left(10-12.5\right)}^2}{12.5}\\ x^2=6.1867\end{array}$

Degree of freedom:

  $df=n-1=4-1=3$

For $x^2=6.1867$ and $df=3$ the $P$ -value is $0.1029$ .

So, the null hypothesis is not rejected when the value of $P$ is not less or equal to the significance level.

  $P>\propto :\text{}0.1029>\text{}0.05$ : fail to reject $H_0$ .

Hence,

The phenotypes occurs in the given ratio $9:3:3:1$

Conclusion:

Therefore, $P>\propto :\text{}0.1029>\text{}0.05$