91影视

Significance level: $5\%=0.05$

Ratio: $R_1:R_2:W_1:W_2=9:3:3:1$

$n=4$

Null hypothesis: $H_0$

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

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: $S=9+3+3+1=16$

Create a table with the observed and expected values:

The expected value is calculated as:

$$\begin{gathered} E1=T脳\frac{R_1}{R}=200脳\frac{9}{16}=112.5 \\ {E}_2=T脳\frac{R_2}{R}=200脳\frac{3}{16}=37.5 \\ {E}_3=T脳\frac{W_1}{R}=200脳\frac{3}{16}=37.5 \\ {E}_4=T脳\frac{W_2}{R}=200脳\frac{1}{16}=12.5 \end{gathered}$$

Now, use the test 鈭抯tatistics:

$$\begin{gathered} x^2=\underset{i=1}{\overset{4}{鈭�}}\frac{{(O_i鈭�E_i)}^2}{E_i} \\ {x}^2=\frac{{(99鈭�112.5)}^2}{112.5}+\frac{{(42鈭�37.5)}^2}{37.5}+\frac{{(49鈭�37.5)}^2}{37.5}+\frac{{(10鈭�12.5)}^2}{12.5} \\ {x}^2=6.1867 \end{gathered}$$

Degree of freedom:

$df=n鈭�1=4鈭�1=3$

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.

$P>鈭�:0.1029>0.05$: fail to reject $H_0$

Hence,

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

Therefore, the$P>鈭�:0.1029>0.05$