# Solved (Free): A cross between white and yellow summer squash gave progeny of the following colors

#### ByDr. Raju Chaudhari

Apr 3, 2021

A cross between white and yellow summer squash gave progeny of the following colors:

Color White Yellow Green
No. of Progeny 155 40 10

Suppose the sample had the same composition but was 10 times as large: 1,550 white, 400 yellow, and 100 green progeny.

Are these data consistent with the 12:3:1 ratio predicted by a certain genetic model? Use a chi-square test at $\alpha = 0.10$.

#### Solution

The proportions are $p{White}=12/16 =0.75, p{Yellow} =3/16=0.1875, p_{Green} = 1/16=0.0625$.

The observed data is

color Obs. Freq.$(O)$ Prop.
White 1550 0.75
Yellow 400 0.1875
Green 100 0.0625
##### Step 1 Setup the hypothesis

The null and alternative hypothesis are as follows:

$H0: p{White}=12/16 =0.75, p{Yellow} =3/16=0.1875, p{Green} = 1/16=0.0625$

(i.e., Data are consistent with the 12:3:1 ratio predicted by a certain genetic model)

$H_1:$ Data are not consistent with the 12:3:1 ratio predicted by a certain genetic model.

##### Step 2 Test statistic

The test statistic for testing above hypothesis is

 $$\begin{equation*} \chi^2= \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ \end{equation*}$$ 

##### Step 3 Level of Significance

The level of significance is $\alpha =0.1$.

##### Step 4 Critical value of $\chi^2$

The level of significance is $\alpha =0.1$. Degrees of freedom $df=k-1=3-1 =2$. chi-square critical region

The critical value of $\chi^2$ for $df=2$ and $\alpha=0.1$ level of significance is $\chi^2 =4.6052$.

##### Step 5 Test Statistic

The expected frequencies can be calculated as

 $$\begin{equation*} E_{i} =N*p_i \end{equation*}$$ 

For example, $E_{1}$ is given by

 $$\begin{eqnarray*} E_{1} & = &N*p_1\\ &=& 2050*0.75\\ &=&1537.5. \end{eqnarray*}$$ 

 $$\begin{eqnarray*} E_{2} & = &N*p_2\\ &=& 2050*0.1875\\ &=&384.375. \end{eqnarray*}$$ 

 $$\begin{eqnarray*} E_{3} & = &N*p_3\\ &=& 2050*0.0625\\ &=&128.125. \end{eqnarray*}$$ 

color Obs. Freq.$(O)$ Prop. $p_i$ Expe.Freq.$(E)$ $(O-E)^2/E$
White 1550 0.75 1537.5 0.102
Yellow 400 0.1875 384.375 0.635
Green 100 0.0625 128.125 6.174

The test statistic is

 $$\begin{eqnarray*} \chi^2&=& \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ &=&\frac{(1550-1537.5)^2}{1537.5}+\cdots + \frac{(100-128.12)^2}{128.12}\\ &=& 0.102 +\cdots + 6.174\\ &=& 6.911. \end{eqnarray*}$$ `

##### Step 6 Decision (Traditional approach)

The test statistic is $\chi^2 =6.911$ which falls $inside$ the critical region bounded by the critical value $4.6052$, we $\textit{reject}$ the null hypothesis.

OR

##### Step 6 Decision ($p$-value approach)

The p-value is $P(\chi^2_{2}>6.911) =0.03157$.

As the p-value $0.0316$ is $\textit{less than}$ the significance level of $\alpha = 0.1$, we $\textit{reject}$ the null hypothesis.

These data are not consistent with the 12:3:1 ratio predicted by a certain genetic model at $\alpha = 0.1$ level of significance.