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

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

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

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 $df=k-1=3-1 =2$.

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

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

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

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

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.

Further Reading

• Statistics
• Descriptive Statistics
• Probability Theory
• Probability Distribution
• Hypothesis Testing
• Confidence interval
• Sample size determination
• Non-parametric Tests
• Correlation Regression
• Statistics Calculators