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:

`$H*0: p*{White}=12/16 =0.75, p*{Yellow} =3/16=0.1875, p*{Green} = 1/16=0.0625$

(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

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

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

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

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.

#### Further Reading

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