The color vision of birds plays a role in their foraging behavior:

Birds use color to select and avoid certain types of food. The authors of the article "Colour Avoidance in Northern Bob whites :Effects of Age,Sex, and Previous Experience" (Animal Behaviour [1995]:519â526)studied the pecking behavior of one-day-old bob whites. In an area painted white, they inserted four pins with different colored beads. The color of the pin chosen on the birdâs first peck was noted for each of 33 bob whites, resulting in the following data:

Color | Blue | Green | Yellow | Red |
---|---|---|---|---|

First Peck Frequency | 16 | 8 | 6 | 3 |

Do the data provide evidence of a color preference? Use $\alpha =0.05$.

#### Solution

The observed data is

Color | Obs. Freq.$(O)$ | Prop. |
---|---|---|

Blue | 16 | 0.25 |

Green | 8 | 0.25 |

Yellow | 6 | 0.25 |

Red | 3 | 0.25 |

##### Step 1 Setup the hypothesis

The null and alternative hypothesis are as follows:

`$H_0:p_{1} =p_{2} =p_{3}=p_4=\frac{1}{4}$`

$H_1:$ At least one of the proportion is different from `$\frac{1}{4}$`

.

##### 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 $\alpha =0.05$.

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

The level of significance is $\alpha =0.05$. Degrees of freedom $df=k-1=4-1 =3$.

The critical value of $\chi^2$ for $df=3$ and $\alpha=0.05$ level of significance is $\chi^2 =7.8147$.

##### 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

` $$ \begin{eqnarray*} E_{1} & = &N*p_1\\ &=& 33*0.25\\ &=&8.25. \end{eqnarray*} $$ `

Color | Obs. Freq.$(O)$ | Prop. $p_i$ | Expe.Freq.$(E)$ | $(O-E)^2/E$ |
---|---|---|---|---|

Blue | 16 | 0.25 | 8.25 | 7.28 |

Green | 8 | 0.25 | 8.25 | 0.008 |

Yellow | 6 | 0.25 | 8.25 | 0.614 |

Red | 3 | 0.25 | 8.25 | 3.341 |

The test statistic is

` $$ \begin{eqnarray*} \chi^2&=& \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ &=&\frac{(16-8.25)^2}{8.25}+\cdots + \frac{(3-8.25)^2}{8.25}\\ &=& 7.28 +\cdots + 3.341\\ &=& 11.243. \end{eqnarray*} $$ `

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

The test statistic is $\chi^2 =11.243$ which falls $inside$ the critical region bounded by the critical value $7.8147$, we $\textit{reject}$ the null hypothesis.

**OR**

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

The p-value is $P(\chi^2_{3}>11.243) =0.01048$.

As the p-value $0.0105$ is $\textit{less than}$ the significance level of $\alpha = 0.05$, we $\textit{reject}$ the null hypothesis.

Thus we conclude that the data provide evidence of a color preference. That is at least one of the proportion is different from the expected proportion.

#### Further Reading

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