# Solved (Free): The color vision of birds plays a role in their foraging behavior: Birds use color to select and avoid certain types of food

#### ByDr. Raju Chaudhari

Apr 3, 2021

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]:519â526)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 birdâs 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.