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 Bobwhites:Effects of Age, Sex, and Previous Experience" (Animal Behaviour [1995]:519-526) studied the pecking behavior of one-day-old bobwhites.In an area painted white,they inserted four pins with different colored beads.The color of the pin chosen on the bird's first peck was noted for each of 33 bobwhites, 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? Test using $\alpha =0.01$.

#### 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 The null and alternative hypothesis are as follows:

$H_0:$ Bobwhites have no preference for one color over the other,

i.e., $H_0: p_1=p_2=p_3=p_4 = \frac{1}{4}=0.25$

$H_1:$ Bobwhites have preference for one color over the other.

#### 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 $\alpha =0.01$.

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

The level of significance is $\alpha =0.01$. Degrees of freedom $df=k-1=4-1 =3$.

The critical value of $\chi^2$ for $df=3$ and $\alpha=0.01$ level of significance is $\chi^2 =11.3449$.

#### 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 $outside$ the critical region bounded by the critical value $11.3449$, we $\textit{fail to reject}$ 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 $\textit{greater than}$ the significance level of $\alpha = 0.01$, we $\textit{fail to reject}$ the null hypothesis.

There is no sufficient evidence to support the alternative hypothesis. That is Bobwhites have no preference for one color over the other.

#### Further Reading

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