Rock Sparrows. Rock Sparrows breeding in northern Italy are the subject of a long-term ecology and conservation study due to their wide variety of breeding patterns. Both males and females have a yellow patch on their breasts that is thought to play a significant role in their sexual behavior. A Pilastro et al. concluded an experiment in which they increased or reduced the size of a female's breast patch by dying feathers at the edge of a patch and then observed several characteristics of the behavior of the male. Their results were published in the paper "Male Rock Sparrows Adjust Their Breeding Strategy According to Female Ornamentation: Parental or Mating Investment?" (Animal Behaviour, Vol. $66$, Issue $2$, pp. $265-271$). Eight mating pairs were observed in each of three groups: a reduced-patch-size group, a control group, and an enlarged-patch-size group. The data on the WeissStats site, based on the results reported by the researchers, give the number of minutes per hour that males sang in the vicinity of the nest after the patch size manipulation was done on the females.

a.

Examine the data to evaluate if there is enough evidence to establish that the three types of breast treatments had different mean singing rates in male Rock Sparrows.

It is necessary to state the null and alternative hypotheses.

Null Hypothesis:

$H_0:{\mu }_{Northeast}={\mu }_{Reduced}={\mu }_{Control}={\mu }_{Enlarged}$

That is, the mean singing rates of male Rock Sparrows treated with the three breast treatments are identical.

Alternative hypothesis:

$H_a$: A minimum of one ${\mu }_1$in comparison to others.

That is, the mean singing rates of male Rock Sparrows treated to three different types of breast treatments varied.

The importance level here is, $\alpha =0.01$

Calculate the value of the test statistic:

MINITAB procedures:

Step 1: Select Stat > ANOVA > One-Way Analysis from the menu bar.

Step 2: Type Rate in the Response section.

Step 3: In the Factor column, type Breast.

Step 4: Press the OK key.

One-way ANOVA: RATE versus BREAST

Method

Null hypothesis All resources are equal.

Alternative hypothesis One of the means is different.

Significance level $\alpha =0.01$

For the sake of the analysis, equal variances were assumed.

Factor Information

Factor Levels Values

Breast $3$ Control, Enlarged, Reduced

Analysis of Variance

Source DF Adj ss Adj MS F-Value P-Value

BREAST $2$ $960.3$ $480.14$ $6.09$ $0.008$

Error $21$ $1655.3$ $78.82$

Total $23$ $2615.6$

Model Summary

S R-sq R-sq(adj) R-sq(pred)

$8.87834$ $36.71\%$ $30.69\%$ $17.34\%$

Means

BREAST N Mean stDev $99\%$CI

Control $8$ $14.84$ $6.56$ $(5.95,23.73)$

Enlarged $8$ $19.78$ $13.05$ $(10.89,28.66)$

Reduced $8$ $4.59$ $4.82$ $(-4.30,13.48)$

Pooled StDev $=8.87834$

The value of F is $6.09$and the p-value is $0.008$according to the MINITAB output.

b.

p-value approach:

MINITAB calculated a p-value of $0.008$.

Rejection Rule:

The null hypothesis must be rejected if $P\le \alpha$an is true.

The p-value is $0.008$, which is below the threshold of significance. That is. $p(=0.008)<a=(0.001)$. As a result, the null hypothesis is rejected at a $1\%$level.

As a result, at the $1\%$level of significance, the test results might be declared statistically significant.

The findings are sufficient to show that there is a difference in mean singing rates among male Rock Sparrows exposed to the three types of breast treatments at a $1\%$level of significance.

c.

The residual and residual versus fits normal probability graphs should be obtained.

MINITAB PROCEDURES:

To begin, go to Stat > ANOVA > One-way ANOVA.

Step 2: Type Rate in the Response section.

Step 3: In the Factor column, type Breast.

Step 4: In the graph, select normal probability of residual and residual versus fits.

Step 5: Press the OK key.

MINITAB output: Residual normal probability plot

MINITAB output: Fits vs. residues

The largest to smallest standard deviation ratio is,

$$\begin{gathered} Ratio=\frac{13.05}{4.82} \\ =2.7 \end{gathered}$$

The highest to smallest deviation ratio is more than 2. This shows that the equal standard deviation assumption has been broken.