Question

6.122 Be Nice to Pigeons, As They Remember Your Face In a study \(^{30}\) conducted in Paris, France, equal amounts of pigeon feed were spread on the ground in two adjacent locations. A person was present in both sites, with one acting hostile and running at the birds to scare them away and the other acting neutral and just observing. The two people were randomly exchanged between the two sites throughout and the birds quickly learned to avoid the hostile person's site and to eat at the site of the neutral person. At the end of the training session, both people behaved neutrally but the birds continued to remember which one was hostile. In the most interesting part of the experiment, when the two people exchanged coats (orange worn by the hostile one and yellow by the neutral one throughout training), the pigeons were not fooled and continued to recognize and avoid the hostile person. The quantity measured is difference in number of pigeons at the neutral site minus the hostile site. With \(n=32\) measurements, the mean difference in number of pigeons is 3.9 with a standard deviation of 6.8 . Test to see if this provides evidence that the mean difference is greater than zero, meaning the pigeons can recognize faces (and hold a grudge!)

Short answer

Yes, the evidence shows that pigeons can recognize faces. Since the calculated t-score (3.37) is greater than the critical t-value (1.696), we reject the null hypothesis that there is no difference in the number of pigeons between the neutral site and the hostile site.

Step-by-step solution

Formulate Hypothesis

The null hypothesis (\(H_0\)) is that the mean difference, \(\mu\), equals zero (there is no difference in the number of pigeons between sites). The alternative hypothesis (\(H_1\)) is that \(\mu\) is greater than zero (there are more pigeons at the neutral site than at the hostile one). So, \(H_0: \mu = 0\) and \(H_1: \mu > 0\).

Calculate t-score

The t-score is calculated as follows: \(t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu_0\) is the mean under the null hypothesis, \(s\) is the standard deviation and \(n\) is the sample size. Substituting the given information gives: \(t = \frac{3.9 - 0}{6.8 / \sqrt{32}} = 3.37.\)

Refer t-distribution Table

Refer a t-distribution table to find the critical t-value. As the sample size is 32, the degrees of freedom is \(n-1=31\), and if we test at a 5% level of significance for a one-tailed test, the critical t-value is approximately 1.696.

Compare t-scores and Draw Conclusion

Since the calculated t-score (3.37) is greater than the critical t-value (1.696), we reject the null hypothesis. Therefore, the provides evidence that pigeons can recognize faces (and hold a grudge!).