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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!)

Step by step solution

01

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 \(H_1: \mu > 0\).

02

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: \(t = \frac{3.9 - 0}{6.8 / \sqrt{32}} = 3.37.\)

03

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.

04

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!).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

T-distribution

The T-distribution is a fundamental concept in hypothesis testing. It is especially used when dealing with small sample sizes, typically less than 30, or when the population standard deviation is unknown. In the pigeon study, we used the T-distribution because the sample size, consisting of 32 measurements, is borderline where a T-distribution often applies.

The T-distribution is similar to a normal distribution but has thicker tails, which means that it is more prone to producing values that fall far from its mean. This makes it an ideal choice in statistical scenarios where small sample sizes create increased uncertainty. This allows researchers to account for the variability due to smaller samples.

When using a T-distribution, we must reference a T-table, which provides the critical t-values needed to make statistical decisions. These critical values vary based on degrees of freedom (calculated as the sample size minus one) and the level of significance chosen for the test. For instance, with 31 degrees of freedom and a 5% level of significance, our pigeon study uses a t-critical value to determine if the test's results are significant.

Null Hypothesis

The null hypothesis, denoted as \(H_0\), is a starting point in hypothesis testing. It represents a statement of no effect or no difference. In the context of the pigeon experiment, the null hypothesis is that the mean difference in the number of pigeons at the neutral site minus the hostile site equals zero. This suggests that pigeons show no preference and possibly do not recognize faces.

Formulating a null hypothesis allows researchers to have a specific claim to test against. If the collected data provides sufficient evidence to reject \(H_0\), it suggests there might indeed be a significant effect or association. However, if it is not rejected, we maintain that our initial assumption of no effect (or difference) holds true.

Rejecting the null hypothesis requires calculating a test statistic, such as a t-score, and comparing it with a critical value to assess the evidence provided by the sample data. In our example, a calculated t-score greater than a critical t-value provides grounds to reject the null hypothesis.

Alternative Hypothesis

The alternative hypothesis, denoted as \(H_1\) or \(H_a\), is what a researcher aims to support. It proposes that there is an effect or a difference. In the pigeon study, the alternative hypothesis claims that the mean difference in the number of pigeons at the neutral versus the hostile site is greater than zero. This would imply pigeons can distinguish between persons, possibly remembering and avoiding hostile faces.

An alternative hypothesis is typically expressed as a one-tailed or two-tailed hypothesis. In our scenario, it's a one-tailed hypothesis because we are specifically investigating if the mean difference is greater than zero, indicating a directional effect.

To establish support for the alternative hypothesis, researchers need evidence that conclusively challenges the null hypothesis. Once a calculated t-score exceeds the critical t-value, the data supports \(H_1\). Indeed, the pigeon study's t-score of 3.37 being higher than the critical value indicates evidence sufficiently strong to assert that pigeons hold grudges against perceived hostile persons.