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Chapter 13: Problem 4

It is hypothesized that when homing pigeons are disoriented in a certain manner, they will exhibit no preference for any direction of flight after takeoff (so that the direction \(X\) should be uniformly distributed on the interval from \(0^{\circ}\) to \(360^{\circ}\) ). To test this, 120 pigeons are disoriented, let loose, and the direction of flight of each is recorded; the resulting data follows. Use the chisquared test at level 10 to see whether the data supports the hypothesis. \begin{tabular}{l|ccc} Direction & \(0-<45^{\circ}\) & \(45-<90^{\circ}\) & \(90-<135^{\circ}\) \\ \hline Frequency & 12 & 16 & 17 \end{tabular} \begin{tabular}{l|ccc} Direction & \(135-<180^{\circ}\) & \(180-<225^{\circ}\) & \(225-<270^{\circ}\) \\ \hline Frequency & 15 & 13 & 20 \end{tabular} \begin{tabular}{l|cc} Direction & \(270-<315^{\circ}\) & \(315-<360^{\circ}\) \\ \hline Frequency & 17 & 10 \end{tabular}

Short Answer

Expert verified
The data supports that the pigeons show no directional preference.

Step by step solution

01

Define the Hypotheses

The null hypothesis (H0) states that the pigeons show no preference for any direction, meaning the directions are uniformly distributed. The alternative hypothesis (H1) is that the pigeons do have a preference, indicating a deviation from uniform distribution.
02

Calculate Expected Frequencies

Since the direction should be uniform, each of the 8 intervals should have the same frequency. Divide total pigeons (120) by the number of intervals (8):\[\text{Expected Frequency} = \frac{120}{8} = 15\]So, each interval is expected to have 15 pigeons.
03

Apply the Chi-Squared Formula

The chi-squared test statistic is calculated using the formula: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \]where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency. Calculate this for each interval and sum them up.
04

Calculate the Test Statistic

Using the observed frequencies (12, 16, 17, 15, 13, 20, 17, 10), and expected frequency (15), calculate for each direction: \((12-15)^2/15 + (16-15)^2/15 + (17-15)^2/15 + \ldots + (10-15)^2/15\) This results in: \[\chi^2 = \frac{9}{15} + \frac{1}{15} + \frac{4}{15} + 0 + \frac{4}{15} + \frac{25}{15} + \frac{4}{15} + \frac{25}{15}\ \approx 9.067\]
05

Determine the Critical Value

For a chi-squared test with \(n = 8\) intervals, the degrees of freedom is \(n - 1 = 7\). At a significance level of 0.10, use chi-squared distribution table to find the critical value: \(\chi^2_{0.10, 7} \approx 12.017\).
06

Compare Test Statistic with Critical Value

Compare the calculated test statistic (9.067) with the critical value (12.017). Since 9.067 is less than 12.017, we fail to reject the null hypothesis.
07

Conclusion

Since the test statistic is less than the critical value, we find insufficient evidence to reject the null hypothesis. Thus, at the 10% significance level, the data supports the hypothesis that pigeons show no preference for any direction.

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Key Concepts

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

Uniform Distribution
When we say something has a uniform distribution, it means that all outcomes are equally likely within a given range. Think of a fair die where each side has an equal chance of showing up. In the context of this exercise, the hypothesis suggests that the pigeons have no directional preference. Hence, each direction (broken into 8 intervals of 45 degrees) should have an equal number of pigeons flying in that direction. For 120 pigeons, this implies each interval should ideally guide 15 pigeons. This setup forms the foundation for conducting the statistical test to verify the uniformity of the distribution.
Statistical Hypothesis Testing
Statistical hypothesis testing is used to make decisions based on data. We start with two hypotheses: the null and the alternative. The null hypothesis (often symbolized as \(H_0\)) is a statement of no effect or no difference; in our exercise, it represents that the pigeons' flight directions are uniformly distributed. Contrarily, the alternative hypothesis (\(H_1\)) suggests that there is a directional preference among pigeons. The aim is to either "reject" or "fail to reject" the null hypothesis based on the data we have. Rejection indicates strong evidence in favor of the alternative hypothesis. Here, the chi-squared test checks if the observed distribution significantly deviates from the expected uniform distribution.
Critical Value
The critical value is a threshold used to judge whether the test statistic is extreme enough to reject the null hypothesis. It is determined based on the chosen significance level (often set at 0.05 or 0.10, as in this case) and the degrees of freedom. The chi-squared distribution, which depends on the degrees of freedom, helps in finding this critical value. In our context, with 7 degrees of freedom and at the 0.10 significance level, the critical value is approximately 12.017. The test statistic, 9.067, is compared against this critical value. Since 9.067 is less than 12.017, we do not reject the null hypothesis, concluding there isn't sufficient evidence that suggests the pigeons prefer any particular flight direction.
Degrees of Freedom
Degrees of freedom is an important concept in hypothesis testing. It relates to the number of values in a calculation that are free to vary without violating any constraints. For chi-squared tests, the degrees of freedom is generally calculated as the number of categories minus one; in this case, there are 8 flight intervals, so degrees of freedom is 7. Crucially, the degrees of freedom impact the shape of the chi-squared distribution used to determine the critical value. Understanding this helps in setting accurate critical bounds for hypothesis testing. Knowing the degrees of freedom, you refer to chi-squared distribution tables to find the critical value tailored to your specific testing scenario.

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Most popular questions from this chapter

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