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Seagulls by the seashore Do seagulls show a preference for where they land? To answer this question, biologists conducted a study in an enclosed outdoor space with a piece of shore whose area was made up of \(56 \%\) sand, \(29 \%\) mud, and \(15 \%\) rocks. The biologists chose 200 seagulls at random. Each seagull was released into the outdoor space on its own and observed until it landed somewhere on the piece of shore. In all, 128 seagulls landed on the sand, 61 landed in the mud, and 11 landed on the rocks. Do these data provide convincing evidence that seagulls show a preference for where they land?

Short Answer

Expert verified
The data provide evidence that seagulls have a preference for where they land.

Step by step solution

01

Formulating Hypotheses

First, we need to set up our null and alternative hypotheses. The null hypothesis (\(H_0\)) is that the seagulls do not have a preference for landing on any type of terrain—sand, mud, or rocks. This implies the observed proportion of landings should match the area proportions given (56% sand, 29% mud, 15% rocks). The alternative hypothesis (\(H_a\)) is that there is a preference, meaning the observed proportions significantly differ from the expected ones.
02

Calculate Expected Frequencies

Calculate the expected frequencies for each type of terrain based on the given percentages and the total number of trials (seagulls). For sand, the expected frequency is \(0.56 \times 200 = 112\); for mud, \(0.29 \times 200 = 58\); for rocks, \(0.15 \times 200 = 30\).
03

Observed Frequencies

Record the observed frequencies from the data: 128 seagulls landed on sand, 61 on mud, and 11 on rocks.
04

Perform Chi-Square Test

Use the chi-square test to determine if there is a significant difference between observed and expected frequencies. The formula is: \[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \] where \(O_i\) is the observed frequency and \(E_i\) is the expected frequency. Plug in the values to get: \[ \chi^2 = \frac{(128-112)^2}{112} + \frac{(61-58)^2}{58} + \frac{(11-30)^2}{30} \] Simplify this to get the chi-square statistic.
05

Compare Chi-Square Statistic with Critical Value

Determine the degrees of freedom, which is the number of categories minus one (3 - 1 = 2 in this case). Use a chi-square table to find the critical value for \(\chi^2\) with 2 degrees of freedom at a typical significance level of 0.05. Compare the computed statistic to this critical value.
06

Conclude the Test

If the chi-square statistic is greater than the critical value from the table, reject the null hypothesis, which would suggest the seagulls show a preference. If not, we do not have enough evidence to claim a preference.

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Key Concepts

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

Chi-Square Test
The Chi-Square Test is a statistical method used to determine if there is a significant difference between observed and expected frequencies. In simple terms, it helps us check whether the actual data (what we see happening) aligns with what we would expect to happen based on a certain hypothesis.

To conduct a Chi-Square Test, you follow these steps:
  • Calculate the expected frequencies, which are derived from the null hypothesis.
  • Determine the observed frequencies, which are the actual counts from your experiment or study.
  • Apply the Chi-Square formula to compute the test statistic: \[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]where \(O_i\) is the observed frequency, and \(E_i\) is the expected frequency.
  • Compare the computed statistic with a critical value from a Chi-Square distribution table, considering the degrees of freedom and chosen significance level.
If the statistic exceeds the critical value, there might be a significant difference, suggesting that the observed outcomes are unlikely due to random chance alone.
Null Hypothesis
In hypothesis testing, the null hypothesis is a default statement that there is no effect or no difference. It is denoted by \(H_0\). In many experiments, like the seagull landing study, the null hypothesis suggests that observed outcomes match the expected distribution.

For the seagulls example, the null hypothesis would assert that seagulls land on different terrains (sand, mud, rocks) in proportion to their respective areas—meaning they show no preference.
The alternative hypothesis, on the other hand, would propose that there is a notable preference, which means that the actual landing frequencies differ significantly from that expected under \(H_0\).

In testing, our goal often is to see if there is enough evidence to reject the null hypothesis. If we can't, we conclude that the data is consistent with \(H_0\), implying no preference.
Expected Frequency
Expected Frequency is a key element in the Chi-Square Test. It represents the number of times we would anticipate an outcome, if the null hypothesis holds true. For this, we rely on the probabilities defined under the null hypothesis.

To determine expected frequencies, multiply the total number of observations by the probability of each outcome. Using the seagulls example:
  • Sand: 56% of the shore area suggests an expected frequency of \(0.56 \times 200 = 112\)
  • Mud: 29% gives us \(0.29 \times 200 = 58\)
  • Rocks: 15% converts to \(0.15 \times 200 = 30\)
These calculations provide a benchmark to compare against the observed frequencies.
The essence of these expected values is to help calculate the chi-square statistic, evaluating whether what we observe is unusually surprising or not.
Degrees of Freedom
Degrees of Freedom (often abbreviated as df) is a concept used in statistical tests including the Chi-Square Test. It refers to the number of values in the final calculation of a statistic that are free to vary. In simple terms, it tells us how many "pieces" of information are available to estimate another aspect of our data.

To find the degrees of freedom for a chi-square test of independence, use the formula:\[\text{df} = k - 1\]where \(k\) is the number of categories or groups.

In the seagulls' study, with three landing terrains (sand, mud, rocks), the degrees of freedom is:\[3 - 1 = 2\]This concept helps in determining the critical chi-square value, which is essential to decide whether we should reject the null hypothesis. The degrees of freedom, together with a significance level, help in reading the Chi-Square distribution table correctly.

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Most popular questions from this chapter

Benford's law Faked numbers in tax returns, invoices, or expense account claims often display patterns that aren't present in legitimate records. Some patterns are obvious and easily avoided by a clever crook. Others are more subtle. It is a striking fact that the first digits of numbers in legitimate records often follow a model known as Benford's law. \({ }^{3}\) Call the first digit of a randomly chosen record \(X\) for short. Benford's law gives this probability model for \(X\) (note that a first digit can't be 0 ): $$ \begin{array}{lccccccccc} \hline \text { First digit: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \text { Probability: } & 0.301 & 0.176 & 0.125 & 0.097 & 0.079 & 0.067 & 0.058 & 0.051 & 0.046 \\ \hline \end{array} $$ A forensic accountant who is familiar with Benford's law inspects a random sample of 250 invoices from a company that is accused of committing fraud. The table below displays the sample data. $$ \begin{array}{lcrrrrrrrr} \hline \text { First digit: } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \text { Count: } & 61 & 50 & 43 & 34 & 25 & 16 & 7 & 8 & 6 \\ \hline \end{array} $$ (a) Are these data inconsistent with Benford's law? Carry out an appropriate test at the \(\alpha=0.05\) level to support your answer. If you find a significant result, perform a follow-up analysis. (b) Describe a Type I error and a Type II error in this setting, and give a possible consequence of each. Which do you think is more serious?

Housing According to the Census Bureau, the distribution by ethnic background of the New York City population in a recent year was Hispanic: \(28 \%\) Black: \(24 \% \quad\) White: \(35 \%\) Asian: \(12 \%\) Others: \(1 \%\) The manager of a large housing complex in the city wonders whether the distribution by race of the complex's residents is consistent with the population distribution. To find out, she records data from a random sample of 800 residents. The table below displays the sample data. \({ }^{4}\) $$ \begin{array}{lccccc} \hline \text { Race: } & \text { Hispanic } & \text { Black } & \text { White } & \text { Asian } & \text { 0ther } \\ \text { Count: } & 212 & 202 & 270 & 94 & 22 \\ \hline \end{array} $$ Are these data significantly different from the city's distribution by race? Carry out an appropriate test at the \(\alpha=0.05\) level to support your answer. If you find a significant result, perform a follow-up analysis.

Refer to the following setting. The National Longitudinal Study of Adolescent Health interviewed a random sample of 4877 teens (grades 7 to 12 ). One question asked was "What do you think are the chances you will be married in the next ten years?" Here is a two-way table of the responses by gender: \({ }^{28}\) $$ \begin{array}{lcc} \hline & \text { Female } & \text { Male } \\ \text { Almost no chance } & 119 & 103 \\ \text { Some chance, but probably not } & 150 & 171 \\ \text { A 50-50 chance } & 447 & 512 \\ \text { A good chance } & 735 & 710 \\ \text { Almost certain } & 1174 & 756 \\ \hline \end{array} $$ The appropriate null hypothesis for performing a chi-square test is that (a) equal proportions of female and male teenagers are almost certain they will be married in 10 years. (b) there is no difference between the distributions of female and male teenagers' opinions about marriage in this sample. (c) there is no difference between the distributions of female and male teenagers' opinions about marriage in the population. (d) there is no association between gender and opinion about marriage in the sample. (e) there is no association between gender and opinion about marriage in the population.

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