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Exercises 48 to 50 refer to the following setting. Do birds learn to time their breeding? Blue titmice eat caterpillars. The birds would like lots of caterpillars around when they have young to feed, but they must breed much earlier. Do the birds learn from one year鈥檚 experience when to time their breeding next year? Researchers randomly assigned 7 pairs of birds to have the natural caterpillar supply supplemented while feeding their young and another 6 pairs to serve as a control group relying on natural food supply. The next year, they measured how many days after the caterpillar peak the birds produced their nestlings.\(^{35}\) Year-to-year comparison Rather than comparing the two groups in each year, we could compare the behavior of each group in the first and second years. The study report says: 鈥淥ur main prediction was that females receiving additional food in the nestling period should not change laying date the next year, whereas controls, which (in our area) breed too late in their first year, were expected to advance their laying date in the second year.鈥 Comparing days behind the caterpillar peak in Years 1 and 2 gave \(t=0.63\) for the control group and \(t=-2.63\) for the supplemented group. (a) What type of t statistic (one-sample, paired, or two-sample) are these? Justify your answer. (b) What are the degrees of freedom for each t? (c) Explain why these t-values do not agree with the prediction.

Short Answer

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(a) Paired t-statistics for both groups; (b) Control: 5 df, Supplemented: 6 df; (c) Control group did not change, supplemented group did change, opposite the prediction.

Step by step solution

01

Determine t-statistic type for control group

The t-statistic for the control group compares measurements made on the same group (the control group) in two different years. Since this involves comparing two sets of related observations from the same subjects, this is a paired t-test.
02

Determine t-statistic type for supplemented group

Similarly, the t-statistic for the supplemented group compares the measurements on the same group (the supplemented group) over two different years. This also falls under a paired t-test because the same subjects are observed in two different conditions (Year 1 and Year 2).
03

Identify degrees of freedom for the control group

In a paired t-test, the degrees of freedom are calculated as the number of pairs minus one. The control group has 6 pairs of birds, so the degrees of freedom is calculated as \(df = 6 - 1 = 5\).
04

Identify degrees of freedom for the supplemented group

Similarly, the supplemented group has 7 pairs of birds. Thus, for the paired t-test, the degrees of freedom is \(df = 7 - 1 = 6\).
05

Compare the t-values with predictions

The prediction states that controls should change their laying dates, yet their t-value is \(t = 0.63\), not statistically significant, indicating little change. The supplemented group was predicted to have no change, yet their t-value \(t = -2.63\), is statistically significant, indicating a notable change. This contrasts with the prediction as the results are opposite to expectations.

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Key Concepts

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

Statistical Comparison
In statistical studies, comparing two different conditions or groups is essential for drawing meaningful conclusions. One common method used for comparison is the paired t-test. This statistical test is used to compare two related but distinct sets of data, such as observations before and after an intervention on the same subjects.
In the context of bird breeding behavior, researchers wanted to investigate whether providing additional food to birds changed their nesting behavior in subsequent years. By comparing the number of days the birds laid eggs before or after the peak caterpillar period in consecutive years, the researchers applied a paired t-test. This is because each bird pair was observed in the same group (either supplemented or control) in both years.
Paired t-tests help determine if the means of the paired observations are significantly different, thereby providing insights into behavioral changes.
Bird Breeding Behavior
Bird breeding is intricately linked to the availability of food resources, especially for species like blue titmice that rely heavily on caterpillars to feed their young. Timing is crucial, as mismatched breeding can lead to food scarcity when it's needed most.
Researchers in the study aimed to see if birds adjust their breeding timing based on past experiences with food availability. They hypothesized that birds receiving extra food one year would not modify their laying dates the next year. Contrarily, birds relying solely on natural food would tend to advance their breeding to better synchronize with the caterpillar peak.
This behavior is a fascinating aspect of animal ecology, as it reflects how wildlife species might adapt to changing environments and food supplies. Understanding these behavioral changes is critical for conserving such species as their habitats alter over time.
Degrees of Freedom
Degrees of freedom are an essential statistical concept used to determine the number of independent values that can vary in an analysis without breaking any constraints. They are vital for calculating statistical significance in tests like the paired t-test.
In the study, each group of bird pairs involved measurements over two consecutive years, which means they each had only one independent variable due to the paired nature of the observations.
  • For the control group, with 6 pairs, the degrees of freedom is calculated as the number of pairs minus one: \(df = 6 - 1 = 5\).
  • For the supplemented group, with 7 pairs, it is \(df = 7 - 1 = 6\).
Knowing the degrees of freedom helps in interpreting the significance of the t-values, which in this case showed unexpected results against the predictions made, highlighting the complexities of predicting ecological behaviors.

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Most popular questions from this chapter

What鈥檚 wrong? A driving school wants to find out which of its two instructors is more effective at preparing students to pass the state鈥檚 driver鈥檚 license exam. An incoming class of 100 students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30 of Instructor A鈥檚 students and 22 of Instructor B鈥檚 students pass the state exam. Do these results give convincing evidence that Instructor A is more effective? Min Jae carried out the significance test shown below to answer this question. Unfortunately, he made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one. State: I want to perform a test of $$H_{0} : p_{1}-p_{2}=0$$ $$H_{a} : p_{1}-p_{2}>0$$ where \(p_{1}=\) the proportion of Instructor A's students that passed the state exam and \(p_{2}=\) the proportion of Instructor B's students that passed the state exam. Since no significance level was stated, I'll use \(\sigma=0.05\) Plan: If conditions are met, I'll do a two-sample \(z\) test for comparing two proportions. \(\bullet\) Random The data came from two random samples of 50 students. \(\bullet\) Normal The counts of successes and failures in the two groups - \(30,20,22\) , and \(28-\) are all at least \(10 .\) \(\bullet\) Independent There are at least 1000 students who take this driving school's class. Do: From the data, \(\hat{p}_{1}=\frac{20}{50}=0.40\) and \(\hat{p}_{2}=\frac{30}{50}=0.60 .\) So the pooled proportion of successes is $$\hat{p}_{C}=\frac{22+30}{50+50}=0.52$$ \(\bullet\) Test statistic $$z=\frac{(0.40-0.60)-0}{\sqrt{\frac{0.52(0.48)}{100}+\frac{0.52(0.48)}{100}}}=-2.83$$ Conclude: The P-value, \(0.9977,\) is greater than \(\alpha=\) \(0.05,\) so we fail to reject the null hypothesis. There is not convincing evidence that Instructor A's pass rate is higher than Instructor B's.

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