/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 Birds in the trees Researchers s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 7

Birds in the trees Researchers studied the behavior of birds that were searching for seeds and insects in an Oregon forest. In this forest, \(54 \%\) of the trees were Douglas firs, \(40 \%\) were ponderosa pines, and \(6 \%\) were other types of trees. At a randomly selected time during the day, the researchers observed 156 red-breasted nuthatches: 70 were seen in Douglas firs, 79 in ponderosa pines, and 7 in other types of trees. \(^{2}\) Do these data provide convincing evidence that nuthatches prefer particular types of trees when they're searching for seeds and insects?

Short Answer

Expert verified
Nuthatches have a preference for certain tree types.

Step by step solution

01

State the Hypotheses

We will use a chi-square goodness-of-fit test to determine if there is a preference among the tree types. The null hypothesis ( H_0 ) is that nuthatches have no preference and the trees are chosen based on their proportional availability: 54% Douglas fir, 40% ponderosa pine, and 6% other trees. The alternative hypothesis ( H_a ) is that nuthatches prefer certain types of trees.
02

Calculate Expected Counts

If nuthatches have no preference, we expect the number of birds caught in each tree type to follow the forest distributions: \( 0.54 \times 156 = 84.24 \) for Douglas firs, \( 0.40 \times 156 = 62.4 \) for ponderosa pines, and \( 0.06 \times 156 = 9.36 \) for other trees.
03

Compute Chi-Square Statistic

To calculate the chi-square statistic, use the formula: \[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]where \(O_i\) is the observed count and \(E_i\) is the expected count. For Douglas firs: \(\frac{(70 - 84.24)^2}{84.24} = 2.446\), for ponderosa pines: \(\frac{(79 - 62.4)^2}{62.4} = 4.193\), and for other trees: \(\frac{(7 - 9.36)^2}{9.36} = 0.595\). Thus, the chi-square statistic is \(\chi^2 = 2.446 + 4.193 + 0.595 = 7.234\).
04

Determine Degrees of Freedom

Degrees of freedom in a chi-square test for goodness of fit are calculated as \(df = \ ext{{number of categories}} - 1\). Here, we have 3 categories (Douglas firs, ponderosa pines, other trees), so \(df = 3 - 1 = 2\).
05

Find the P-Value

Using a chi-square distribution table or calculator with \(df = 2\) and \(\chi^2 = 7.234\), we find the p-value associated with this statistic. The p-value is approximately 0.026 when comparing to the critical value for a 5% significance level (\(\chi^2 = 5.99\)).
06

Conclusion

Since the p-value of 0.026 is less than the 0.05 significance level, we reject the null hypothesis. There is sufficient evidence to suggest that nuthatches prefer certain types of trees when searching for seeds and insects.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Hypothesis Testing
Hypothesis testing is like making an educated guess and then checking if evidence supports it. When researchers are curious about whether birds like particular trees, they set up a hypothesis test. In this process, two opposing hypotheses are made:
  • Null Hypothesis (H_0): Assumes that there is no special preference among the types of trees. The birds choose trees in proportion to their availability.
  • Alternative Hypothesis (H_a): Suggests that the birds do have tree type preferences.
By calculating probabilities and comparing them, researchers decide which hypothesis is more likely based on the evidence. This helps in understanding bird behavior better.
Chi-Square Statistic
The chi-square statistic is a fancy way of measuring how far off your observations are from what you'd expect if everything were random. In our bird example, researchers first observed how many birds were in each type of tree. Then, they calculated how many birds should have been in each tree, assuming no preference.To compute the chi-square statistic, the formula used is:\[\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}\]where \(O_i\) is the observed number of birds in a type of tree, and \(E_i\) is the expected number. The larger the chi-square statistic, the more evidence there might be of a preference.
Expected Counts
Expected counts predict how many birds would be found on each type of tree if they had no preferences. These numbers are calculated based on the proportion of each tree type in the forest.In our example:
  • For Douglas firs, we expect \(0.54 \times 156 = 84.24\) birds.
  • For ponderosa pines, \(0.40 \times 156 = 62.4\) birds.
  • For other trees, \(0.06 \times 156 = 9.36\) birds.
These expected counts form a basis for comparison 鈥� if the actual counts are wildly different, it may suggest a preference.
P-Value
The p-value is like a percentage that tells us how surprising our results are, assuming no preference. It helps answer the question: "If the null hypothesis were true, how likely are these results?" A small p-value (typically less than 0.05) indicates that the observed results are rare under the null hypothesis, suggesting that tree preference might exist. In our example, the calculated p-value was 0.026, below the usual significance level of 0.05. Consequently, the null hypothesis was rejected, indicating that the birds might have a preference.
Degrees of Freedom
Degrees of freedom are like the number of options available for the data to vary. They're crucial for interpreting the chi-square statistic correctly. In general, the formula for determining degrees of freedom in a chi-square test is:\[df = \text{number of categories} - 1\]In this example, since there are three types (categories) of trees, degrees of freedom would be \(3 - 1 = 2\). This number helps in finding the critical values from chi-square distribution tables, which are then used to calculate the p-value.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Skittles Statistics teacher Jason Molesky contacted Mars, Inc., to ask about the color distribution for Skittles candies. Here is an excerpt from the response he received: "The original flavor blend for the SKITTLES BITE SIZE CANDIES is lemon, lime, orange, strawberry and grape. They were chosen as a result of consumer preference tests we conducted. The flavor blend is 20 percent of each flavor." (a) State appropriate hypotheses for a significance test of the company's claim. (b) Find the expected counts for a bag of Skittles with 60 candies. (c) How large a \(\chi^{2}\) statistic would you need to have significant evidence against the company's claim at the \(\alpha=0.05\) level? At the \(\alpha=0.01\) level? (d) Create a set of observed counts for a bag with 60 candies that gives a \(P\) -value between 0.01 and \(0.05 .\) Show the calculation of your chi-square statistic.

Sorry, no chi-square How do U.S. residents who travel overseas for leisure differ from those who travel for business? The following is the breakdown by occupation: \({ }^{17}\) $$ \begin{array}{lcc} \hline \text { 0ccupation } & \text { Leisure } & \text { Business } \\ \text { Professional/technical } & \text { travelers (\%) } & \text { travelers (\%) } \\ \text { Manager/executive } & 36 & 39 \\ \text { Retired } & 23 & 48 \\ \text { Student } & 14 & 3 \\ \text { 0ther } & 7 & 3 \\ \text { Total } & 20 & 7 \\ \hline \end{array} $$ Explain why we can't use a chi-square test to learn whether these two distributions differ significantly.

Aw, nuts! A company claims that each batch of its deluxe mixed nuts contains \(52 \%\) cashews, \(27 \%\) almonds, \(13 \%\) macadamia nuts, and \(8 \%\) brazil nuts. To test this claim, a quality-control inspector takes a random sample of 150 nuts from the latest batch. The one-way table below displays the sample data. $$ \begin{array}{lcccc} \hline \text { Nut: } & \text { Cashew } & \text { Almond } & \text { Macadamia } & \text { Brazil } \\ \text { Count: } & 83 & 29 & 20 & 18 \\ \hline \end{array} $$ (a) State appropriate hypotheses for performing a test of the company's claim. (b) Calculate the expected counts for each type of nut. Show your work.

How to quit smoking It's hard for smokers to quit. Perhaps prescribing a drug to fight depression will work as well as the usual nicotine patch. Perhaps combining the patch and the drug will work better than either treatment alone. Here are data from a randomized, double-blind trial that compared four treatments. \({ }^{19} \mathrm{~A}\) "success" means that the subject did not smoke for a year following the beginning of the study. $$ \begin{array}{llcc} \hline \text { Group } & \text { Treatment } & \text { Subjects } & \text { Successes } \\ 1 & \text { Nicotine patch } & 244 & 40 \\ 2 & \text { Drug } & 244 & 74 \\ 3 & \text { Patch plus drug } & 245 & 87 \\ 4 & \text { Placebo } & 160 & 25 \\ \hline \end{array} $$ (a) Summarize these data in a two-way table. Then compare the success rates for the four treatments. (b) Explain in words what the null hypothesis \(H_{0}: p_{1}=\) \(p_{2}=p_{3}=p_{4}\) says about subjects' smoking habits. (c) Do the data provide convincing evidence of a difference in the effectiveness of the four treatments at the \(\alpha=0.05\) significance level?

Exercises 59 to 60 refer to the following setting. For their final project, a group of AP \(^{\otimes}\) Statistics students investigated the following question: "Will changing the rating scale on a survey affect how people answer the question?" To find out, the group took an SRS of 50 students from an alphabetical roster of the school's just over 1000 students. The first 22 students chosen were asked to rate the cafeteria food on a scale of 1 (terrible) to 5 (excellent). The remaining 28 students were asked to rate the cafeteria food on a scale of 0 (terrible) to 4 (excellent). Here are the data: $$ \begin{array}{lcccrc} &{1 \text { to 5 scale }} \\ \text { Rating } & 1 & 2 & 3 & 4 & 5 \\ \text { Frequency } & 2 & 3 & 1 & 13 & 3 \\ \hline & {0 \text { to 4 scale }} \\ \text { Rating } & 0 & 1 & 2 & 3 & 4 \\ \text { Frequency } & 0 & 0 & 2 & 18 & 8 \\ \hline \end{array} $$ Average ratings (1.3,10.2) The students decided to compare the average ratings of the cafeteria food on the two scales. (a) Find the mean and standard deviation of the ratings for the students who were given the 1 -to- 5 scale. (b) For the students who were given the 0 -to- 4 scale, the ratings have a mean of 3.21 and a standard deviation of \(0.568 .\) Since the scales differ by one point, the group decided to add 1 to each of these ratings. What are the mean and standard deviation of the adjusted ratings? (c) Would it be appropriate to compare the means from parts (a) and (b) using a two-sample \(t\) test? Justify your answer.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.