Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Null and alternative hypotheses for a test are given. Give the notation \((\bar{x},\) for example) for a sample statistic we might record for each simulated sample to create the randomization distribution. \(H_{0}: \rho=0 \quad\) vs \(H_{a}: \rho \neq 0\)

Short Answer

Expert verified
The sample statistic we might record for each simulated sample to create the randomization distribution in this scenario is the sample correlation coefficient, \( r \).

Step by step solution

01

Understanding the Hypotheses

Here, we have two hypotheses about the population correlation coefficient \( \rho \). The null hypothesis \( H_0 \) is that \( \rho = 0 \), which means there is no correlation between the two variables. The alternative hypothesis \( H_a \) is that \( \rho \neq 0 \), indicating there is a correlation, either positive or negative.
02

Decide on the Sample Statistic

For the sample statistic, we need something that can help us estimate the correlation in our sample data and compare it to the hypothesized value. The appropriate statistic in this case is the sample correlation coefficient, \( r \). This statistic can provide a measure of the strength and direction of the relationship between the two variables in the sample data.
03

Creating the Randomization Distribution

We would use this \( r \) value as our test statistic and record its value for each simulated sample. By doing this, we can create a randomization distribution of \( r \) values under the assumption that the null hypothesis is true (i.e., that there is no correlation). This distribution can then be used to determine the probability of observing a result as extreme as our sample data if the null hypothesis is true, which forms the basis for deciding whether to reject the null hypothesis.

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 Vaia!

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

Interpreting a P-value In each case, indicate whether the statement is a proper interpretation of what a p-value measures. (a) The probability the null hypothesis \(H_{0}\) is true. (b) The probability that the alternative hypothesis \(H_{a}\) is true. (c) The probability of seeing data as extreme as the sample, when the null hypothesis \(H_{0}\) is true. (d) The probability of making a Type I error if the null hypothesis \(H_{0}\) is true. (e) The probability of making a Type II error if the alternative hypothesis \(H_{a}\) is true.

For each situation described, indicate whether it makes more sense to use a relatively large significance level (such as \(\alpha=0.10\) ) or a relatively small significance level (such as \(\alpha=0.01\) ). A pharmaceutical company is testing to see whether its new drug is significantly better than the existing drug on the market. It is more expensive than the existing drug. Which significance level would the company prefer? Which significance level would the consumer prefer?

Flaxseed and Omega-3 Exercise 4.30 on page 271 describes a company that advertises that its milled flaxseed contains, on average, at least \(3800 \mathrm{mg}\) of ALNA, the primary omega-3 fatty acid in flaxseed, per tablespoon. In each case below, which of the standard significance levels, \(1 \%\) or \(5 \%\) or \(10 \%,\) makes the most sense for that situation? (a) The company plans to conduct a test just to double-check that its claim is correct. The company is eager to find evidence that the average amount per tablespoon is greater than 3800 (their alternative hypothesis), and is not really worried about making a mistake. The test is internal to the company and there are unlikely to be any real consequences either way. (b) Suppose, instead, that a consumer organization plans to conduct a test to see if there is evidence against the claim that the product contains at least \(3800 \mathrm{mg}\) per tablespoon. If the organization finds evidence that the advertising claim is false, it will file a lawsuit against the flaxseed company. The organization wants to be very sure that the evidence is strong, since if the company is sued incorrectly, there could be very serious consequences.

Approval Rating for Congress In a Gallup poll \(^{51}\) conducted in December 2015 , a random sample of \(n=824\) American adults were asked "Do you approve or disapprove of the way Congress is handling its job?" The proportion who said they approve is \(\hat{p}=0.13,\) and a \(95 \%\) confidence interval for Congressional job approval is 0.107 to 0.153 . If we use a 5\% significance level, what is the conclusion if we are: (a) Testing to see if there is evidence that the job approval rating is different than \(14 \%\). (This happens to be the average sample approval rating from the six months prior to this poll.) (b) Testing to see if there is evidence that the job approval rating is different than \(9 \%\). (This happens to be the lowest sample Congressional approval rating Gallup ever recorded through the time of the poll.)

Mating Choice and Offspring Fitness Does the ability to choose a mate improve offspring fitness in fruit flies? Researchers have studied this by taking female fruit flies and randomly dividing them into two groups; one group is put into a cage with a large number of males and able to freely choose who to mate with, while flies in the other group are each put into individual vials, each with only one male, giving no choice in who to mate with. Females are then put into egg laying chambers, and a certain number of larvae collected. Do the larvae from the mate choice group exhibit higher survival rates? A study \(^{44}\) published in Nature found that mate choice does increase offspring fitness in fruit flies (with p-value \(<0.02\) ), yet this result went against conventional wisdom in genetics and was quite controversial. Researchers attempted to replicate this result with a series of related experiments, \({ }^{45}\) with data provided in MateChoice. (a) In the first replication experiment, using the same species of fruit fly as the original Nature study, 6067 of the 10000 larvae from the mate choice group survived and 5976 of the 10000 larvae from the no mate choice group survived. Calculate the p-value. (b) Using a significance level of \(\alpha=0.05\) and \(\mathrm{p}\) -value from (a), state the conclusion in context. (c) Actually, the 10,000 larvae in each group came from a series of 50 different runs of the experiment, with 200 larvae in each group for each run. The researchers believe that conditions dif- fer from run to run, and thus it makes sense to treat each \(\mathrm{run}\) as a case (rather than each fly). In this analysis, we are looking at paired data, and the response variable would be the difference in the number of larvae surviving between the choice group and the no choice group, for each of the 50 runs. The counts (Choice and NoChoice and difference (Choice \(-\) NoChoice) in number of surviving larva are stored in MateChoice. Using the single variable of differences, calculate the p-value for testing whether the average difference is greater than \(0 .\) (Hint: this is a single quantitative variable, so the corresponding test would be for a single mean.) (d) Using a significance level of \(\alpha=0.05\) and the p-value from (c), state the conclusion in context. (e) The experiment being tested in parts (a)-(d) was designed to mimic the experiment from the original study, yet the original study yielded significant results while this study did not. If mate choice really does improve offspring fitness in fruit flies, did the follow-up study being analyzed in parts (a)-(d) make a Type I, Type II, or no error? (f) If mate choice really does not improve offspring fitness in fruit flies, did the original Nature study make a Type I, Type II, or no error?

See all solutions

Recommended explanations on Math Textbooks

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.

Sign-up for free