/*! 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 143 Clicker Questions A statistics i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Clicker Questions A statistics instructor would like to ask "clicker" questions that about \(80 \%\) of her students in a large lecture class will get correct. A higher proportion would be too easy and a lower proportion might discourage students. Suppose that she tries a sample of questions and receives 76 correct answers and 24 incorrect answers among 100 responses. The hypotheses of interest are \(H_{0}: p=0.80\) vs \(H_{a}: p \neq 0.80 .\) Discuss whether or not the methods described below would be appropriate ways to generate randomization samples in this setting. Explain your reasoning in each case. (a) Sample 100 answers (with replacement) from the original student responses. Count the number of correct responses. (b) Sample 100 answers (with replacement) from a set consisting of 8 correct responses and 2 incorrect responses. Count the number of correct responses.

Short Answer

Expert verified
Method (a) is not an appropriate means of generating randomization samples due to potential leaning towards observed data. Method (b), however, is deemed suitable since it mirrors the null hypothesis, leading to a balanced and fair comparison for hypothesis testing.

Step by step solution

01

Evaluation of Method (a)

Method (a) suggests sampling 100 answers with replacement from the original student responses and counting the number of correct responses. This method would follow the original student response proportions (76% correct, 24% incorrect), which may not be an appropriate reflection of the null hypothesis of 80% correctness, since it might lean towards the observed data. This method may thus not be an appropriate way to generate randomization samples.
02

Evaluation of Method (b)

Method (b) proposes sampling 100 answers with replacement from a set with 8 correct responses and 2 incorrect responses and then counting the correct responses. This method mirrors the null hypothesis that 80% of responses are correct and 20% are incorrect, providing a better reflection of the expected distributions under the null hypothesis than Method (a). This method would provide a balanced and fair comparison for hypothesis testing and can therefore be considered an appropriate way to generate randomization samples.

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 91Ó°ÊÓ!

Key Concepts

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

Statistics Education
Statistics education plays a critical role in helping students understand how to interpret data and make informed decisions based on evidence. It is not just about learning formulas or performing calculations; it involves a deeper understanding of the principles that guide data analysis. In the context of the exercise, it demonstrates how hypothesis testing can be applied in an educational setting.

Teaching students about hypothesis testing involves guiding them to formulating a hypothesis, an essential skill in statistics.
  • They need to understand the null hypothesis ( $H_0$ ), which often assumes no effect or no difference.
  • Equally important is the alternative hypothesis ( $H_a$ ), which is typically what researchers are trying to prove.
Hypothesis testing helps students learn how to systematically compare observed results with expected outcomes, fostering critical thinking. This can then be applied across various fields of study, illustrating the versatility and power of statistical analysis.
Randomization Techniques
Randomization techniques are crucial when conducting statistical experiments or sampling. These techniques ensure that every sample has an equal chance of being selected, which helps eliminate bias and increase the validity of the results.

In the given exercise, the instructor employs randomization to test if the proportion of correct answers aligns with the expected 80%.
  • Method (a) involves sampling from real student responses, repeating data collection.
  • Method (b) involves sampling from a synthetic dataset that reflects the hypothesized distribution.
Random sampling strives to ensure that every sample is a true representation of the population. Method (b) effectively mirrors the null hypothesis by using a fair mix of correct and incorrect responses, which facilitates a better assessment of the hypothesis being tested.
Null Hypothesis
The null hypothesis is a foundational concept in statistics, central to hypothesis testing. It represents a statement, usually of no effect or no change, that we seek to reject or fail to reject based on the data.

In the exercise, the null hypothesis is that 80% of the instructor’s students would answer correctly ( $H_0: p = 0.80$ ). It is essential to determine if the observed 76% reflects this hypothesis or if it suggests a significant difference.
  • The null hypothesis provides a baseline for comparison.
  • By comparing the null with the observed data, statisticians can infer whether data supports or refutes the null.
Evaluating the null hypothesis involves using statistical tests to see if there is enough evidence to suggest that observed data significantly deviates from the expected distribution. This involves calculating sample means, standard deviations, and other statistical measures. If the data shows a significant difference, the null hypothesis is rejected in favor of the alternative hypothesis.

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

Flaxseed and Omega-3 Studies have shown that omega-3 fatty acids have a wide variety of health benefits. Omega- 3 oils can be found in foods such as fish, walnuts, and flaxseed. A company selling milled flaxseed advertises that one tablespoon of the product contains, on average, at least \(3800 \mathrm{mg}\) of ALNA, the primary omega-3. (a) The company plans to conduct a test to ensure that there is sufficient evidence that its claim is correct. To be safe, the company wants to make sure that evidence shows the average is higher than \(3800 \mathrm{mg}\). What are the null and alternative hypotheses? (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 an average of \(3800 \mathrm{mg}\) per tablespoon. The consumer organization will only take action if it finds evidence that the claim made by the company is false and the actual average amount of omega- 3 is less than \(3800 \mathrm{mg}\). What are the null and alternative hypotheses?

Testing for a Home Field Advantage in Soccer In Exercise 3.108 on page \(215,\) we see that the home team was victorious in 70 games out of a sample of 120 games in the FA premier league, a football (soccer) league in Great Britain. We wish to investigate the proportion \(p\) of all games won by the home team in this league. (a) Use StatKey or other technology to find and interpret a \(90 \%\) confidence interval for the proportion of games won by the home team. (b) State the null and alternative hypotheses for a test to see if there is evidence that the proportion is different from 0.5 . (c) Use the confidence interval from part (a) to make a conclusion in the test from part (b). State the confidence level used. (d) Use StatKey or other technology to create a randomization distribution and find the p-value for the test in part (b). (e) Clearly interpret the result of the test using the p-value and using a \(10 \%\) significance level. Does your answer match your answer from part (c)? (f) What information does the confidence interval give that the p-value doesn't? What information does the p-value give that the confidence interval doesn't? (g) What's the main difference between the bootstrap distribution of part (a) and the randomization distribution of part (d)?

Determining Statistical Significance How small would a p-value have to be in order for you to consider results statistically significant? Explain. (There is no correct answer! This is just asking for your personal opinion. We'll study this in more detail in the next section.)

Does Massage Really Help Reduce Inflammation in Muscles? In Exercise 4.132 on page \(279,\) we learn that massage helps reduce levels of the inflammatory cytokine interleukin-6 in muscles when muscle tissue is tested 2.5 hours after massage. The results were significant at the \(5 \%\) level. However, the authors of the study actually performed 42 different tests: They tested for significance with 21 different compounds in muscles and at two different times (right after the massage and 2.5 hours after). (a) Given this new information, should we have less confidence in the one result described in the earlier exercise? Why? (b) Sixteen of the tests done by the authors involved measuring the effects of massage on muscle metabolites. None of these tests were significant. Do you think massage affects muscle metabolites? (c) Eight of the tests done by the authors (including the one described in the earlier exercise) involved measuring the effects of massage on inflammation in the muscle. Four of these tests were significant. Do you think it is safe to conclude that massage really does reduce inflammation?

In Exercises 4.146 to \(4.149,\) hypotheses for a statistical test are given, followed by several possible confidence intervals for different samples. In each case, use the confidence interval to state a conclusion of the test for that sample and give the significance level used. Hypotheses: \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2} .\) In addition, in each case for which the results are significant, state which group ( 1 or 2 ) has the larger mean. (a) \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}:\) $$ 0.12 \text { to } 0.54 $$ (b) \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) : -2.1 to 5.4 (c) \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) : -10.8 to -3.7

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.