/*! 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 98 Refer to Exercise \(8.97 .\) Sup... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Refer to Exercise \(8.97 .\) Suppose 14 out of 20 voters in Pennsylvania report having voted for an independent candidate. The null hypothesis is that the population proportion is \(0.50 .\) What value of the test statistic should you report?

Short Answer

Expert verified
The test statistic to be reported is 2.

Step by step solution

01

Understanding the Problem

The null hypothesis \(H_0\) states that the population proportion \(p\) is 0.50. We have a sample of 20 voters where 14 have voted for an independent candidate. Let's denote the number of successes (votes for an independent candidate) as \(X = 14\) and the sample size as \(n = 20\).
02

Calculate the Sample Proportion

We can calculate the sample proportion \(\hat{p}\) using the formula \(\hat{p} = X/n\). Substituting the given values, \(\hat{p} = 14/20 = 0.70\).
03

Calculate the Test Statistic

The test statistic \(z\) for testing a proportion is calculated as \(z = (\hat{p} - p_0) / \sqrt{(p_0(1-p_0))/n}\), where \(p_0\) is the null hypothesis proportion. Substituting the values, \(z = (0.70 - 0.50) / \sqrt{(0.50(1-0.50))/20} = 2\).

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!

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

In each case. choose whether the appropriate test is a one-proportion \(z\) -test or a two-proportion z-test. Name the population(s). a. A researcher takes a random sample of 4 -year-olds to find out whether girls or boys are more likely to know the alphabet. b. A pollster takes a random sample of all U.S. adult voters to see whether more than \(50 \%\) approve of the performance of the current U.S. president. c. A researcher wants to know whether a new heart medicine reduces the rate of heart attacks compared to an old medicine. d. A pollster takes a poll in Wyoming about homeschooling to find out whether the approval rate for men is equal to the approval rate for women. e. A person is studied to see whether he or she can predict the results of coin flips better than chance alone.

In a 2018 study reported in The Lancet, Molina et al. reported on a study for treatment of patients with HIV-1. The study was a randomized, controlled, double-blind study that compared the effectiveness of ritonavir-boosted darunavir (rbd), the drug currently used to treat HIV-1, with dorovirine, a newly developed drug. Of the 382 subjects taking ritonavir-boosted darunavir, 306 achieved a positive result. Of the 382 subjects taking dorovirine, 321 achieved a positive outcome. See page 430 for guidance. a. Find the sample percentage of subjects who achieved a positive outcome in each group. b. Perform a hypothesis test to test whether the proportion of patients who achieve a positive outcome with the current treatment (ritonavir-boosted darunavir) is different from the proportion of patients who achieve a positive outcome with the new treatment (dorovirine). Use a significance level of \(0.01\). Based on this study, do you think dorovirine might be a more effective treatment option for HIV-1 than ritonavir-boosted darunavir? Why or why not?

The label on a can of mixed nuts says that the mixture contains \(40 \%\) peanuts. After opening a can of nuts and finding 22 peanuts in a can of 50 nuts, a consumer thinks the proportion of peanuts in the mixture differs from \(40 \%\). The consumer writes these hypotheses: \(\mathrm{H}_{0}: \mathrm{p} \neq 0.40\) and \(\mathrm{H}_{\mathrm{a}}: \mathrm{p}=0.44\) where \(p\) represents the proportion of peanuts in all cans of mixed nuts from this company. Are these hypotheses written correctly? Correct any mistakes as needed.

A 20-question multiple choice quiz has five choices for each question. Suppose that a student just guesses, hoping to get a high score. The teacher carries out a hypothesis test to determine whether the student was just guessing. The null hypothesis is \(p=0.20\), where \(p\) is the probability of a correct answer. a. Which of the following describes the value of the \(z\) -test statistic that is likely to result? Explain your choice. i. The \(z\) -test statistic will be close to 0 . ii. the \(z\) -test statistic will be far from 0 . b. Which of the following describes the \(\mathrm{p}\) -value that is likely to result? Explain your choice. i. The p-value will be small. ii. The p-value will not be small.

According to a 2015 University of Michigan poll, \(71.5 \%\) of high school seniors in the United States had a driver's license. A sociologist thinks this rate has declined. The sociologist surveys 500 randomly selected high school seniors and finds that 350 have a driver's license. a. Pick the correct null hypothesis. i. \(p=0.715\) ii. \(p=0.70\) iii. \(\hat{p}=0.715\) iv. \(\hat{p}=0.70\) b. Pick the correct alternative hypothesis. i. \(p>0.715\) ii. \(p<0.715\) iii. \(\hat{p}<0.715\) iv. \(p \neq 0.715\) c. In this context, the symbol \(p\) represents (choose one) i. the proportion of high school seniors in the entire United States that have a driver's license. ii. the proportion of high school seniors in the sociologist's random sample that have a driver's license.

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.