Elementary Statistics ( 3rd International Edition ) Isbn:9781260092561
3rd Edition
ISBN: 9781259969454
Author: William Navidi Prof.; Barry Monk Professor
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Videos
Textbook Question
Chapter 10.2, Problem 17E
Defective electronics: A team of designers was given the task of reducing the defect rate in the manufacture of a certain printed circuit board. The team decided to reconfigure the cooling system. A total of 973 boards produced the week before the reconfiguration was implemented, and 254 of these were defective. A total of 847 boards were produced the week after reconfiguration, and 95 of these were defective.
- Construct a 90% confidence inters-al for the decrease in the defective rate after the reconfiguration.
- A quality control engineer claims that the reconfiguration has decreased the proportion of defective parts by more than 0.15. Does the confidence interval contradict this claim?
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
In a particular year, approximately 136,943,000 people visited an emergency room in the United States. In a random sample of 100,000 visits, 12,462 were subsequently admitted to the hospital. Provide a 99% confidence interval for the proportion of people visiting an emergency room who were subsequently admitted to the hospital.
a) 0.121745 to 0.127495
b) 0.121930 to 0.127310
c) 0.122573 to 0.126667
d) 0.122902 to 0.126338
In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and the United Kingdom. They found that a substantially greater percentage of U.K. ads use humor.a. Suppose that a random sample of 400 television ads in the United Kingdom reveals that 142 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.K. television ads that use humor.b. Suppose a random sample of 500 television ads in the United States reveals that 122 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.S. television ads that use humor.c. Do the confidence intervals you computed in parts a and b suggest that a greater percentage of U.K. ads use humor? Explain.
A U.S. Food Survey showed that Americans routinely eat beef in their diet. Suppose that in a study of 49 consumers in Illinois and 64 consumers in Texas the following results were obtained from two samples regarding average yearly beef consumption:
Illinois Texas
= 49 = 64
= 54.1lb = 60.4lb
S1 = 7.0 S2 = 8.0
Develop a 95% confidence Interval Estimate for the difference between the two population means.
Chapter 10 Solutions
Elementary Statistics ( 3rd International Edition ) Isbn:9781260092561
Ch. 10.1 - In Exercises 5 and 6, fill in each blank with e...Ch. 10.1 - In Exercises 5 and 6, fill in each blank with e...Ch. 10.1 - Prob. 7ECh. 10.1 - Prob. 8ECh. 10.1 - In Exercises 9-14, construct the confidence...Ch. 10.1 - Prob. 10ECh. 10.1 - In Exercises 9-14, construct the confidence...Ch. 10.1 - Prob. 12ECh. 10.1 - Prob. 13ECh. 10.1 - Prob. 14E
Ch. 10.1 - Does this diet help? A group of 78 people enrolled...Ch. 10.1 - Contaminated water: The concentration of benzene...Ch. 10.1 - Fertilizer: In an agricultural experiment, the...Ch. 10.1 - Computer crashes: A computer system administrator...Ch. 10.1 - Are you smarter than your older brother? In a...Ch. 10.1 - Effectiveness of distance learning: A study was...Ch. 10.1 - Boys and girls: The National Health Statistics...Ch. 10.1 - Body mass index: In a survey of adults with...Ch. 10.1 - Energy drinks: A survey of college students...Ch. 10.1 - Low-fat or low-carb? Are low-fat diets or low-carb...Ch. 10.1 - Online testing: Do you prefer taking tests on...Ch. 10.1 - Drive safely: How often does the average driver...Ch. 10.1 - Interpret calculator display: The following TI-84...Ch. 10.1 - Prob. 28ECh. 10.1 - Prob. 29ECh. 10.1 - Prob. 30ECh. 10.1 - Prob. 31ECh. 10.2 - In Exercises 3 and 4, fill in each blank with the...Ch. 10.2 - Prob. 4ECh. 10.2 - Prob. 5ECh. 10.2 - Prob. 6ECh. 10.2 - In Exercises 7-12, construct the confidence...Ch. 10.2 - In Exercises 7-12, construct the confidence...Ch. 10.2 - In Exercises 7-12, construct the confidence...Ch. 10.2 - In Exercises 7-12, construct the confidence...Ch. 10.2 - In Exercises 7-12, construct the confidence...Ch. 10.2 - Prob. 12ECh. 10.2 - Traffic accidents: Traffic engineers compared...Ch. 10.2 - Computers in the classroom: In a new experimental...Ch. 10.2 - Pain after surgery: In a random sample of 50...Ch. 10.2 - Pretzels: In order to judge the effectiveness of...Ch. 10.2 - Defective electronics: A team of designers was...Ch. 10.2 - Satisfied? A poll taken by the General Social...Ch. 10.2 - Cancer prevention: Colonoscopy is a medical...Ch. 10.2 - Social media: A Pew poll found that in a sample of...Ch. 10.2 - Interpret calculator display: The following TI-84...Ch. 10.2 - Prob. 22ECh. 10.2 - Prob. 23ECh. 10.2 - Prob. 24ECh. 10.2 - Finding the sample size: Polls are to be conducted...Ch. 10.3 - In Exercises 3 and 4, fill in each blank with the...Ch. 10.3 - Prob. 4ECh. 10.3 - Prob. 5ECh. 10.3 - In Exercises 5 and 6, determine whether the...Ch. 10.3 - Fast computer: microprocessors are compared on a...Ch. 10.3 - Brake wear: For a sample of 9 automobiles, the...Ch. 10.3 - Strength of concrete: The compressive strength, m...Ch. 10.3 - Truck pollution: In an experiment to determine the...Ch. 10.3 - High cholesterol: A group of eight individuals...Ch. 10.3 - Tires and fuel economy: A tire manufacturer is...Ch. 10.3 - Growth spurt: It is generally known that boys grow...Ch. 10.3 - SAT coaching: A sample of 32 students took a class...Ch. 10.3 - Interpret calculator display: The following TI-84...Ch. 10.3 - Interpret calculator display: The following TI-84...Ch. 10.3 - Prob. 17ECh. 10.3 - Prob. 18ECh. 10.3 - Advantage of matched pairs: Refer to Exercise 12....Ch. 10.3 - Paired or independent? To construct a confidence...Ch. 10 - In Exercises 1 and 2, determine whether the...Ch. 10 - Prob. 2CQCh. 10 - Prob. 3CQCh. 10 - Prob. 4CQCh. 10 - Prob. 5CQCh. 10 - Prob. 6CQCh. 10 - Prob. 7CQCh. 10 - Prob. 8CQCh. 10 - Prob. 9CQCh. 10 - Prob. 10CQCh. 10 - Prob. 11CQCh. 10 - Refer to Exercise 11. Find the critical value for...Ch. 10 - Prob. 13CQCh. 10 - Prob. 14CQCh. 10 - Prob. 15CQCh. 10 - Prob. 1RECh. 10 - Prob. 2RECh. 10 - Prob. 3RECh. 10 - Prob. 4RECh. 10 - Prob. 5RECh. 10 - Prob. 6RECh. 10 - Prob. 7RECh. 10 - Prob. 8RECh. 10 - Prob. 9RECh. 10 - Prob. 10RECh. 10 - Prob. 11RECh. 10 - Prob. 12RECh. 10 - Prob. 13RECh. 10 - Prob. 14RECh. 10 - Prob. 15RECh. 10 - Prob. 1WAICh. 10 - Prob. 2WAICh. 10 - Prob. 3WAICh. 10 - Prob. 4WAICh. 10 - Prob. 1CSCh. 10 - Prob. 2CSCh. 10 - Prob. 3CS
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.Similar questions
- What is meant by the sample space of an experiment?arrow_forward1. A recent survey showed that from a sample of 500 packages delivered by a Postal Service, 480were delivered on time. a) Construct a 95% confidence interval for the proportion of all packages that are deliveredon time by the Postal Service.arrow_forwardIn an experiment of normal body temperatures in degrees Fahrenheit, 6 males ages 25-40 were given a random temperature check. The results were as follows: 97.1 99.0 98.2 98.9 98.1 99.4 a. Construct and interpret a 95% Confidence Interval for the average normal body temperature of males ages 25-40. b. Construct and interpret a 95% Prediction Interval for the next temperature taken. c. The average normal body temperature is generally accepted as 98.6 °F. Given the above data, should scientists reconsider this general statement? Use ? = 0.05.arrow_forward
- In a recent questionnaire about homeownership, a random sample of 545 homeowners were asked about fixed and adjustable-rate mortgages, and 268 reported that they have an adjustable-rate mortgage. What value of z should be used to calculate a confidence interval with a 95% confidence level? z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576arrow_forwardBulletproof Vests. In the article “A Common Police Vest Fails the Bulletproof Test,” E. Lichtblau reported on a U.S. Department of Justice study of 103 bulletproof vests containing a fiber known as Zylon. In ballistics tests, only 4 of these vests produced acceptable safety outcomes (and resulted in immediate changes in federal safety guidelines). Find a 95% confidence interval for the proportion of all such vests that would produce acceptable safety outcomes by using the a. one-proportion z-interval procedure. b. one-proportion plus-four z-interval procedure. (See page 556 for the details of this procedure.) c. Explain the large discrepancy between the two methods. d. Which confidence interval would you use? Explain your answer.arrow_forwardThe figure to the right shows the results of a survey in which 1005 adults from Country A, 1023 adults from Country B, 995 adults from Country C, 1020 adults from Country D, and 1012 adults from Country E were asked whether national identity is strongly tied to birthplace. Country A37%22%27%49%Country BCountry CCountry DCountry E10% Construct a 99% confidence interval for the population proportion of adults who say national identity is strongly tied to birthplace for each country listed. The 99% confidence interval for the proportion of adults from Country B who say national identity is strongly tied to birthplace is enter your response here ,enter your response here . (Round to three decimal places as needed.)arrow_forward
- The figure to the right shows the results of a survey in which 1005 adults from Country A, 1023 adults from Country B, 995 adults from Country C, 1020 adults from Country D, and 1012 adults from Country E were asked whether national identity is strongly tied to birthplace. Country A37%22%27%49%Country BCountry CCountry DCountry E10% Construct a 99% confidence interval for the population proportion of adults who say national identity is strongly tied to birthplace for each country listed. Question content area bottom Part 1 The 99% confidence interval for the proportion of adults from Country A who say national identity is strongly tied to birthplace is (enter your response here ,enter your response here ). (Round to three decimal places as needed.)arrow_forwardIn a random sample of 850 consumers. 385 reported that they were able to purchase a Playstation 5. Which of the following is a 98% confidence interval for the population proportion of consumers that were able to purchase a Playstation 5? A) (0.4131, 0.4927) B)(0.4089, 0.4969) C)(0.4248, 0.4810) D) (0.4194, 0.4864)arrow_forward3) A random sample of 100 students at a high school was asked whether they would ask their father or mother for help with a financial problem. A second sample of 100 different students was asked the same question regarding a dating problem. Forty-three students in the first sample and 47 students in the second sample replied that they turned to their mother rather than their father for help. Construct a 98% confidence interval for p1-p2arrow_forward
- in a random sample of 800 austin automobile owners, it was found that 480 would like to see the size of the cars reduced. A 95% confidence interval for the proportion of all austin car owners who would like to see another car is: a) (0.532, 0.667) b) (0.572, 0.628) c) (0.566, 0.634)arrow_forwardSuppose that a safety group surveyed 1,100 drivers. Among those surveyed, 65% said that careless or aggressive driving was the biggest threat on the road, and 25% said that cell phone usage by other drivers was the driving behavior that annoyed them the most. Based on these data and assuming that the sample was a simple random sample, construct and interpret a 95% confidence interval estimate for the true proportion in the population of all drivers who are annoyed by cell phone users. The confidence interval estimate is ??––––––??> (Round to three decimal places as needed. Use ascending order.) Interpret the confidence interval estimate. A. There is a 0.95 probability that the population proportion of drivers who are annoyed by cell phone users is in the interval. B. There is 95% confidence that the population proportion of drivers who are annoyed by cell phone users is in the interval. C. There is 95% confidence that the population proportion…arrow_forwardDeterminations of saliva pH levels were made in two independent random samples of seventh-grade schoolchildren. Sample A children were caries-free while sample B children had a high incidence of caries. The results were as follows: A 7.14 7.11 7.61 7.98 7.21 7.16 7.89 7.24 7.86 B 7.36 7.04 7.19 7.41 7.10 7.15 7.36 7.57 7.19 Construct and interpret a 90% confidence interval for the difference between the population means.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
College Algebra (MindTap Course List)AlgebraISBN:9781305652231Author:R. David Gustafson, Jeff HughesPublisher:Cengage Learning
College Algebra (MindTap Course List)
Algebra
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:Cengage Learning
Hypothesis Testing using Confidence Interval Approach; Author: BUM2413 Applied Statistics UMP;https://www.youtube.com/watch?v=Hq1l3e9pLyY;License: Standard YouTube License, CC-BY
Hypothesis Testing - Difference of Two Means - Student's -Distribution & Normal Distribution; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=UcZwyzwWU7o;License: Standard Youtube License