We wish to estimate what percent of adult residents in a certain county are parents. Out of 100 adult residents sampled, 15 had kids. Based on this, construct a 99% confidence interval for the proportion, p, of adult residents who are parents in this county

We wish to estimate what percent of adult residents in a certain county are parents. Out of 100 adult residents sampled, 15 had kids. Based on this, construct a 99% confidence interval for the proportion, p, of adult residents who are parents in this county

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: In a survey of 3001 adults, 1486 say they have started paying bills online in the last year.…

A: Assuming that the number of people paying bills online has a binomial distribution with n=3001, and…

Q: Of 125 adults selected randomly from one town, 28 of them smoke. Construct a 99% confidence interval…

A: It is given that out of 125 adults in one town, 28 of them smoke. That is the value of n is 125 and…

Q: In a survey of 4026 adults, 708 oppose allowing transgender students to use the bathrooms of the…

A: Given: n=4026 x=708

Q: n a survey of 2558 adults, 1413 say they have started paying bills online in the last year.…

A: Solution: Given information: n= 2558 Sample size of adults x = 1413 adults started paying bills…

Q: In a survey of 3254 adults, 1433 say they have started paying bills online in the last year.…

A: Given X=1433 n=3254

Q: In a survey of 3478 adults, 1462 say they have started paying bills online in the last year.…

A: From the provided information,Confidence level = 99%

Q: adults in a recent year, 742 made a New Year's resolution to eat healthier. Construct 90% and…

A: The value of z at a 90% confidence level is 1.645, which can be found by using the standard normal…

Q: In a survey of 2000 adults in a recent year, 708 made a New Year's resolution to eat healthier.…

A: Confidence interval for population proportion is given by, p^ +- z* √(pq/n) At 90% z = 1.645 At 95%…

Q: In a survey of 2113 adults in a recent year, 741 made a New Year's resolution to eat healthier.…

Q: In a survey of 2703 adults, 1438 say they have started paying bills online in the last year.…

A: We have given that, Number of favorable cases(X) = 1438, Sample size(n) = 2703 Sample proportion…

Q: In a survey of 3985 adults, 711 oppose allowing transgender students to use the bathrooms of the…

A: Given, n=3985 x=711 To find a confidence interval for a population proportion.

Q: In a survey of 4008 adults, 728 oppose allowing transgender students to use the bathrooms of the…

Q: In a survey of 4052 adults, 723 oppose allowing transgender students to use the bathrooms of the…

A: It is given that x = 723 and n= 4,052.

Q: In a survey of 3185 adults, 1455 say they have started paying bills online in the last year.…

Q: In a survey of 2002 adults in a recent year, 701 made a New Year's resolution to eat healthier.…

A: Given:

Q: In a survey of 2004 adults in a recent year, 717 made a New Year's resolution to eat healthier.…

Q: Out of 180 randomly selected adults in the United States who were surveyed, 81 exercise on a regular…

Q: According to a report on consumer fraud and identity theft, Indiana had 1671 complaints of identity…

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 500…

Q: in a survey of 2042 adults in a recent year, 715 made a New Year's resolution to eat healthier.…

Q: In a survey of 2552 adults, 1423 say they have started paying bills online in the last year.…

A: Given that, In a survey of 2552 adults, 1423 say they have started paying bills online in the last…

Q: Of 380 randomly selected medical students, 21 said that they planned to work in a rural community…

A: Given that, Out Of 380 randomly selected medical students, 21 said that they planned to work in a…

Q: In a survey of 3260 adults, 1456 say they have started paying bills online in the last year.…

A: Givenn=3260x=1456p^=xn=14563260=0.4466

Q: a survey of 3409 adults, 1445 say they have started paying bills online in the last year. Construct…

A: The confidence interval = 99%The faovarble cases [x] = 1445The sample size [n] = 3409

Q: Construct a 99% confidence interval for the population proportion. Interpret the results.

Q: n a survey of 3932 adults, 718 oppose allowing transgender students to use the bathrooms of the…

Q: In a survey of 3983 adults, 747 oppose allowing transgender students to use the bathrooms of the…

Q: In a sample of 300 people, 48 identified as dog owners. Construct a 99% confidence interval for the…

A: Obtain the 99% confidence interval for the true proportion of people who are dog owners. The 99%…

Q: In a survey of 2782 adults, 1411 say they have started paying bills online in the last year.…

Q: In a survey of 2745 adults, 1477 say they have started paying bills online in the last year.…

A: Given, n= 2745 x=1477 The sample proportion is given by: p^=xn=14772745=0.5381 Also, α=0.99…

Q: In a survey of 2693 adults, 1495 say they have started paying bills online in the last year.…

A: Given that, In a survey of 2693 adults, 1495 say they have started paying bills online in the last…

Q: In a survey of 2334 adults in a recent year,1370 say they have made a New Year's resolution.…

A: Given n=2334 X=1370

Q: In a survey of 2646 adults, 1401 say they have started paying bills online in the last year.…

A: Number of adults (n) = 2646 Number of adults who say they have started paying bills online in the…

Q: Out of 157 randomly selected adults in the United States who were surveyed, 77 exercise on a regular…

Q: n a survey of 2847 adults, 1408 say they have started paying bills online in the last year.…

A: Given that Sample size (n) =2847 People started paying bills online (x) = 1408 Proportion (p) =…

Q: An internet service provider sampled 540 customers and found that 55 of them experienced an…

Q: Out of 100 people sampled 48 had kids. Based on this construct a 90% confidence interval for the…

Q: In a survey of 3342 adults, 1489 say they have started paying bills online in the last year.…

A: Given Information: To construct a 99% confidence interval for the population proportion and…

Q: In a survey of 2013 adults in a recent year, 731 made a New Year's resolution to eat healthier.…

Q: In a survey of 3426 adults, 1487 say they have started paying bills online in the last year.…

A: From the provided information, Sample size (n) = 3426 Out of which 1487 say they have started paying…

Q: n a survey of 3196 adults, 1405 say they have started paying bills online in the last year.…

A: Given data, n=3196 x=1405 p=x/n=1405/3196 =0.4396 z-value at 99% confidence is Zc=2.576

Q: In a survey of 2588 adults, 1466 say they have started paying bills online in the last year.…

A: Construct 99% confidence interval for population proportion:The sample of 2,588 adults among that…

Q: We wish to estimate what percent of 18 to 25-year old's that own a Samsung phone. Out of 400…

A: Solution: Let X be the number of adults own a Samsung phone and n be the sample number of adults.…

Q: In a survey of 3246 adults, 1418 say they have started paying bills online in the last year.…

A: Given information Number of adults surveyed(n) = 3246 Number of people who started paying bills…

Q: Out of 42738 freshmen entering the State University system, 27298 of them need to take remedial…

Question

100%

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1: introduction

VIEW

Step 2: Construction of confidence interval

VIEW

Solution

VIEW

Step by step

Solved in 3 steps with 1 images

SEE SOLUTION Check out a sample Q&A here

Similar questions

After an advertisement campaign of polish, the manufacturers wanted to know how many of the possible costumers had read the advertisement. The select a random sample of 95 costumers and found that only 25 of them had read the advertisement. Construct a 90% confidence interval for the percentage of customers who have read the advertisement.
In a survey of 3038 adults, 1448 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.
In a survey of 2947 adults, 1430 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.
In a survey of 3344 adults, 1499 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.
In a survey of 2567 adults, 1431 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.
6) When 415 college students were surveyed, 175 said they own their car. Construct a 95% confidence interval for the proportion of college students who say they own their cars.
In a survey of 2024 adults in a recent year, 709 made a New Year's resolution to eat healthier. Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.
Dr. Soria surveyed 34,000+ college students about their rates of housing insecurity during the pandemic. She found that 22% of students experienced housing insecurity (SE = .50). What is the 95% confidence interval for the true proportion of all college students experiencing housing insecurity at that time?
In a certain state, a survey of 500 workers showed that 38% belonged to a union. Find the 93% confidence interval for the overall proportion of workers who belong to a union
In a survey of 500 women, 120 said that Valentine’s Day was their favorite occasion to receive flowers. Construct an 88% confidence interval for the proportion of all women who feel that Valentine’s Day is their favorite occasion to receive flowers.
In a survey of 2033 adults in a recent year, 735 made a New Year's resolution to eat healthier. Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.
In a survey of 3500 adults, 1443 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.