Suppose an insurance company randomly surveyed 853 drivers and found that 725 claimed they always buckled up. Construct a 87% confidence interval for the fraction of all drivers whockle up. p' = .849941 .065 1.51 || Margin of Error: E = .01847 We are 87% confident that the fraction of drivers who always buckle up is between and .8689

Suppose an insurance company randomly surveyed 853 drivers and found that 725 claimed they always buckled up. Construct a 87% confidence interval for the fraction of all drivers whockle up. p' = .849941 .065 1.51 || Margin of Error: E = .01847 We are 87% confident that the fraction of drivers who always buckle up is between and .8689

Name: Suppose an insurance company randomly surveyed 853 drivers and found
Uploaded: 2024-03-20T07:07:00.000Z
Channel: Bartleby
Description: Suppose an insurance company randomly surveyed 853 drivers and found that 725 claimed theyalways buckled up. Construct a 87% confidence interval for the fraction of all drivers whockle up.p' = .849941.0651.51Margin of Error: E.01847We are 87% confid

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

Similar questions

We wish to estimate what percent of adult residents in a certain county are parents. Out of 200 adult residents sampled, 134 had kids. Based on this, construct a 99% confidence interval for the proportion, p, of adult residents who are parents in this county.Give your answers as decimals, to three places.
A random sample of 100 employees at a company shows that 62 are female. Form a 90% confidence interval for the proportion of all employees who are female. The interval is 0.62 what margin of error? Round your answer to 3 decimal places.
We wish to estimate what percent of adult residents in a certain county are parents. Out of 100 adult residents sampled, 48 had kids. Based on this, construct a 90% confidence interval for the proportion ππ of adult residents who are parents in this county.
Suppose a new treatment for a certain disease is given to a sample of 180 patients. The treatment was successful for 153of the patients. Assume that these patients are representative of the population of individuals who have this disease. Calculate a 98% confidence interval for the proportion successfully treated. (Round the answers to three decimal places.)
You want to get a 90% confidence interval for the difference between the percentage of females with red hair compared to males with red hair. You sample 76 females and find 34 of them have red hair. You sample 65 males and find 33 have red hair. < p1−p2 <
Suppose the population standard deviation is = 5, an SRS of n = 100 is obtained, and the confidence level is chosen to be 99.5%. The margin of error for estimating a mean is given by: a. 1.576. b. 1.165. c. 1.404. d. 1.223.
We wish to estimate what percent of adult residents in a certain county are parents. Out of 200 adult residents sampled, 34 had kids. Based on this, construct a 90% confidence interval for the proportion, p, of adult residents who are parents in this county. Give your answers as decimals, to three places.
A sample of 1100 observations taken from a population produced a sample proportion of 0.38. Make a 90 % confidence interval for Round your answers to three decimal places. = ( ) to ( )
If n = 420 and p (p-hat) = 0.2, construct a 95% confidence interval. Give your answers to three decimals < p <
In a sample of 300 adults, 198 had children. Construct a 90% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three places Rectan
From a sample of 500 Pepperdine students, 205 said they commuted to school. Find the 95% confidence interval using 2 decimal places for the proportion of Pepperdine students who commute. If you use a calculator, indicate the function used and the parameters you entered for the function.
It is not taking (0.1404, 0.3596) as the confidence interval