In a random sample of 1250 adult drivers, 450 said that they would cut their driving by 10% if this significantly helped the environment.Find a 95% confidence interval estimate of the proportion of adult drivers who are willing to cut their driving by 10% to help theenvironment.O 0.36 + 1.96 (0.36)(0.64)V1250(0.90)(0.10)O 0.36 +1.961250(0.36)(0.64)O 0.36 + 1.96,O 0.36 +1.96(0.10)(0.90)V12303 4NextOv A 9:Co

Answered: Image /qna-images/answer/ed4a8899-bd9a-40a7-b318-b08f79ee352e.jpg

Answered: In a random sample of 1250 adult…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A pollster is going to sample a number of voters in a large city and construct a 99% confidence interval for the proportion who support the incumbent candidate for mayor. Find a sample size so that the margin of error will be no larger than 0.05. Be sure to round up to the next whole number. The required sample size is X Ś
4. The lower value of a 90% confidence interval for the variance in maze-running times of a sample of 31 rats is 4.62. Based on this, form a 98% confidence interval for 02 Answer: 3.973 to 13.5219
A study was done to determine whether there is a relationship between snoring and the risk of heart disease. Among 1,105 snorers in the study, 89 had heart disease, while only 29 of 1,379 nonsnorers had heart disease. Determine a 95% confidence interval that estimates p1 − p2 = difference in proportions of snorers and nonsnorers who have heart disease. (Let p1 be the proportion of snorers and p2 be the proportion of nonsnorers. Round your answers to three decimal places.) ____to ____
A student was asked to find a 90% confidence interval for widget width using data from a random sample of size n = 20. Which of the following is a correct interpretation of the interval (11.5,21.6)? There is a 90% chance that the mean of a sample of 20 widgets will be between 11.5 and 21.6. The mean width of all widgets is between 11.5 and 21.6, 90% of the time. We know this is true because the mean of our sample is between 11.5 and 21.6. With 90% confidence, the mean width of a randomly selected widget will be between 11.5 and 21.6. We are 90% confident the mean width of all widgets is between 11.5 and 21.6. There is a 90% chance that the mean of the population is between 11.5 and 21.6.
A sample of 1200 females and 1000 males found that 78.3% of females and 35.1% males supported increased funding of arts. Construct a 90% confidence interval around the difference in proportions.
A research company has carried out a poll to find out about people’s economic outlook. 894 of 1854 young adults (aged between 20–29 years) said that the economic situation is likely to improve over the next year. By contrast, 1292 of 2243 mature adults (aged between 30–39 years) agreed with this statement.) Calculate the upper limit of an 80% confidence interval for the difference between the proportions of people holding this optimistic view in the two age groups in the population. Form the confidence interval around a positive point estimate for the difference between the proportions. In order to obtain a positive point estimate, make sure you subtract the lower value from the higher value when calculating the difference between the sample proportions. Provide your answer as a number between 0 and 1 rounded to 4 decimal places.
Assume that a sample is used to estimate a poplulation mean. Find the 98% confidence interval for a sample of size 863 with a mean of 65.7 and a standard deviation of 19.2. Enter your answer as a tri-linear inequaliy accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).
An internet service provider sampled 540 customers and found that 75 of them had experienced an interruption in high-speed service during the previous month. Compute a 99% confidence interval for the proportion of customers who have experienced a service interruption during the previous month.
21.20 Count Buffon’s coin. The eighteenth-century French naturalist Count Buffon tossed a coin 4040 times. He got 2048 heads. Give a 95% confidence interval for the proportion of all Buffon’s coin tosses that land head side up. Are you confident that this proportion is not 0.5? Why?
A trade magazine routinely checks the drive-through service times upper bound of 163.1 seconds. What does this mean? fast-food restaurants. An 80% confidence interval that results from examining 563 customers in one fast-food chain's drive-through has a lower bound of 159.5 seconds and an Choose the correct answer below. O A. One can be % confident that the mean drive-through service time of this fast-food chain is seconds O B. One can be % confident that the mean drive-through service time of this fast-food chain is between seconds and seconds. O C. The mean drive-through service time of this fast-food chain is seconds % of the time. O D. There is a % probability that the mean drive-through service time of this fast-food chain is between seconds and seconds.
Refer to Exercise 7. Someone suggests that the paired design be replaced with a design in which 18 cars are sampled, the lifetime of the front brakes is measured on 9 of them, and the lifetime of the rear brakes is measured on the other 9. A confidence interval for the difference between the means would then be constructed by using expression (5.21) (in Section 5.6). He claims that this design will produce a more precise confidence interval, since 18 cars will be used instead of 9. a) Will the new design produce a valid confidence interval? Explain. b) Is it likely that the confidence interval produced by the new design will be more precise than, less precise than, or about equally precise as the confidence interval produced by the paired design? Explain.
A study was done to determine whether there is a relationship between snoring and the risk of heart disease. Among 1,105 snorers in the study, 85 had heart disease, while only 21 of 1,379 nonsnorers had heart disease. In USE SALT (a) Determine a 95% confidence interval that estimates p, - p2 difference in proportions of snorers and nonsnorers who have heart disease. (Let p, be the proportion of snorers and p, be the proportion of nonsnorers. Round your answers to three decimal places.) |-0.079 X to -0.045

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