A business was interested in estimating the proportion of customers that came back within oneweek. They took a random sample of 100 customers and found that the 95% confidence intervalwas (0.15, 0.24). Complete the following conclusion."We are 95% confident that the [Select]back within one week is [Select [Select]populationsampleproportion of customers who came

Answered: A business was interested in estimating…

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 psychologist is studying the effects of lack of sleep on the performance of various perceptual-motor tasks. After a given period of sleep deprivation, a measurement of reaction time to an auditory stimulus was taken for each of 36 adult male subjects. The reaction times (in seconds) are summarized as follows:x = 1.82, S=0.22. Find a 99% confidence interval for the mean reaction time for sleep-deprived male subject
From a random sample of 120 students, 95 rated a case of plagiarism as unethical. Using this information, you computed a confidence interval extending from 0.68 to 0.76 for the population proportion. What is the confidence coefficient of the interval?
In a survey funded by the UW school of medicine, 750 of 1000 adult Seattle residents said they did not believe they could come down with a sexually transmitted infection (STI). Construct a 95% confidence interval estimage of the proportion of adult Seattle residents who don't believe they can contract an STI. (Use a z score of 1.96 for your computations.) (.728, .772) (.723, .777) (.718, .782) (.713, .878) (.665, .835)
A beverage company would like to determine if their drink is preferred to a rival company’s drink. They perform an experiment with 150 randomly selected individuals and have them taste test the two different drinks in a controlled environment so that each person's preference is independent of others. A 90% confidence interval for the proportion preferring their drink is (0.453, 0.587). Which of the following is the correct interpretation of this confidence interval? 90% of all consumers prefer this beverage company’s drink to their rival’s between 45.3% and 58.7% of the time. We’re 90% confident that between 45.3% and 58.7% of all consumers would prefer this beverage company’s drink to their rival’s drink. We’re 90% confident that there’s a 45.3% to 58.7% chance that a randomly selected person would prefer this beverage company’s drink to their rival’s drink. We’re 90% confident that the proportion of people in the sample who prefer this beverage company’s drink over their…
2 different treatments were used to speed the relief time (in minutes) of migraines. 8 subjects received each treatment. The recovery times are listed below. Treatment 1: 20 19 29 25 34 32 39 38 Treatment 2: 18 18 16 17 29 28 21 37 Construct a 90% confidence interval for the difference in the relief time between the 2 treatments.
Suppose we wish to estimate the average number of hours per week that Americans watch TV. Use the following data from an SRS to find a 90% confidence interval: 30 64 49 40 21 58 9 43 39 43
In a random sample of 500 people eating lunch at a cafeteria on various Fridays, it was found that 160 preferred seafood. A 99% confidence interval for the actual proportion of people eating seafood on Fridays (rounded off to two decimals) is: A. [0.28; 0.36] B. [0.29; 0.35] C. [0.27; 0.37] D. [0.26; 0.36]
The following data represent a random sample of five measurements from a normally distributed population: 7,4,2,5,7 a. Find 90% confidence interval for u. b. Find a 99% confidence interval for u.
In a survey of 31 students at a local high school, 15 support changing the school's mascot. Find the 95% confidence interval for the true population proportion of students who support changing school's mascot.Choose the correct confidence interval and critical value used in the calculation. A. (0.254,0.714) (Zc=2.576) B. (0.304,0.664)(Zc=1.96) C. (0.304,0.664)(tc=2.0423) D. 0.334,0.634)(Zc=1.645) E. (0.274,0.694) (Zc=2.326)
you conducted additional studies with these men and found the following: a. The RR for those who drank one drink per day was 1.50 (95% CI:1.32-1.67) in comparison to the abstainers. b. For those men who drank three or more drinks per day the RR compared to the abstainers was 4.20 (95% CI 2.68-5.85). Question: We can tell from the Confidence Intervals provided in #7 whether there are likely more people in the “one drink a day” group or the “3+ drinks per day” group. Which one likely has more people and how can we tell?
An insurance company collects data on seat-belt use among drivers in a country. Of 1500 drivers 30-39 years old, 33% said that they buckle up, whereas 458 of 1800 drivers 55-64 years old said that they did. Find a 95% confidence interval for the difference between the proportions of seat-belt users for drivers in the age groups 30-39 years and 55-64 years. ..... Construct a 95% confidence interval. The 95% confidence interval for p, - p2 is from to (Round to three decimal places as needed.)
You want to find a 92% confidence interval for the proportion of PLM students who get a student scholarship allowance from the government of the City of Manila. You found an information from the city administrator that states that around 75% of PLM students are to get the allowance. At a maximum error of 5%, how many samples do you need?

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