A study of the consultants in a particular industry has determined that the standard deviation of the hourly fee of the consultants is $21. A randomsample of 50 consultants in the industry has a mean hourly fee of $115. Find a 90% confidence interval for the true mean hourly fee of allconsultants in the industry. Then give its lower limit and upper limit.Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list offormulas.)Lower limit: IUpper limit: |Save For LaterSubmit AssignmenCheck© 2022 McGraw Hill LLC. AIl Rights Reserved. Terms of Use | Privacy Center| AccessibilityétvDIIF10F8F7F6F5F4SCF3F2

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Answered: A study of the consultants in a…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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A statistics practitioner took a random sample of 42 observations from a population whose standard deviation is 30 and computed the sample mean to be 102. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval = B. Estimate the population mean with 90% confidence. Confidence Interval = C. Estimate the population mean with 99% confidence. Confidence Interval =
A statistics practitioner took a random sample of 55 observations from a population whose standard deviation is 28 and computed the sample mean to be 110. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval =| B. Estimate the population mean with 90% confidence. Confidence Interval = C. Estimate the population mean with 99% confidence. Confidence Interval =
A Surprising fact indicates that a small proportion of teens use a password on their mobile device. A poll chooses a simple random sample of 120 teens and asks if they use a password on their mobile device. The results showed that 56 of teens in our sample used a password on their mobile device. Find a 90 % confidence interval for the proportion of teens who use a password on their mobile device. Interpret your confidence interval.
According to recent data by the American Marketing Association, 76.5%of people in the 18-24 age group used social media in 2017. You want to find out what the social media proportion is among your classmates who are all between 18 and 24 years old. Suppose you obtain a random sample of 229 students and find that 202 students in this sample are active social media users. Complete parts a through c below. a. Construct a 99% confidence interval to estimate the actual proportion of social media users. A 99% confidence interval to estimate the actual proportion has a lower limit of------ and an upper limit of------- (Round to three decimal places as needed.) b. What is the margin of error for this sample? c. Is there evidence that this proportion matches that of the overall population based on the sample?
Please answer the 4 items.
solve correctly. I will rate accordingly.
Answer each question and justify all work for full You will need your z and t charts for this quiz. 1. Death penalty: A group claims that the overall feelings of the general population on the . death penalty is changing. 643 people were interviewed, and of them, 258 believed in using the death penalty. Construct a 95% confidence interval for the proportion of people that believe in the death penalty. Follow the steps below: What is the best point estimate for the proportion of people who believe in the death penalty? › (round to two decimal places)
A survey of 32,782 adults in 2004 indicated that 24.6% of adults were current smokers. A similar study conducted in 1995 of 40,000 adults indicated that 29.7% were current smokers. a. Find and interpret a point estimate of the difference between the proportion of current smokers in 1995, and the proportion of current smokers in 2004. b. A 95% confidence interval for the true difference is (-0.058,-0.045). Interpret. c. What assumptions must you make for the confidence interval i part b to be valid? (…) a. Find a point estimate of the difference between the proportion of current smokers in 1995 and the proportion of current smokers in 2004. (Round to three decimal places as needed.)
A researcher wishes to estimate the proportion of college students who cheat on exams. A poll of 480 college students showed that 23% of them had, or intended to, cheat on examinations. Find the 95% confidence interval for the population proportion. Round to four decimal places.
A random sample of 64 shoppers showed that they spend an average of $27.31 per visit at the Sunday Mornings Bookstore. The standard deviation of the population is $2.14. Find the 90% confidence interval of the true mean. Please include all work showing how you found your answer.
If the analyst decides to increase the sample size (n) from 75 to 100, while maintaining the same confidence level, the width of the confidence interval would then: fluctuate up and down. remain unchanged. decrease. increase.
A statistics practitioner took a random sample of 48 observations from a population whose standard deviation is 32 and computed the sample mean to be 107. Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits. A. Estimate the population mean with 95% confidence. Confidence Interval = B. Estimate the population mean with 90% confidence. Confidence Interval = C. Estimate the population mean with 99% confidence. Confidence lInterval =

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