The average salary of 25 random employees in a certain financial sector was 3400 USD. The sample standard deviation was360 USD. We are interested in the 90% confidence interval for the population mean. What is the value of the margin of error?Select one:O a. 184.79O b. 92.39O c. 153.99O d. 123.19

The Margin of error for 90% confidence interval for the population mean when the population standard…

Answered: The average salary of 25 random…

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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mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. A random sample of 40 home theater systems has a mean price of $138.00. Assume the population standard deviation is $17.80. Construct a 90% confidence interval for the population mean.
According to the College Board, scores on the math section of the SAT Reasoning college entrance test for the class of 2010 had a mean of 516 and a standard deviation of 116. Assume that they are roughly normal.One of the quartiles of the scores from the math section of the SAT Reasoning test is 438. The other quartile is _______.
Use the standard normal distribution or the t-distribution to construct a 90% confidence interval for the population mean. Justify your decision. If neither distribution can be used, explain why. Interpret the results. In a random sample of 40 people, the mean body mass index (BMI) was 26.6 and the standard deviation was 6.13.
A survey of several 9 to 11 year olds recorded the following amounts spent on a trip to the mall: $19.26, $21.87, $15.83, $20.57, $19.31, $24.74, $22.82 Construct the 95% confidence interval for the average amount spent by 9 to 11 year olds on a trip to the mall. Assume the population is approximately normal. Copy Step 2 of 4: Calculate the sample standard deviation for the given sample data. Round your answer to two decimal places.
Suppose we wish to study the average income of Cactus Club customers, so we plan to conduct a research project with a 95% confidence level. We need our range of error (E) to be no greater than $500. Based on our past experience, we estimate the sample standard deviation to be $5,000. What sample size do you need in order to conduct the research? Select one: a. 372 b. 784 c. 824 d. 385 e. 196
Please answer a,b,c,d. Ty!
A study of 4 bolts of carpet showed that their average length was 76.2 yards and the sample standard deviation was 5.6 yards. Which of the following is the 80% confidence interval for the mean length per bolt of carpet? O A. 75.8 to 76.6 O B. 72.62 to 79.78 O C. 71.61 to 80.79 D. 75.40 to 77.00
Salaries of 25 college graduates who took a statistics course in college have a mean of $79,350 and a standard deviation of $1,280. Construct a 98% confidence interval for estimating the population standard deviation. Enter the upper bound of the confidence interval. (Round your answer to nearest whole dollar.)
MC Qu. 10-64 Sales at a fast-food restaurant average... Sales at a fast-food restaurant average $6,000 per day. The restaurant decided to introduce an advertising campaign to Increase daily sales. In order to determine the effectiveness of the advertising campaign, a sample of 49 day's sales were taken. The sample showed average daily sales of $6,300. From past history, the restaurant knew that its population standard deviation is about $1,000. If the level of significance is 0.01, have sales increased as a result of the advertising campaign? Multiple Choice Fail to reject the null hypothesis. Reject the null hypothesis and conclude the mean is higher than $6,000 per day. Reject the null hypothesis and conclude the mean is lower than $6,000 per day. Reject the null hypothesis and conclude that mean equal to $6,000 per day.
College tuition: A simple random sample of 40 colleges and universities in the United States has a mean tuition of S18,600 with a standard deviation of S10,800. Construct a 95% confidence interval for the mean tultion for all colleges and universities in the United States. Round the answers to the nearest whole number. A 95% confidence interval for the mean tuition for all colleges and universities is
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.
You measure 37 textbooks' weights, and find they have a mean weight of 69 ounces. Assume the population standard deviation is 12.9 ounces. Based on this, construct a 80% confidence interval for the true population mean textbook weight.Report your answers to three decimal places.

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