a) For 30 randomly selected Rolling Stones concerts, the mean gross earnings is 2.82 million dollars. Assuming a population standard deviation gross earnings of 0.53 million dollars, obtain a 99confidence interval for the mean gross earnings of all Rolling Stones concerts (in millions).Confidence interval: (b) Which of the following is the correct interpretation for your answer in part (a)?O A. We can be 99% confident that the mean gross earnings of all Rolling Stones concerts lies in the intervalB. There is a 99% chance that the mean gross earnings of all Rolling Stones concerts lies in the intervalO C. We can be 99% confident that the mean gross earnings for this sample of 30 Rolling Stones concerts lies in the intervalO D. None of the above

Answered: Image /qna-images/answer/baa9c6fc-9573-459e-b958-af36d6119d70.jpg

Answered: a) For 30 randomly selected Rolling…

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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The dean randomly selected 33 students who took the calculus final exam and looked at their scores on it. In this sample the mean was 80.3 and the standard deviation was 3.7. No information about the population's distribution is available. Construct a 98% confidence interval for the mean score of all students who took this exam. Round your answers to 2 places after the decimal point, if necessary. If it is not appropriate to construct a confidence interval in this situation, then enter "0" in both answer boxes below. 98% confidence interval:
The distribution of study times for first-year students follow a Normal distribution with a standard deviation σ = 78 minutes. Suppose a sample of 431 students found that the mean study time for these students was x¯= 195 minutes per week. Use this to find a 95% confidence interval to estimate the mean study time for all students. We are 95% confident that the mean study time for first-year students is between (Answer) and minutes ( Answer) minutes.
Please answer the following question using *****EXCEL****
How can we also find normality?
The heights of 9 men were taken with sample mean of 165 cm and standard deviation 5 cm. The heights of 7 Japanese women were taken with sample mean of 150 cm and standard deviation 3 cm. Assuming that the population standard deviations for heights of men and women are both the same, construct the 95% confidence interval for the difference in mean heights between men and women.
A random sample of 10 children found that their average growth for the first year was 9.8 inches, with a standard deviation of 0.96 inches. Find the 85% confidence interval of the mean growth of all American children in the first year of life.What is the name of the confidence interval?
Please help with 1 a,b,c,d,e, and f.
Dr. Higney, an academic researcher in an English department, wants to learn about the average length of a novel. In order to do this, Dr. Higney collects data on the word counts of 200 randomly selected novels. The sample mean of this data set is = 85252 with a sample standard deviation of 8413 words. Using this data, construct a 90% confidence interval for the true mean length of novels. (84,273.50, 86,230.51) (82719.67,89784.33) (94086.04, 96417.96) (80937.53, 90566.47) S
In a random sample of 91 college students, the average number of Discord messages handled in one day is 216. The 95% confidence interval for the mean number of messages handled by students daily is given as 210 to 222. a. What is the population standard deviation? b. If the number of samples were doubled, what would be the new confidence interval (keeping the same confidence level?) Please include any equations used so I can fully learn this.
A sample of 30 NBA players had a sample mean of their heights as 80.5 inches. If the standard deviation of the population of NBA players is 3.5 inches, find a 90% confidence interval for the average height of all NBA players.

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