A survey of 25 randomly selected customers found the ages shown (in years). Themean is 33.56 years and the standard deviation is 9.42 yearsa) Construct a 80% confidence interval for the mean age of all customers, assumingthat the assumptions and conditions for the confidence interval have been metb) How large is the margin of error?c) How would the confidence interval change if you had assumed that the populationstandard deviation was known to be 10.0 years?a) What is the confidence interval?(Round to two decimal places as needed.)37 39 21 31 2627 36 35 38 4448 21 20 38 2817 29 48 40 3734 19 48 34 44

Sample mean=33.56Sample standard deviations=9.42New sample standard deviations =10.0Sample size= 25

Answered: A survey of 25 randomly selected…

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 distribution of study times for first-year students follow a Normal distribution with a standard deviation σ = 48 minutes. Suppose a sample of 344 students found that the mean study time for these students was x¯= 170 minutes per week. Use this to find a 94.38% confidence interval to estimate the mean study time for all students. We are 94.38% confident that the mean study time for first-year students is between the following ( answer) minutes and ( answer) minutes. The answer needs to be the mean study time for first-year students is between what amount of minutes and the following minutes.
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