results in a mean daily leisure time of 4.09 hours, with a standard deviation of 1.53 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no childre children (H - 42) Let i, represent the mean leisure hours of adults with no children under the age of 18 and uz represent the mean leisure hours of adults with children under the age of 18. The 90% confidence interval for (H1- H2) is the range from hours to hours.
results in a mean daily leisure time of 4.09 hours, with a standard deviation of 1.53 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no childre children (H - 42) Let i, represent the mean leisure hours of adults with no children under the age of 18 and uz represent the mean leisure hours of adults with children under the age of 18. The 90% confidence interval for (H1- H2) is the range from hours to hours.
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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Transcribed Image Text:A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.03 hours, with a standard deviation of 2.36 hours. A random sample of 40 adults with children under the age of 18
results in a mean daily leisure time of 4.09 hours, with a standard deviation of 1.53 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with
children (H1 - H2):
.....
Let u, represent the mean leisure hours of adults with no children under the age of 18 and u, represent the mean leisure hours of adults with children under the age of 18.
The 90% confidence interval for (u1 - H2) is the range from hours to
hours.
(Round to two decimal places as needed.)
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