A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.06hours, with a standard deviation of 2.36 hours. A random sample of 40 adults with children under the age of 18 results ina mean daily leisure time of 4.06 hours, with a standard deviation of 1.97 hours. Construct and interpret a 95%confidence interval for the mean difference in leisure time between adults with no children and adults with children(H₁-H₂).Let μ₁ represent the mean leisure hours of adults with no children under the age of 18 and µ2 represent the meanleisure hours of adults with children under the age of 18.from-The 95% confidence interval for (μ₁ −μ₂) is the range(Round to two decimal places as needed.)hours tohours.

Here, mean1 is 5.06 and standard deviation1 is 2.36. Mean2 is 4.06 and standard deviation2 is 1.97.…

Answered: A random sample of 40 adults with no…

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 most recent American Time Use Survey found that many Americans barely spend any time reading for fun. People ages 15 to 19 average only 7.8 minutes of leisurely reading per day with a standard deviation of 5.4 minutes. However, people ages 75 and over read for an average of 43.8 minutes per day with a standard deviation of 35.5 minutes. A 95% confidence interval constructed for the true difference in mean amount of time in minutes spent reading per day (15 to 19 - 75 and older) was −36±2.1762−36±2.1762. Does the interval provide convincing evidence that those 75 and older read more than those 15 to 19?
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