A random sample of 40 adults with no children under the age of 18 years results

Question

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5 67 hours, with a standard deviation of 2.29 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.48 hours, with a standard deviation of 1.56 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children (mu_1 - mu_2). Let mu_1 represent the mean leisure hours of adults with no children under the age of 18 and mu_2 represent the mean leisure hours of adults with children under the age of 18. The 90% confidence interval for f (mu_1 - mu_2) is the range from hours to hours.

Explanation / Answer

Formula to take out the CI are:

Mu1-Mu2 +/- Z*sqrt(sigma1^2/n1 + sigma2^2/n2))

= (5.67-4.48) +/- 1.28*sqrt( ((2.29^2) / 40) + ((1.56^2)/40))

=1.19 +/- .5607

=.629 to 1.7507

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.