A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.55 hours, with a standard deviation of 2.28 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.14 hours, with a standard deviation of 1.86 hours.
Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no children and adults with children (μ1 − μ2). Let μ1 represent the mean leisure hours of adults with no children under the age of 18 and μ2 represent the mean leisure hours of adults with children under the age of 18. Round to six decimal places as needed.
Since population SD is unknown, we will use t score for the confidence interval.
90% CI for u1-u2 is given as:
x1-x2 +- t0.05,n1+n2-2*√s1^2/n1 + s2^2/n2
= (5.55-4.14) +- 1.665*√2.28^2/40 + 1.86^2/40
= 1.41 +- 0.774628
= (0.635372, 2.184628)
We are 90% confident that the true difference in population means lies in the above interval.
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