Question

A random sample of 4040 adults with no children under the age of 18 years results in a mean daily leisure time of 5.345.34 ​hours, with a standard deviation of 2.372.37 hours. A random sample of 4040 adults with children under the age of 18 results in a mean daily leisure time of 4.124.12 ​hours, with a standard deviation of 1.641.64 hours. Construct and interpret a 9090​% confidence interval for the mean difference in leisure time between adults with no children and adults with children left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2.

Let mu 1μ1 represent the mean leisure hours of adults with no children under the age of 18 and mu 2μ2 represent the mean leisure hours of adults with children under the age of 18.

The 9090​% confidence interval for left parenthesis mu 1 minus mu 2 right parenthesisμ1−μ2 is the range from nothing hours to nothing hours.

​(Round to two decimal places as​ needed.)

What is the interpretation of this confidence​ interval?

A.

There is 9090​% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of ainsufficient evidence of a significant difference in the number of leisure hours.

B.

There is a 9090​% probability that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours.

C.

There is 9090​% confiden

Answer 1

Solution:-

90​% confidence interval for the mean difference in leisure time between adults with no children and adults with children is C.I = ( 0.46, 1.98).

C.I = (5.34 - 4.12) + 1.662*0.4557

C.I = 1.22 + 0.75737

C.I = 1.22 + 0.75737

C.I = ( 0.463, 1.977)

B) There is a 90​% probability that the difference of the means is in the interval. Since the confidence interval contains all the positive values, hence there is a significant difference in the number of leisure hours.