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 5.22 hours, with a standard deviation of 2.31 hours. A random sample of 40 adults with children under the age of 51 results in a mass daily hours time of 4.21 hours, with a standard deviation of 1.62 hours. Construct and interpret a 95% 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_4 represent the mean leisure hours of adults with children under the age of 18. The 95% confidence interval for (mu_1 - mu_2) is the range from hours to hours. (Round to two decimal places as needed.)

Explanation / Answer

Solution:

We have
The 95% confidence interval for 12 is
( x- ) ± Z/2 sqrt(S12/n + S22/m)
= (5.22 - 4.21 ) ± Z0.05/2 . sqrt(2.31^2/40+1.62^2/40)
= (1.01) ± 1.96 * 0.4461
= [0.1356, 1.8844]
The difference are lies in confidence interval at 0.05 level of significance.
The 95%confidence interval for 1-2 is the range from 0.1356 hours to 1.8844 hours.

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