Question

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.64 ​hours, with a standard deviation of 2.41 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.24 ​hours, with a standard deviation of 1.52 hours. Construct and interpret a 95​% confidence interval for the mean difference in leisure time between adults with no children and adults with children (u1-u2).
The 95% confidence interval for (u1-u2) is the range from _ hours to _ hours. (Round to two decimal places as needed.)

Answer 1

x1            =	5.64	x2            =	4.24
n1           =	40	n2           =	40
σ1           =	2.41	σ2           =	1.52
std error σ1-2=√(σ21/n1+σ22/n2)    =			0.451

Point estimate of differnce '=x1-x2 =			1.400
for 95 % CI value of z=			1.960
margin of error E=z*std error =			0.883
lower bound=(x1-x2)-E =			0.52
Upper bound=(x1-x2)+E =			2.28
from above 95% confidence interval for population mean =(0.52 , 2.28)

(since interval is above 0, there appear to be significant difference between population mean at 5% level)