Question

11.3.21 Question Help * A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.03 hours, with a standard deviation of 2.42 hours, A random sample of 40 adults with children under the age of 18 results in a mean da y e sure time of 4.14 hours with a standard de ation o 1.86 hours. Construct and interpret a 95% confidence interval or the mean difference in leisure me between adults with no children and adults with children Let μ 1 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 The 95% confidence interval for (μ-μ2) is the range from hours to hours. μ 1- 2 (Round to two decimal places as needed.)

Answer 1

Ans:

pooled standard deviation=sqrt(((40-1)*2.42^2+(40-1)*1.86^2)/(40+40-2))

=2.158

standard error for difference=2.158*sqrt((1/40)+(1/40))=0.4825

df=40+40-2=78

critical t value=tinv(0.05,78)=1.991

95% confidence interval for difference in means

=(5.03-4.14)+/-1.991*0.4825

=0.89+/-0.96

=(-0.07, 1.85)