Answer
Confidence interval: $0.71\lt µ_1-µ_2\lt2.33$
We are 90% confident that the mean difference in leisure time between adults with no children and adults with children is between 0.71 and 2.33.
Work Step by Step
$n=40$, so:
$d.f.=n-1=39$
$level~of~confidence=(1-α).100$%
$90$% $=(1-α).100$%
$0.90=1-α$
$α=0.10$
$t_{\frac{α}{2}}=t_{0.05}=1.685$
(According to Table VI, for d.f. = 39 and area in right tail = 0.05)
$Lower~bound=(x ̅_1-x ̅_2)-t_{\frac{α}{2}}\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=(5.62-4.10)-1.685\sqrt {\frac{2.43^2}{40}+\frac{1.82^2}{40}}=0.71$
$Upper~bound=(x ̅_1-x ̅_2)+t_{\frac{α}{2}}\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=(5.62-4.10)+1.685\sqrt {\frac{2.43^2}{40}+\frac{1.82^2}{40}}=2.33$
