Answer
Confidence interval: $1.1\lt µ_1-µ_2\lt12.9$
We are 95% confident that $µ_1-µ_2$ is between 1.1 and 12.9.
Work Step by Step
$n=20$, so:
$d.f.=n-1=19$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.093$
(According to Table VI, for d.f. = 19 and area in right tail = 0.025)
$Lower~bound=(x ̅_1-x ̅_2)-t_{\frac{α}{2}}\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=(111-104)-2.093\sqrt {\frac{8.6^2}{20}+\frac{9.2^2}{20}}=1.1$
$Upper~bound=(x ̅_1-x ̅_2)+t_{\frac{α}{2}}\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=(111-104)+2.093\sqrt {\frac{8.6^2}{20}+\frac{9.2^2}{20}}=12.9$
