Answer
Confidence interval: $-12.44\lt µ_1-µ_2\lt0.04$
We are 95% confident that $µ_1-µ_2$ is between -12.44 and 0.04.
Work Step by Step
$n=24$ (use the smaller value of $n$), so:
$d.f.=n-1=23$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.069$
(According to Table VI, for d.f. = 23 and area in right tail = 0.025)
$Lower~bound=(x ̅_1-x ̅_2)-t_{\frac{α}{2}}\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=(104.2-110.4)-2.069\sqrt {\frac{12.3^2}{24}+\frac{8.7^2}{27}}=-12.44$
$Upper~bound=(x ̅_1-x ̅_2)+t_{\frac{α}{2}}\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=(104.2-110.4)+2.069\sqrt {\frac{12.3^2}{24}+\frac{8.7^2}{27}}=0.04$
