Answer
Confidence interval: $3.57\lt µ_1-µ_2\lt12.83$
We are 90% confident that $µ_1-µ_2$ is between 3.57 and 12.83.
Work Step by Step
$n=18$ (use the smaller value of $n$), so:
$d.f.=n-1=17$
$level~of~confidence=(1-α).100$%
$90$% $=(1-α).100$%
$0.90=1-α$
$α=0.10$
$t_{\frac{α}{2}}=t_{0.05}=1.740$
(According to Table VI, for d.f. = 17 and area in right tail = 0.05)
$Lower~bound=(x ̅_1-x ̅_2)-t_{\frac{α}{2}}\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=(50.2-42.0)-1.740\sqrt {\frac{6.4^2}{25}+\frac{9.9^2}{18}}=3.57$
$Upper~bound=(x ̅_1-x ̅_2)+t_{\frac{α}{2}}\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=(50.2-42.0)+1.740\sqrt {\frac{6.4^2}{25}+\frac{9.9^2}{18}}=12.83$
