Answer
Confidence interval: $0.73\lt µ_1-µ_2\lt7.67$
We are 90% confident that the difference between the populations, $µ_1-µ_2$, is between 0.73 and 7.67.
Work Step by Step
$n=8$ (use the smaller value of $n$), so:
$d.f.=n-1=7$
$level~of~confidence=(1-α).100$%
$90$% $=(1-α).100$%
$0.90=1-α$
$α=0.10$
$t_{\frac{α}{2}}=t_{0.05}=1.895$
(According to Table VI, for d.f. = 7 and area in right tail = 0.05)
$Lower~bound=(x ̅_1-x ̅_2)-t_{\frac{α}{2}}\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=(32.4-28.2)-1.895\sqrt {\frac{4.5^2}{13}+\frac{3.8^2}{8}}=0.73$
$Upper~bound=(x ̅_1-x ̅_2)+t_{\frac{α}{2}}\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=(32.4-28.2)+1.895\sqrt {\frac{4.5^2}{13}+\frac{3.8^2}{8}}=7.67$
