Answer
Confidence interval: $1.093\lt µ_{women}-µ_{men}\lt3.607$
We are 95% confident that the difference between the mean measure of flexibility of women and the mean measure of flexibility of men is between 1.093 and 3.607 inches.
Work Step by Step
$n=31$ (use the smaller value of $n$), so:
$d.f.=n-1=30$
$level~of~confidence=(1-α).100$%
$95$% $=(1-α).100$%
$0.95=1-α$
$α=0.05$
$t_{\frac{α}{2}}=t_{0.025}=2.042$
(According to Table VI, for d.f. = 30 and area in right tail = 0.025)
$Lower~bound=(x ̅_1-x ̅_2)-t_{\frac{α}{2}}\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=(20.99-18.64)-2.042\sqrt {\frac{2.07^2}{31}+\frac{3.29^2}{45}}=1.093$
$Upper~bound=(x ̅_1-x ̅_2)+t_{\frac{α}{2}}\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}=(20.99-18.64)+2.042\sqrt {\frac{2.07^2}{31}+\frac{3.29^2}{45}}=3.607$
