Answer
$-t_{\frac{α}{2}}\lt t_0\lt t_{\frac{α}{2}}$: null hypothesis is not rejected.
There is not enough evidence to conclude that the mean feeling-at-ease score is different for males than females.
Work Step by Step
$x ̅_{Female}=\frac{∑x_{{Female}_i}}{n_{Female}}=18.244$
$s_{Female}=\sqrt {\frac{∑(x_{{Female}_i}-x ̅_{Female})^2}{n_{Female}-1}}=0.747$
$x ̅_{Male}=\frac{∑x_{{Male}_i}}{n_{Male}}=18.462$
$s_{Male}=\sqrt {\frac{∑(x_{{Male}_i}-x ̅_{Male})^2}{n_{Male}-1}}=1.029$
$H_0:~µ_{Female}=µ_{Male}$ versus $H_1:~µ_{Female}\ne µ_{Male}$
$t_0=\frac{(x ̅_{Female}-x ̅_{Male})}{\sqrt {\frac{s^2_{Female}}{n_{Female}}+\frac{s^2_{Male}}{n_{Male}}}}=\frac{(18.244-18.462)}{\sqrt {\frac{0.747^2}{16}+\frac{1.029^2}{16}}}=-0.686$
$n=16$, so:
$d.f.=n-1=15$
Two-tailed test:
$t_{\frac{α}{2}}=t_{0.025}=2.131$
(According to Table VI, for d.f. = 15 and area in right tail = 0.025)
Also, $-t_{\frac{α}{2}}=-2.131$
Since $-t_{\frac{α}{2}}\lt t_0\lt t_{\frac{α}{2}}$, we do not reject the null hypothesis.
