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 there is a difference in the reaction times of males and females.
Work Step by Step
$x ̅_1,n_1~and~s_1$ refer to female students and $x ̅_2,n_2~and~s_2$ refer to male students.
$x ̅_1=\frac{∑x_{1_i}}{n_1}=0.45785$
$s_1=\sqrt {\frac{∑(x_{1_i}-x ̅_1)^2}{n_1}}=0.1226$
$x ̅_2=\frac{∑x_{2_i}}{n_2}=0.43167$
$s_2=\sqrt {\frac{∑(x_{2_i}-x ̅_2)^2}{n_2}}=0.1246$
$H_0:~µ_1=µ_2$ versus $H_1:~µ_1\ne µ_2$
$t_0=\frac{(x ̅_1-x ̅_2)-(µ_1-µ_2)}{\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}=\frac{(0.45785-0.43167)-0}{\sqrt {\frac{0.1226^2}{20}+\frac{0.1246^2}{15}}}=0.619$
$n=15$ (use the smaller value of $n$), so:
$d.f.=n-1=14$
Two-tailed test:
$t_{\frac{α}{2}}=t_{0.025}=2.145$
(According to Table VI, for d.f. = 14 and area in right tail = 0.025)
Also, $-t_{\frac{α}{2}}=-2.145$
Since $-t_{\frac{α}{2}}\lt t_0\lt t_{\frac{α}{2}}$, we do not reject the null hypothesis.
